Linear Transformations 线性变换

回顾一下，当我们用一个 $m\times n$ 矩阵乘以一个 $n\times 1$ 列向量时，结果是一个 $m\times 1$ 列向量。本节我们将讨论，通过矩阵乘法，一个 $m \times n$ 矩阵如何把一个 $n\times 1$ 列向量变换为一个 $m \times 1$ 列向量。

回顾一下，由 $\vec{x} = \begin{pmatrix}x_1 \\ x_2\\ \vdots \\ x_n\end{pmatrix} $ 给出的 $n \times 1$ 向量称为属于 $\mathbb{R}^n$，即所有 $n \times 1$ 向量的集合。本节我们将讨论 $\mathbb{R}^n$ 中向量的变换。

考虑下面的例子。

考虑矩阵 $A = \begin{pmatrix}1 & 2 & 0 \\ 2 & 1 & 0\end{pmatrix} .$ 证明通过矩阵乘法，$A$ 把 $\mathbb{R}^3$ 中的向量变换为 $\mathbb{R}^2$ 中的向量。

首先，回顾 $\mathbb{R}^3$ 中的向量是大小为 $3 \times 1$ 的向量，而 $\mathbb{R}^{2}$ 中的向量大小为 $2 \times 1$。如果我们用一个 $2 \times 3$ 矩阵 $A$ 乘以一个 $3 \times 1$ 向量，结果将是一个 $2 \times 1$ 向量。这就是我们说 $A$ 变换向量时的含义。

现在，对 $\mathbb{R}^3$ 中的 $\begin{pmatrix}x \\ y \\ z\end{pmatrix}$，用给定矩阵左乘得到新向量。这个乘积形如 $\begin{pmatrix}1 & 2 & 0 \\ 2 & 1 & 0\end{pmatrix} \begin{pmatrix}x \\ y \\ z\end{pmatrix} = \begin{pmatrix}x+2y \\ 2x+y\end{pmatrix} $ 所得乘积是一个 $2 \times 1$ 向量，由 $x$ 和 $y$ 的取值决定。下面是一些数值例子。$\begin{pmatrix}1 & 2 & 0 \\ 2 & 1 & 0\end{pmatrix} \begin{pmatrix}1 \\ 2 \\ 3\end{pmatrix} = $ \begin{pmatrix}5 \\ 4\end{pmatrix} $ 这里，$\mathbb{R}^3$ 中的向量 $\begin{pmatrix}1 \\ 2 \\ 3\end{pmatrix}$ 被该矩阵变换为 $\mathbb{R}^2$ 中的向量 $\begin{pmatrix}5 \\ 4\end{pmatrix}$。

下面是另一个例子：$\begin{pmatrix}1 & 2 & 0 \\ 2 & 1 & 0\end{pmatrix} \begin{pmatrix}10 \\ 5 \\ -3\end{pmatrix} = $ \begin{pmatrix}20 \\ 25\end{pmatrix} $

我们的想法是定义一个函数，它接收 $\mathbb{R}^{3}$ 中的向量，并产生 $\mathbb{R}^{2}$ 中的新向量。在此情形下，该函数就是与矩阵 $A$ 相乘。

令 $T$ 表示这样一个函数。记号 $T:\mathbb{R}^{n}\mapsto \mathbb{R}^{m}$ 表示函数 $T$ 把 $\mathbb{R}^{n}$ 中的向量变换为 $\mathbb{R}^{m}$ 中的向量。记号 $T(\vec{x})$ 表示将变换 $T$ 作用于向量 $\vec{x}$。上面的例子展示了由矩阵乘法实现的变换。在此情形下，我们常写作 $T\_{A}\left( \vec{x}\right) =A \vec{x} $ 因此，$T\_{A}$ 是由矩阵 $A$ 确定的变换。这种情况下，我们说 $T$ 是一个矩阵变换。

回顾矩阵乘法的性质：对 $k$ 和 $p$ 两个标量，有 $A\left( kB+pC\right) =kAB+pAC $ 特别地，对于 $m\times n$ 矩阵 $A$ 以及 $\mathbb{R}^{n}$ 中的 $n\times 1$ 向量 $B$ 和 $C$，该公式成立。

换句话说，这意味着矩阵乘法给出了线性变换的一个例子，我们现在就来定义它。

考虑下面的例子。

设 $T:\mathbb{R}^{n}\mapsto \mathbb{R}^{m}$ 为一个函数，其中对每个 $\vec{x} \in \mathbb{R}^{n}$，都有 $T\left(\vec{x}\right)\in \mathbb{R}^{m}$。则 $T$ 是一个线性变换，当且仅当对任意标量 $k ,p$ 以及 $\mathbb{R}^{n}$ 中的向量 $\vec{x}\_1$ 和 $\vec{x}\_2$（$n\times 1$ 向量），都有 $T\left( k \vec{x}\_1 + p \vec{x}\_2 \right) = kT\left(\vec{x}\_1\right)+ pT\left(\vec{x}\_{2} \right) $

考虑下面的例子。

令 $T$ 为一个由 $T:\mathbb{R}^3\to\mathbb{R}^2$ 给出的变换，定义为 $T\begin{pmatrix}x \\ y \\ z\end{pmatrix} = \begin{pmatrix}x+y \\ x-z\end{pmatrix} \mbox{ for all } \begin{pmatrix}x \\ y \\ z\end{pmatrix} \in\mathbb{R}^3 $ 证明 $T$ 是一个线性变换。

由 定义 1 可知，我们需要证明对所有标量 $k,p$ 以及向量 $\vec{x}\_1, \vec{x}\_2$，都有 $T\left( k \vec{x}\_1 + p \vec{x}\_2 \right) = kT\left(\vec{x}\_1\right)+ pT\left(\vec{x}\_{2} \right)$。令 $\vec{x}\_1 = \begin{pmatrix}x_1 \\ y_1 \\ z_1\end{pmatrix}, \vec{x}\_2 = \begin{pmatrix}x_2 \\ y_2 \\ z_2\end{pmatrix} $ 则 $\begin{aligned}T\left( k \vec{x}\_1 + p \vec{x}\_2 \right) &= T \left( k \begin{pmatrix}x_1 \\ y_1 \\ z_1\end{pmatrix} + p \begin{pmatrix}x_2 \\ y_2 \\ z_2\end{pmatrix} \right) \\ &= T \left( \begin{pmatrix}kx_1 \\ ky_1 \\ kz_1\end{pmatrix} + \begin{pmatrix}px_2 \\ py_2 \\ pz_2\end{pmatrix} \right) \\ &= T \left( \begin{pmatrix}kx_1 + px_2 \\ ky_1 + py_2 \\ kz_1 + pz_2\end{pmatrix} \right) \\ &= \begin{pmatrix}(kx_1 + px_2) + (ky_1 + py_2) \\ (kx_1 + px_2)- (kz_1 + pz_2)\end{pmatrix} \\ &= \begin{pmatrix}(kx_1 + ky_1) + (px_2 + py_2) \\ (kx_1 - kz_1) + (px_2 - pz_2)\end{pmatrix} \\ &= \begin{pmatrix}kx_1 + ky_1 \\ kx_1 - kz_1\end{pmatrix} + \begin{pmatrix}px_2 + py_2 \\ px_2 - pz_2\end{pmatrix} \\ &= k \begin{pmatrix}x_1 + y_1 \\ x_1 - z_1\end{pmatrix} + p \begin{pmatrix}x_2 + y_2 \\ x_2 - z_2\end{pmatrix} \\ &= k T(\vec{x}\_1) + p T(\vec{x}\_2)\end{aligned} $ 因此 $T$ 是一个线性变换。

线性变换的两个重要例子是零变换和恒等变换。对一切 $\vec{x}$ 由 $T\left( \vec{x} \right) = \vec{0}$ 定义的零变换是线性变换的一个例子。类似地，由 $T\left( \vec{x} \right) = \vec{x}$ 定义的恒等变换也是线性的。请运用 例 2 中演示的方法自行证明这些。

我们在本节开头讨论了矩阵变换，即通过与矩阵相乘来变换向量。这些矩阵变换实际上都是线性变换。

设 $T:\mathbb{R}^{n}\mapsto \mathbb{R}^{m}$ 为由 $T(\vec{x}) = A\vec{x}$ 定义的变换。则 $T$ 是一个线性变换。

事实上，每一个线性变换都可以表示为一个矩阵变换，因此线性变换与矩阵变换完全是一回事。

在上述例子中，线性变换的作用都是乘以一个矩阵。事实证明，线性变换总是如此。如果 $T$ 是任意一个将 $\mathbb{R}^{n}$ 映射到 $\mathbb{R}^{m}$ 的线性变换，那么总存在一个 $m\times n$ 矩阵 $A$，使得对所有 $\vec{x} \in \mathbb{R}^{n}$ 都有 $T\left(\vec{x}\right) = A\vec{x} \label{matrixoftransf}$。

设 $T:\mathbb{R}^{n}\mapsto \mathbb{R}^{m}$ 是一个线性变换。则我们可以找到一个矩阵 $A$，使得 $T(\vec{x}) = A\vec{x}$。此时，我们说 $T$ 由矩阵 $A$ 确定或诱导。

理由如下。设 $T:\mathbb{R}^{n}\mapsto \mathbb{R}^{m}$ 是一个线性变换，你想求出由该线性变换所定义的矩阵（如 $\eqref{matrixoftransf}$ 所述）。注意到 $\vec{x} =\begin{pmatrix}x\_{1} \\ x\_{2} \\ \vdots \\ x\_{n}\end{pmatrix} = x\_{1}\begin{pmatrix}1 \\ 0 \\ \vdots \\ 0\end{pmatrix} + x\_{2}\begin{pmatrix}0 \\ 1 \\ \vdots \\ 0\end{pmatrix} +\cdots + x\_{n}\begin{pmatrix}0 \\ 0 \\ \vdots \\ 1\end{pmatrix} = \sum\_{i=1}^{n}x\_{i}\vec{e}\_{i} $，其中 $\vec{e}\_{i}$ 是 $I_n$ 的第 $i^{th}$ 列，即除第 $i^{th}$ 个位置为 1、其余位置全为 0 的 $n \times 1$ 向量。

由于 $T$ 是线性的，$\begin{aligned}T\left( \vec{x} \right)&=\sum\_{i=1}^{n}x\_{i}T\left( \vec{e}\_{i}\right) \\ &=\begin{pmatrix}\| & & \| \\ T\left( \vec{e}\_{1}\right) & \cdots & T\left( \vec{e}\_{n}\right) \\ \| & & \|\end{pmatrix} \begin{pmatrix}x\_{1} \\ \vdots \\ x\_{n}\end{pmatrix} \\ &= A\begin{pmatrix}x\_{1} \\ \vdots \\ x\_{n}\end{pmatrix}\end{aligned}$ 所求矩阵的第 $i^{th}$ 列由 $T\left( \vec{e}\_{i}\right)$ 构成。注意集合 $\left\{ \vec{e}\_1, \vec{e}\_2, \cdots, \vec{e}\_n \right\}$ 称为 $\mathbb{R}^n$ 的标准基。因此，$T$ 的矩阵可通过将 $T$ 作用于标准基得到。我们将其正式表述为以下定理。

设 $T: \mathbb{R}^{n} \mapsto \mathbb{R}^{m}$ 是一个线性变换。则满足 $T\left(\vec{x}\right)=A\vec{x}$ 的矩阵 $A$ 由下式给出：$A= \begin{pmatrix}\| & & \| \\ T\left( \vec{e}\_{1}\right) & \cdots & T\left( \vec{e}\_{n}\right) \\ \| & & \|\end{pmatrix} $，其中 $\vec{e}\_{i}$ 是 $I_n$ 的第 $i^{th}$ 列，而 $T\left( \vec{e}\_{i} \right)$ 即为 $A$ 的第 $i^{th}$ 列。

以下推论是一个本质性的结果。

变换 $T:\mathbb{R}^n\rightarrow \mathbb{R}^m$ 是线性变换，当且仅当它是一个矩阵变换。

考虑下面的例子。

设 $T$ 是一个线性变换，$T:\mathbb{R}^{3}\rightarrow \mathbb{ R}^{2}$，其中 $T\begin{pmatrix}1 \\ 0 \\ 0\end{pmatrix} =\begin{pmatrix}1 \\ 2\end{pmatrix} ,$ T\begin{pmatrix}0 \\ 1 \\ 0\end{pmatrix} =\begin{pmatrix}9 \\ -3\end{pmatrix} ,$ T\begin{pmatrix}0 \\ 0 \\ 1\end{pmatrix} =\begin{pmatrix}1 \\ 1\end{pmatrix} $ 求 $T$ 的矩阵 $A$，使得对所有 $\vec{x}$ 都有 $T \left( \vec{x} \right)=A\vec{x}$。

解答。 根据 定理 2，我们如下构造 $A$：$A = \begin{pmatrix}\| & & \| \\ T\left( \vec{e}\_{1}\right) & \cdots & T\left( \vec{e}\_{n}\right) \\ \| & & \|\end{pmatrix} $ 此时 $A$ 将是一个 $2 \times 3$ 矩阵，因此我们需要求 $T \left(\vec{e}\_1 \right), T \left(\vec{e}\_2 \right),$ 和 $T \left(\vec{e}\_3 \right)$。幸运的是，这些值已经给出，因此我们可以按需填入 $A$，将这些向量作为 $A$ 的列。于是 $A=\begin{pmatrix}1 & 9 & 1 \\ 2 & -3 & 1\end{pmatrix}$

在这个例子中，我们已给出 $T \left(\vec{e}\_1 \right), T \left(\vec{e}\_2 \right),$ 和 $T \left(\vec{e}\_3 \right)$ 的结果向量。构造矩阵 $A$ 很简单，因为我们只需将这些向量作为 $A$ 的列即可。下一个例子展示当 $T \left(\vec{e}\_i \right)$ 没有给出得这么明确时，如何求 $A$。

设 $T$ 是一个线性变换，$T:\mathbb{R}^{2}\rightarrow \mathbb{R}^{2}$，且 $T\begin{pmatrix}1 \\ 1\end{pmatrix} =\begin{pmatrix}1 \\ 2\end{pmatrix} ,$ $T\begin{pmatrix}0 \\ -1\end{pmatrix} =\begin{pmatrix}3 \\ 2\end{pmatrix} $ 求 $T$ 的矩阵 $A$，使得对所有 $\vec{x}$ 都有 $T \left( \vec{x} \right)=A\vec{x}$。

根据 定理 2，为求此矩阵，我们需要确定 $T$ 对 $\vec{e}\_{1}$ 和 $\vec{e}\_{2}$ 的作用。在 示例 9.9.2 中，这些结果向量是已知的。然而在本例中，我们已知的是 $T$ 对两个不同向量的作用。我们如何求出 $T$ 对 $\vec{e}\_{1}$ 和 $\vec{e}\_{2}$ 的作用？特别地，对 $\vec{e}\_{1}$，设存在 $x$ 和 $y$ 使得 $\begin{pmatrix}1 \\ 0\end{pmatrix} = x\begin{pmatrix}1\\ 1\end{pmatrix} +y\begin{pmatrix}0 \\ -1\end{pmatrix} \label{matrixvalues}$

由于 $T$ 是线性的，有 $T\begin{pmatrix}1 \\ 0\end{pmatrix} = x T\begin{pmatrix}1 \\ 1\end{pmatrix} +y T\begin{pmatrix}0 \\ -1\end{pmatrix} $

代入数值后，该和变为 $T\begin{pmatrix}1 \\ 0\end{pmatrix} = x\begin{pmatrix}1 \\ 2\end{pmatrix} +y\begin{pmatrix}3 \\ 2\end{pmatrix} \label{matrixvalues2}$

因此，如果我们知道满足 $\eqref{matrixvalues}$ 的 $x$ 和 $y$ 的值，就可以将其代入方程 $\eqref{matrixvalues2}$。由此可求得 $T\left(\vec{e}\_1\right)$，即矩阵 $A$ 的第一列。

我们继续求 $x$ 和 $y$。方法是求解 $\eqref{matrixvalues}$，而这可通过解如下方程组完成：$\begin{array}{c} x = 1 x - y = 0 \end{array}$

我们看到 $x=1$ 且 $y=1$ 是该方程组的解。将这些值代入方程 $\eqref{matrixvalues2}$，得到 $T\begin{pmatrix}1 \\ 0\end{pmatrix} = 1 \begin{pmatrix}1 \\ 2\end{pmatrix} + 1 \begin{pmatrix}3 \\ 2\end{pmatrix} = \begin{pmatrix}1 \\ 2\end{pmatrix} + \begin{pmatrix}3 \\ 2\end{pmatrix} = \begin{pmatrix}4 \\ 4\end{pmatrix} $

因此 $\begin{pmatrix}4 \\ 4\end{pmatrix}$ 是 $A$ 的第一列。

第二列的计算方式相同，留作练习。

所得矩阵 $A$ 由下式给出：$A = \begin{pmatrix}4 & -3 \\ 4 & -2\end{pmatrix}$

这个例子展示了一种求 $A$ 的矩阵的很长的步骤。虽然这种方法可靠且总能得到正确的矩阵 $A$，但下面的步骤提供了一种替代方法。

设 $T:\mathbb{R}^{n}\rightarrow \mathbb{R}^{m}$ 是一个线性变换。设存在向量 $\left\{ \vec{a}\_{1},\cdots ,\vec{a}\_{n}\right\}$，属于 $\mathbb {R}^{n}$，使得 $\begin{pmatrix}\vec{a}\_{1} & \cdots & \vec{a}\_{n}\end{pmatrix} ^{-1}$ 存在，且 $T \left(\vec{a}\_{i}\right)=\vec{b}\_{i} $，则 $T$ 的矩阵必为如下形式：$\begin{pmatrix}\vec{b}\_{1} & \cdots & \vec{b}\_{n}\end{pmatrix} \begin{pmatrix}\vec{a}\_{1} & \cdots & \vec{a}\_{n}\end{pmatrix} ^{-1}$

我们将在下面的例子中说明这一步骤。你或许也会发现，用这一步骤来演练 示例 2 会很有帮助。

设 $T:\mathbb{R}^{3}\rightarrow \mathbb{R}^{3}$ 是一个线性变换，且 $T\begin{pmatrix}1 \\ 3 \\ 1\end{pmatrix} =\begin{pmatrix}0 \\ 1 \\ 1\end{pmatrix} ,T\begin{pmatrix}0 \\ 1 \\ 1\end{pmatrix} =\begin{pmatrix}2 \\ 1 \\ 3\end{pmatrix} ,T\begin{pmatrix}1 \\ 1 \\ 0\end{pmatrix} =\begin{pmatrix}0 \\ 0 \\ 1\end{pmatrix} $ 求该线性变换的矩阵。

根据 步骤 1，$A= \begin{pmatrix}1 & 0 & 1 \\ 3 & 1 & 1 \\ 1 & 1 & 0\end{pmatrix} ^{-1}$，且 $B=\begin{pmatrix}0 & 2 & 0 \\ 1 & 1 & 0 \\ 1 & 3 & 1\end{pmatrix}$

接着，步骤 1 断言 $T$ 的矩阵为 $C= BA^{-1} =\begin{pmatrix}2 & -2 & 4 \\ 0 & 0 & 1 \\ 4 & -3 & 6\end{pmatrix} $

事实上，你可以先验证对上面这 3 个向量都有 $T(\vec{x})=C\vec{x}$：

$$\begin{pmatrix}2 & -2 & 4 \\ 0 & 0 & 1 \\ 4 & -3 & 6\end{pmatrix} \begin{pmatrix}1 \\ 3 \\ 1\end{pmatrix} =\begin{pmatrix}0 \\ 1 \\ 1\end{pmatrix} ,\\ \begin{pmatrix}2 & -2 & 4 \\ 0 & 0 & 1 \\ 4 & -3 & 6\end{pmatrix} \begin{pmatrix}0 \\ 1 \\ 1\end{pmatrix} =\begin{pmatrix}2 \\ 1 \\ 3\end{pmatrix} \\ \\\begin{pmatrix}2 & -2 & 4 \\ 0 & 0 & 1 \\ 4 & -3 & 6\end{pmatrix} \begin{pmatrix}1 \\ 1 \\ 0\end{pmatrix} =\begin{pmatrix}0 \\ 0 \\ 1\end{pmatrix}$$

但更一般地，对任意 $\vec{x}$ 都有 $T(\vec{x})= C\vec{x}$。为看出这一点，令 $\vec{y}=A^{-1}\vec{x}$，然后利用 $T$ 的线性性：$T(\vec{x})= T(A\vec{y}) = T \left( \sum_i \vec{y}\_i\vec{a}\_i \right) = \sum \vec{y}\_i T(\vec{a}\_i) \sum \vec{y}\_i \vec{b}\_i = B\vec{y} = BA^{-1}\vec{x} = C\vec{x} $

回想前面讨论的点积。考虑映射 $\vec{v}$$\mapsto$ $\mathrm{proj}\_{\vec{u}}\left( \vec{v}\right)$，它把一个向量变换为其在给定向量 $\vec{u}$ 上的投影。事实证明该映射是线性的，这一结果可由点积的性质推出。证明如下。$\begin{aligned}\mathrm{proj}\_{\vec{u}}\left( k \vec{v}+ p \vec{w}\right) &=\left( \frac{(k \vec{v}+ p \vec{w})\bullet \vec{u}}{ \vec{u}\bullet \vec{u}}\right) \vec{u} \\ &= k \left( \frac{ \vec{v}\bullet \vec{u}}{\vec{u}\bullet \vec{u}}\right) \vec{u}+p \left( { 0.05in}\frac{\vec{w}\bullet \vec{u}}{\vec{u}\bullet \vec{u}}\right) \vec{u} \\ &= k \\ \mathrm{proj}\_{\vec{u}}\left( \vec{v}\right) +p \\ \mathrm{proj} \_{\vec{u}}\left( \vec{w}\right)\end{aligned}$

考虑下面的例子。

令 $\vec{u} = \begin{pmatrix}1 \\ 2 \\ 3\end{pmatrix}$，并令 $T$ 为投影映射 $T: \mathbb{R}^3 \mapsto \mathbb{R}^3$，定义为对任意 $\vec{v} \in \mathbb{R}^3$ 有 $T(\vec{v}) = \mathrm{proj}\_{\vec{u}}\left( \vec{v}\right) $。

设 $T: \mathbb{R}^n \mapsto \mathbb{R}^m$ 为线性变换。则 $T$ 有一些重要的性质将在本节中考察。考虑如下定理。

线性变换的性质　设 $T: \mathbb{R}^n \mapsto \mathbb{R}^m$ 为线性变换，且设 $\vec{x} \in \mathbb{R}^n$。

这些性质有助于确定变换对给定向量的作用。考虑如下示例。

设 $T:\mathbb{R}^3 \mapsto \mathbb{R}^4$ 为一线性变换，满足 $T \begin{pmatrix}1 \\ 3 \\ 1\end{pmatrix} = \begin{pmatrix}4 \\ 4 \\ 0 \\ -2\end{pmatrix}, T \begin{pmatrix}4 \\ 0 \\ 5\end{pmatrix} = \begin{pmatrix}4 \\ 5 \\ -1 \\ 5\end{pmatrix}$。求 $T \begin{pmatrix}-7 \\ 3 \\ -9\end{pmatrix}$。

利用定理 9.6.1中的第三个性质，可把 $\begin{pmatrix}-7 \\ 3 \\ -9\end{pmatrix}$ 写成 $\begin{pmatrix}1 \\ 3 \\ 1\end{pmatrix}$ 与 $\begin{pmatrix}4 \\ 0 \\ 5\end{pmatrix}$ 的线性组合来求 $T \begin{pmatrix}-7 \\ 3 \\ -9\end{pmatrix}$。

因此我们要找 $a,b \in \mathbb{R}$ 使得 $\begin{pmatrix}-7 \\ 3 \\ -9\end{pmatrix} = a \begin{pmatrix}1 \\ 3 \\ 1\end{pmatrix} + b \begin{pmatrix}4 \\ 0 \\ 5\end{pmatrix} $

所需的增广矩阵及其化简行阶梯形式如下：$\left[\begin{array}{rr\|r}1 & 4 & -7 \\ 3 & 0 & 3 \\ 1 & 5 & -9\end{array}\right] \rightarrow \cdots \rightarrow \left[\begin{array}{rr\|r}1 & 0 & 1 \\ 0 & 1 & -2 \\ 0 & 0 & 0\end{array}\right] $

故 $a = 1,\ b = -2$，且 $\begin{pmatrix}-7 \\ 3 \\ -9\end{pmatrix} = 1 \begin{pmatrix}1 \\ 3 \\ 1\end{pmatrix} + (-2) \begin{pmatrix}4 \\ 0 \\ 5\end{pmatrix} $

现在，利用上面的第三个性质，有 $\begin{aligned}T \begin{pmatrix}-7 \\ 3 \\ -9\end{pmatrix} &=T \left( 1 \begin{pmatrix}1 \\ 3 \\ 1\end{pmatrix} + (-2) \begin{pmatrix}4 \\ 0 \\ 5\end{pmatrix} \right) \\ &= 1T \begin{pmatrix}1 \\ 3 \\ 1\end{pmatrix} -2T \begin{pmatrix}4 \\ 0 \\ 5\end{pmatrix} \\ &= \begin{pmatrix}4 \\ 4 \\ 0 \\ -2\end{pmatrix} -2 \begin{pmatrix}4 \\ 5 \\ -1 \\ 5\end{pmatrix} \\ &= \begin{pmatrix}-4 \\ -6 \\ 2 \\ -12\end{pmatrix}\end{aligned}$

因此，$T \begin{pmatrix}-7 \\ 3 \\ -9\end{pmatrix} = \begin{pmatrix}-4 \\ -6 \\ 2 \\ -12\end{pmatrix}$.$

假设两个线性变换对所有向量在 $\vec{x}$ 上的作用方式相同。则称这两个变换相等。

设 $S$ 与 $T$ 为从 $\mathbb{R}^n$ 到 $\mathbb{R}^m$ 的线性变换。则 $S = T$ 当且仅当对每个 $\vec{x} \in \mathbb{R}^n$，都有 $S \left( \vec{x} \right) = T \left( \vec{x} \right) $

假设两个线性变换作用于同一向量 $\vec{x}$，先作用变换 $T$，再作用由 $S$ 给出的第二个变换。我们可求得先后施加两者而得到的复合变换。

设 $T: \mathbb{R}^k \mapsto \mathbb{R}^n$ 与 $S: \mathbb{R}^n \mapsto \mathbb{R}^m$ 为线性变换。则 $S$ 与 $T$ 的复合为 $S \circ T: \mathbb{R}^k \mapsto \mathbb{R}^m$。$S \circ T$ 的作用由 \$S \circ T) (\vec{x}) = S(T(\vec{x})) $ \mbox{for all} $ \vec{x} \in \mathbb{R}^k$$ 给出。

注意所得向量将属于 $\mathbb{R}^m$。须仔细看清变换的顺序。我们记作 $S \circ T$，但先施加变换 $T$，再施加 $S$。

设 $T: \mathbb{R}^k \mapsto \mathbb{R}^n$ 与 $S: \mathbb{R}^n \mapsto \mathbb{R}^m$ 为线性变换，且 $T$ 由矩阵 $A$ 诱导，$S$ 由矩阵 $B$ 诱导。则 $S \circ T$ 是由矩阵 $BA$ 诱导的线性变换。

考虑如下示例。

设 $T$ 为由矩阵 $A = \begin{pmatrix}1 & 2 \\ 2 & 0\end{pmatrix} $ 诱导的线性变换，$S$ 为由矩阵 $B = \begin{pmatrix}2 & 3 \\ 0 & 1\end{pmatrix} $ 诱导的线性变换。求复合变换 $S \circ T$ 的矩阵。然后，对 $\vec{x} = \begin{pmatrix}1 \\ 4\end{pmatrix}$ 求 \$S \circ T)(\vec{x})$。

由定理 2，$S \circ T$ 的矩阵由 $BA$ 给出。$BA = \begin{pmatrix}2 & 3 \\ 0 & 1\end{pmatrix} \begin{pmatrix}1 & 2 \\ 2 & 0\end{pmatrix} = \begin{pmatrix}8 & 4 \\ 2 & 0\end{pmatrix} $$

为求 \$S \circ T)(\vec{x})$，将 $\vec{x}$ 乘以 $BA$：$\begin{pmatrix}8 & 4 \\ 2 & 0\end{pmatrix} \begin{pmatrix}1 \\ 4\end{pmatrix} = \begin{pmatrix}24 \\ 2\end{pmatrix} $

验证：先求 $T(\vec{x})$：$\begin{pmatrix}1 & 2 \\ 2 & 0\end{pmatrix} \begin{pmatrix}1 \\ 4\end{pmatrix} = \begin{pmatrix}9 \\ 2\end{pmatrix} $$

然后，计算 $S(T(\vec{x}))$：$\begin{pmatrix}2 & 3 \\ 0 & 1\end{pmatrix} \begin{pmatrix}9 \\ 2\end{pmatrix} = \begin{pmatrix}24 \\ 2\end{pmatrix} $$

考虑复合变换 $S \circ T$，并设其作用满足 \$S \circ T) (\vec{x}) = \vec{x}$。即变换 $S$ 把向量 $T(\vec{x})$ 变回 $\vec{x}$。此时，$S$ 与 $T$ 互为逆。考虑如下定义。$

设 $T: \mathbb{R}^n \mapsto \mathbb{R}^n$ 与 $S:\mathbb{R}^n \mapsto \mathbb{R}^n$ 为线性变换。假设对每个 $\vec{x} \in \mathbb{R}^n$，有 \$S \circ T)(\vec{x}) = \vec{x} $ 且 \$T \circ S)(\vec{x}) = \vec{x} $。则 $S$ 称为 $T$ 的一个逆，$T$ 称为 $S$ 的一个逆。从几何上看，它们互为反向作用。

如下定理至关重要，它断言上述逆变换是唯一的。

设 $T:\mathbb{R}^n \mapsto \mathbb{R}^n$ 为线性变换，由矩阵 $A$ 诱导。则 $T$ 存在逆变换当且仅当矩阵 $A$ 可逆。此时逆变换唯一，记作 $T^{-1}: \mathbb{R}^n \mapsto \mathbb{R}^n$。$T^{-1}$ 由矩阵 $A^{-1}$ 诱导。

考虑如下示例。

设 $T: \mathbb{R}^2 \mapsto \mathbb{R}^2$ 为线性变换，由矩阵 $A = \begin{pmatrix}2 & 3 \\ 3 & 4\end{pmatrix}$ 诱导。证明 $T^{-1}$ 存在，并求其诱导矩阵 $B$。$

由于矩阵 $A$ 可逆，故变换 $T$ 可逆。因此 $T^{-1}$ 存在。

可验证 $A^{-1}$ 为：$A^{-1} = \begin{pmatrix}-4 & 3 \\ 3 & -2\end{pmatrix} $，因此线性变换 $T^{-1}$ 由矩阵 $A^{-1}$ 诱导。$

本节我们将考察 $\mathbb{R}^2$ 中线性变换的一些特例，包括旋转与反射。我们将利用前面讨论的向量加法与标量乘法的几何描述，说明将向量旋转一个角度以及对向量关于一条直线作反射都是线性变换的例子。

更一般地，记由旋转给出的变换为 $T$。为什么这样的变换是线性的？考虑下面说明旋转的图像。记 $\vec{u},\vec{v}$ 为向量。

（图 1）

考虑如何得到 $T\left( \vec{u}+\vec{v}\right)$。很简单，你把 $T (\vec{u})$ 与 $T(\vec{v})$ 相加。理由如下：将 $T(\vec{u})$ 与 $T(\vec{v})$ 相加，得到由 $T(\vec{u})$ 和 $T(\vec{v})$ 所确定的平行四边形的对角线，因为这一运算就是通常的向量加法。现在，假设我们先把 $\vec{u}$ 与 $\vec{v}$ 相加，再将变换 $T$ 作用于 $\vec{u}+\vec{v}$，于是得到 $T( \vec{u}+\vec{v})$。如图所示，这将得到同一个向量。换言之，$T(\vec{u}+\vec{v}) = T(\vec{u}) + T(\vec{v})$。

这是因为旋转保持了向量之间的所有夹角以及它们的长度。特别地，它保持了该平行四边形的形状。因此 $T\left( \vec{u}\right)+T\left( \vec{v}\right)$ 与 $T\left( \vec{u}+\vec{v}\right)$ 给出同一个向量。由此可知，$T$ 对 $\mathbb{R}^{2}$ 中向量的加法满足分配律。

类似地，若 $k$ 为标量，则有 $T\left( k\vec{u}\right)=kT\left( \vec{u}\right)$。因此旋转是线性变换的例子的结论由 定义 9.6.1 给出。

下面的定理给出将所有向量旋转角度 $\theta$ 的线性变换的矩阵。

$$ \begin{pmatrix}\cos \left( \theta \right) & -\sin \left( \theta \right) \\ \sin \left( \theta \right) & \cos \left( \theta \right) \end{pmatrix} $$

令 $\vec{e}\_{1} = \begin{pmatrix}1 \\ 0\end{pmatrix}$，$\vec{e}\_{2} = \begin{pmatrix}0 \\ 1\end{pmatrix}$。它们表示沿正 $x$ 轴与正 $y$ 轴的几何向量，如图所示。

（图 2）

由 定理 5.2.2，我们需要求 $R\_{\theta}(\vec{e}\_{1})$ 与 $R\_{\theta}(\vec{e}\_{2})$，并将它们作为矩阵 $A$（即 $T$ 的矩阵）的列。如上图所示，可利用角度 $\theta$ 的 $\cos, \sin$ 求出 $R\_{\theta}(\vec{e}\_{1})$ 的坐标。$R\_{\theta}(\vec{e}\_{2})$ 的坐标同样可由三角学得到。于是 $R\_{\theta}(\vec{e}\_{1})=\begin{pmatrix}\cos \theta \\ \sin \theta\end{pmatrix}$，$R\_{\theta}(\vec{e}\_{2})=\begin{pmatrix}-\sin \theta \\ \cos \theta\end{pmatrix}$。因此，由 定理 5.2.2 得 $$A=\begin{pmatrix}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{pmatrix} $$

$\left( \cos \left( \theta \right) ,\sin \left( \theta \right) \right)$ 的定义即为点 $R\_{\theta}(\vec{e}\_{1})$ 的坐标。现在，向量 $\vec{e}\_{2}$ 的点恰好在单位圆上比 $\vec{e}\_{1}$ 的点再前进 $\pi /2$，因此经角度 $\theta$ 旋转后，点 $R\_{\theta}(\vec{e}\_{2})$ 的坐标 $x$ 与 $y$ 由下式给出：$\left( x,y\right) =\left( \cos \left( \theta +\pi /2\right) ,\sin \left( \theta +\pi /2\right) \right) =\left( -\sin \theta ,\cos \theta \right) $

考虑下面的例子。

设 $R\_{\frac{\pi}{2}}: \mathbb{R}^2 \to \mathbb{R}^2$ 表示旋转 $\pi/2$。求 $R\_{\frac{\pi}{2}}$ 的矩阵。进而，求 $R\_{\frac{\pi}{2}} (\vec{x})$，其中 $\vec{x} = \begin{pmatrix}1 \\ -2\end{pmatrix}$。

由 定理 1，$R\_{\frac{\pi}{2}}$ 的矩阵为 $$ \begin{pmatrix}\cos \left( \theta \right) & -\sin \left( \theta \right) \\ \sin \left( \theta \right) & \cos \left( \theta \right) \end{pmatrix} = \begin{pmatrix}\cos \left( \pi/2 \right) & -\sin \left( \pi/2 \right) \\ \sin \left( \pi/2 \right) & \cos \left( \pi/2 \right) \end{pmatrix} = \begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix} $$

为求 $R\_{\frac{\pi}{2}} (\vec{x})$，将 $R\_{\frac{\pi}{2}}$ 的矩阵与 $\vec{x}$ 相乘，如下： $$ \begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix} \begin{pmatrix}1 \\ -2\end{pmatrix} = \begin{pmatrix}2 \\ 1\end{pmatrix} $$

现在我们来看一个涉及两个角度的线性变换的例子。

求如下线性变换的矩阵：先将所有向量旋转角度 $\phi$，再旋转角度 $\theta$。因此该线性变换将所有向量旋转角度 $\theta +\phi$。

记 $R\_{\theta +\phi }$ 为将每个向量旋转角度 $\theta +\phi$ 的线性变换。为得到 $R\_{\theta +\phi }$，我们先施加 $R\_{\phi }$ 再施加 $R\_{\theta }$，其中 $R\_{\phi }$ 是旋转角度 $\phi$ 的线性变换，$R\_{\theta }$ 是旋转角度 $\theta$ 的线性变换。记对应的矩阵分别为 $A\_{\theta +\phi }$、$A\_{\phi }$ 与 $A\_{\theta }$，则对每个 $\vec{u}$ 有 $R\_{\theta +\phi }\left( \vec{u}\right)=A\_{\theta +\phi }\vec{u}=A\_{\theta }A\_{\phi }\vec{u} = R\_{\theta }R\_{\phi } \left( \vec{u}\right)$。注意此处矩阵的顺序！

因此有 $$ \begin{aligned}A\_{\theta +\phi } &=\begin{pmatrix}\cos \left( \theta +\phi \right) & -\sin \left( \theta +\phi \right) \\ \sin \left( \theta +\phi \right) & \cos \left( \theta +\phi \right) \end{pmatrix}\\ &=\begin{pmatrix}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{pmatrix} \begin{pmatrix}\cos \phi & -\sin \phi \\ \sin \phi & \cos \phi\end{pmatrix} =A\_{\theta }A\_{\phi }\end{aligned} $$

通常的矩阵乘法给出 $$ \begin{aligned}A\_{\theta +\phi } &= \begin{pmatrix}\cos \left( \theta +\phi \right) & -\sin \left( \theta +\phi \right) \\ \sin \left( \theta +\phi \right) & \cos \left( \theta +\phi \right) \end{pmatrix} \\ &=\begin{pmatrix}\cos \theta \cos \phi -\sin \theta \sin \phi & -\cos \theta \sin \phi -\sin \theta \cos \phi \\ \sin \theta \cos \phi +\cos \theta \sin \phi & \cos \theta \cos \phi -\sin \theta \sin \phi\end{pmatrix} \\ &= A\_{\theta }A\_{\phi }\end{aligned} $$

这些看起来是不是很眼熟？它们正是两角和的常见三角恒等式，这里用线性代数的概念推导得出。

这里我们聚焦于二维旋转。然而，你可以在任意维数中考虑旋转及其他几何概念，这是线性代数的主要优点之一。你可以把一个困难的几何过程分解为若干小步骤，每一步对应与一个适当矩阵的乘法。然后将这些矩阵相乘，就能得到一个单一矩阵，它可以给出施加给定简单步骤序列后结果的数值信息。

将向量关于一条直线作反射的线性变换是 $\mathbb{R}^2$ 中第二类重要的变换。考虑下面的定理。

设 $Q_m: \mathbb{R}^2 \to \mathbb{R}^2$ 为将向量关于直线 $\vec{y}=m\vec{x}$ 反射的线性变换。则 $Q_m$ 的矩阵为 $$ \frac{1}{1+m^2} \begin{pmatrix}1-m^2 & 2m \\ 2m & m^2-1\end{pmatrix} $$

考虑下面的例子。

设 $Q_2: \mathbb{R}^2 \to \mathbb{R}^2$ 表示关于直线 $\vec{y}=2\vec{x}$ 的反射。则 $Q_2$ 是一个线性变换。求 $Q_2$ 的矩阵。进而，求 $Q_2 (\vec{x})$，其中 $\vec{x} = \begin{pmatrix}1 \\ -2\end{pmatrix}$。

由 定理 2，$Q_2$ 的矩阵为 $$ \frac{1}{1+m^2} \begin{pmatrix}1-m^2 & 2m \\ 2m & m^2-1\end{pmatrix} = \frac{1}{1+(2)^2} \begin{pmatrix}1-(2)^2 & 2(2) \\ 2(2) & (2)^2-1\end{pmatrix} = \frac{1}{5} \begin{pmatrix}-3 & 8 \\ 8 & 3\end{pmatrix} $$

为求 $Q_2(\vec{x})$，将 $\vec{x}$ 与 $Q_2$ 的矩阵相乘，如下： $$ \frac{1}{5} \begin{pmatrix}-3 & 8 \\ 8 & 3\end{pmatrix} \begin{pmatrix}1 \\ -2\end{pmatrix} = \begin{pmatrix}- \frac{19}{5} \\ \frac{2}{5}\end{pmatrix} $$

考虑下面同时包含反射与旋转的向量的例子。

求如下线性变换的矩阵：先将所有向量旋转角度 $\pi /6$，再关于 $x$ 轴反射。

由 定理 1，旋转角度 $\pi /6$ 的变换的矩阵为 $$ \begin{pmatrix}\cos \left( \pi /6\right) & -\sin \left( \pi /6\right) \\ \sin \left( \pi /6\right) & \cos \left( \pi /6\right) \end{pmatrix} =\begin{pmatrix}\frac{1}{2}\sqrt{3} & - \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2}\sqrt{3}\end{pmatrix} $$

关于 $x$ 轴反射，等价于以 $m=0$ 将向量关于直线 $\vec{y}=m\vec{x}$ 反射。由 定理 2，将所有向量关于 $x$ 轴反射的变换的矩阵为 $$ \frac{1}{1+m^2} \begin{pmatrix}1-m^2 & 2m \\ 2m & m^2-1\end{pmatrix} = \frac{1}{1+(0)^2} \begin{pmatrix}1-(0)^2 & 2(0) \\ 2(0) & (0)^2-1\end{pmatrix} = \begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix} $$

Therefore, the matrix of the linear transformation which first rotates through $\pi /6$ and then reflects through the $x$ axis is given by $$ \begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix} \begin{pmatrix}\frac{1}{2}\sqrt{3} & - \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2}\sqrt{3}\end{pmatrix} = \begin{pmatrix}\frac{1}{2}\sqrt{3} & - \frac{1}{2} \\ - \frac{1}{2} & - \frac{1}{2}\sqrt{3}\end{pmatrix} $$

Outcomes 学习目标

1. Determine if a linear transformation is onto or one to one.

1. 判断一个线性变换是满射还是单射。

Let $T: \mathbb{R}^n \mapsto \mathbb{R}^m$ be a linear transformation. We define the range or image of $T$ as the set of vectors of $\mathbb{R}^{m}$ which are of the form $T \left(\vec{x}\right)$ (equivalently, $A\vec{x}$ for some $\vec{x}\in \mathbb{R}^{n}$. It is common to write $T\mathbb{R}^{n}$, $T\left( \mathbb{R}^{n}\right)$, or $\mathrm{Im}\left( T\right)$ to denote these vectors.

Lemma 1: Range of a Matrix Transformation 引理 1：矩阵变换的值域

Let $A$ be an $m\times n$ matrix where $A\_{1},\cdots , A\_{n}$ denote the columns of $A.$ Then, for a vector $\vec{x}=\begin{pmatrix}x\_{1} \\ \vdots \\ x\_{n}\end{pmatrix}$ in $\mathbb{R}^n$,

$$A\vec{x}=\sum\_{k=1}^{n}x\_{k}A\_{k}$$

Therefore, $A \left( \mathbb{R}^n \right)$ is the collection of all linear combinations of these products.

Proof

This follows from the definition of matrix multiplication.

This section is devoted to studying two important characterizations of linear transformations, called one to one and onto. We define them now.

Definition 1: One to One 定义 1：单射

Suppose $\vec{x}\_1$ and $\vec{x}\_2$ are vectors in $\mathbb{R}^n$. A linear transformation $T: \mathbb{R}^n \mapsto \mathbb{R}^m$ is called one to one (often written as $1-1)$ if whenever $\vec{x}\_1 \neq \vec{x}\_2$ it follows that : $T\left( \vec{x}\_1 \right) \neq T \left(\vec{x}\_2\right) $

Equivalently, if $T\left( \vec{x}\_1 \right) =T\left( \vec{x}\_2\right) ,$ then $\vec{x}\_1 = \vec{x}\_2$. Thus, $T$ is one to one if it never takes two different vectors to the same vector.

The second important characterization is called onto.

Definition 2: Onto 定义 2：满射

Let $T: \mathbb{R}^n \mapsto \mathbb{R}^m$ be a linear transformation. Then $T$ is called onto if whenever $\vec{x}\_2 \in \mathbb{R}^{m}$ there exists $\vec{x}\_1 \in \mathbb{R}^{n}$ such that $T\left( \vec{x}\_1\right) = \vec{x}\_2.$

We often call a linear transformation which is one-to-one an injection. Similarly, a linear transformation which is onto is often called a surjection.

The following proposition is an important result.

Proposition 1: One to One 命题 1：单射

Let $T:\mathbb{R}^n \mapsto \mathbb{R}^m$ be a linear transformation. Then $T$ is one to one if and only if $T(\vec{x}) = \vec{0}$ implies $\vec{x}=\vec{0}$.

Proof

We need to prove two things here. First, we will prove that if $T$ is one to one, then $T(\vec{x}) = \vec{0}$ implies that $\vec{x}=\vec{0}$. Second, we will show that if $T(\vec{x})=\vec{0}$ implies that $\vec{x}=\vec{0}$, then it follows that $T$ is one to one. Recall that a linear transformation has the property that $T(\vec{0}) = \vec{0}$.

Suppose first that $T$ is one to one and consider $T(\vec{0})$. $T(\vec{0})=T\left( \vec{0}+\vec{0}\right) =T(\vec{0})+T(\vec{0}) $ and so, adding the additive inverse of $T(\vec{0})$ to both sides, one sees that $T(\vec{0})=\vec{0}$. If $T(\vec{x})=\vec{0}$ it must be the case that $\vec{x}=\vec{0}$ because it was just shown that $T(\vec{0})=\vec{0}$ and $T$ is assumed to be one to one.

Now assume that if $T(\vec{x})=\vec{0},$ then it follows that $\vec{x}=\vec{0}.$ If $T(\vec{v})=T(\vec{u}),$ then $T(\vec{v})-T(\vec{u})=T\left( \vec{v}-\vec{u}\right) =\vec{0} $ which shows that $\vec{v}-\vec{u}=0$. In other words, $\vec{v}=\vec{u}$, and $T$ is one to one.

Note that this proposition says that if $A=\begin{pmatrix}A\_{1} & \cdots & A\_{n}\end{pmatrix}$ then $A$ is one to one if and only if whenever $0 = \sum\_{k=1}^{n}c\_{k}A\_{k} $ it follows that each scalar $c\_{k}=0$.

We will now take a look at an example of a one to one and onto linear transformation.

Example 1: A One to One and Onto Linear Transformation 示例 1：一个单射且满射的线性变换

Suppose $T\begin{pmatrix}x \\ y\end{pmatrix} =\begin{pmatrix}1 & 1 \\ 1 & 2\end{pmatrix} \begin{pmatrix}x \\ y\end{pmatrix} $. Then, $T:\mathbb{R}^{2}\rightarrow \mathbb{R}^{2}$ is a linear transformation. Is $T$ onto? Is it one to one?

Solution

Recall that because $T$ can be expressed as matrix multiplication, we know that $T$ is a linear transformation. We will start by looking at onto. So suppose $\begin{pmatrix}a \\ b\end{pmatrix} \in \mathbb{R}^{2}.$ Does there exist $\begin{pmatrix}x \\ y\end{pmatrix} \in \mathbb{R}^2$ such that $T\begin{pmatrix}x \\ y\end{pmatrix} =\begin{pmatrix}a \\ b\end{pmatrix} ?$ If so, then since $\begin{pmatrix}a \\ b\end{pmatrix}$ is an arbitrary vector in $\mathbb{R}^{2},$ it will follow that $T$ is onto.

This question is familiar to you. It is asking whether there is a solution to the equation $\begin{pmatrix}1 & 1 \\ 1 & 2\end{pmatrix} \begin{pmatrix}x \\ y\end{pmatrix} =\begin{pmatrix}a \\ b\end{pmatrix} $ This is the same thing as asking for a solution to the following system of equations. $\begin{array}{c} x+y=a \\ x+2y=b \end{array} $ Set up the augmented matrix and row reduce. $\left[\begin{array}{rr\|r}1 & 1 & a \\ 1 & 2 & b\end{array}\right] \rightarrow \left[\begin{array}{rr\|r}1 & 0 & 2a-b \\ 0 & 1 & b-a\end{array}\right] \label{ontomatrix}$ You can see from this point that the system has a solution. Therefore, we have shown that for any $a, b$, there is a $\begin{pmatrix}x \\ y\end{pmatrix}$ such that $T\begin{pmatrix}x \\ y\end{pmatrix} =\begin{pmatrix}a \\ b\end{pmatrix}$. Thus $T$ is onto.

Now we want to know if $T$ is one to one. By Proposition 1 it is enough to show that $A\vec{x}=0$ implies $\vec{x}=0$. Consider the system $A\vec{x}=0$ given by: $\begin{pmatrix}1 & 1 \\ 1 & 2\\\end{pmatrix} \begin{pmatrix}x\\ y\end{pmatrix} = \begin{pmatrix}0 \\ 0\end{pmatrix} $

This is the same as the system given by

$$\begin{array}{c} x + y = 0 \\ x + 2y = 0 \end{array}$$

We need to show that the solution to this system is $x = 0$ and $y = 0$. By setting up the augmented matrix and row reducing, we end up with $\left[\begin{array}{rr\|r}1 & 0 & 0 \\ 0 & 1 & 0\end{array}\right]$

This tells us that $x = 0$ and $y = 0$. Returning to the original system, this says that if

$$\begin{pmatrix}1 & 1 \\ 1 & 2\\\end{pmatrix} \begin{pmatrix}x\\ y\end{pmatrix} = \begin{pmatrix}0 \\ 0\end{pmatrix}$$

then $\begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}0 \\ 0\end{pmatrix} $

In other words, $A\vec{x}=0$ implies that $\vec{x}=0$. By Proposition 1, $A$ is one to one, and so $T$ is also one to one.

We also could have seen that $T$ is one to one from our above solution for onto. By looking at the matrix given by $\eqref{ontomatrix}$, you can see that there is a unique solution given by $x=2a-b$ and $y=b-a$. Therefore, there is only one vector, specifically $\begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}2a-b\\ b-a\end{pmatrix}$ such that $T\begin{pmatrix}x \\ y\end{pmatrix} =\begin{pmatrix}a \\ b\end{pmatrix}$. Hence by Definition 1, $T$ is one to one.

Example 2: An Onto Transformation 示例 2：一个满射变换

Let $T: \mathbb{R}^4 \mapsto \mathbb{R}^2$ be a linear transformation defined by $T \begin{pmatrix}a \\ b \\ c \\ d\end{pmatrix} = \begin{pmatrix}a + d \\ b + c\end{pmatrix} \mbox{ for all } \begin{pmatrix}a \\ b \\ c \\ d\end{pmatrix} \in \mathbb{R}^4 $ Prove that $T$ is onto but not one to one.

Solution

You can prove that $T$ is in fact linear.

To show that $T$ is onto, let $\begin{pmatrix}x \\ y\end{pmatrix}$ be an arbitrary vector in $\mathbb{R}^2$. Taking the vector $\begin{pmatrix}x \\ y \\ 0 \\ 0\end{pmatrix} \in \mathbb{R}^4$ we have $T \begin{pmatrix}x \\ y \\ 0 \\ 0\end{pmatrix} = \begin{pmatrix}x + 0 \\ y + 0\end{pmatrix} = \begin{pmatrix}x \\ y\end{pmatrix} $ This shows that $T$ is onto.

By Proposition 1 $T$ is one to one if and only if $T(\vec{x}) = \vec{0}$ implies that $\vec{x} = \vec{0}$. Observe that $T \begin{pmatrix}1 \\ 0 \\ 0 \\ -1\end{pmatrix} = \begin{pmatrix}1 + -1 \\ 0 + 0\end{pmatrix} = \begin{pmatrix}0 \\ 0\end{pmatrix} $ There exists a nonzero vector $\vec{x}$ in $\mathbb{R}^4$ such that $T(\vec{x}) = \vec{0}$. It follows that $T$ is not one to one.

The above examples demonstrate a method to determine if a linear transformation $T$ is one to one or onto. It turns out that the matrix $A$ of $T$ can provide this information.

Theorem 1: Matrix of a One to One or Onto Transformation 定理 1：单射或满射变换的矩阵

Let $T: \mathbb{R}^n \mapsto \mathbb{R}^m$ be a linear transformation induced by the $m \times n$ matrix $A$. Then $T$ is one to one if and only if the rank of $A$ is $n$. $T$ is onto if and only if the rank of $A$ is $m$.

Consider Example 2. Above we showed that $T$ was onto but not one to one. We can now use this theorem to determine this fact about $T$.

Example 3: An Onto Transformation 示例 3：一个满射变换

Let $T: \mathbb{R}^4 \mapsto \mathbb{R}^2$ be a linear transformation defined by $T \begin{pmatrix}a \\ b \\ c \\ d\end{pmatrix} = \begin{pmatrix}a + d \\ b + c\end{pmatrix} \mbox{ for all } \begin{pmatrix}a \\ b \\ c \\ d\end{pmatrix} \in \mathbb{R}^4 $ Prove that $T$ is onto but not one to one.

Solution

Using Theorem 1 we can show that $T$ is onto but not one to one from the matrix of $T$. Recall that to find the matrix $A$ of $T$, we apply $T$ to each of the standard basis vectors $\vec{e}\_i$ of $\mathbb{R}^4$. The result is the $2 \times 4$ matrix A given by $A = \begin{pmatrix}1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0\end{pmatrix} $ Fortunately, this matrix is already in reduced row-echelon form. The rank of $A$ is $2$. Therefore by the above theorem $T$ is onto but not one to one.

Recall that if $S$ and $T$ are linear transformations, we can discuss their composite denoted $S \circ T$. The following examines what happens if both $S$ and $T$ are onto.

Example 4: Composite of Onto Transformations 示例 4：满射变换的复合

Let $T: \mathbb{R}^k \mapsto \mathbb{R}^n$ and $S: \mathbb{R}^n \mapsto \mathbb{R}^m$ be linear transformations. If $T$ and $S$ are onto, then $S \circ T$ is onto.

Solution

Let $\vec{z}\in \mathbb{R}^m$. Since $S$ is onto, there exists a vector $\vec{y}\in \mathbb{R}^n$ such that $S(\vec{y})=\vec{z}$. Furthermore, since $T$ is onto, there exists a vector $\vec{x}\in \mathbb{R}^k$ such that $T(\vec{x})=\vec{y}$. Thus $\vec{z} = S(\vec{y}) = S(T(\vec{x})) = (ST)(\vec{x}), $ showing that for each $\vec{z}\in \mathbb{R}^m$ there exists and $\vec{x}\in \mathbb{R}^k$ such that $(ST)(\vec{x})=\vec{z}$. Therefore, $S \circ T$ is onto.

The next example shows the same concept with regards to one-to-one transformations.

Example 5: Composite of One to One Transformations 示例 5：单射变换的复合

Let $T: \mathbb{R}^k \mapsto \mathbb{R}^n$ and $S: \mathbb{R}^n \mapsto \mathbb{R}^m$ be linear transformations. Prove that if $T$ and $S$ are one to one, then $S \circ T$ is one-to-one.

Solution

To prove that $S \circ T$ is one to one, we need to show that if $S(T (\vec{v})) = \vec{0}$ it follows that $\vec{v} = \vec{0}$. Suppose that $S(T (\vec{v})) = \vec{0}$. Since $S$ is one to one, it follows that $T (\vec{v}) = \vec{0}$. Similarly, since $T$ is one to one, it follows that $\vec{v} = \vec{0}$. Hence $S \circ T$ is one to one.

Outcomes 学习目标

1. Determine if a linear transformation is an isomorphism.
2. Determine if two subspaces of $\mathbb{R}^n$ are isomorphic.

1. 判断一个线性变换是否为同构。
2. 判断 $\mathbb{R}^n$ 的两个子空间是否同构。

Recall the definition of a linear transformation. Let $V$ and $W$ be two subspaces of $\mathbb{R}^{n}$ and $\mathbb{R}^{m}$ respectively. A mapping $T:V\rightarrow W$ is called a linear transformation or linear map if it preserves the algebraic operations of addition and scalar multiplication. Specifically, if $a,b$ are scalars and $\vec{x},\vec{y}$ are vectors,

$$T\left( a\vec{x}+b\vec{y}\right) =aT(\vec{x})+bT(\vec{y})$$

Consider the following important definition.

Definition 1: Isomorphism 同构

A linear map $T$ is called an isomorphism if the following two conditions are satisfied.

• $T$ is one to one. That is, if $T(\vec{x})=T(\vec{y}),$ then $\vec{x}=\vec{y}.$
• $T$ is onto. That is, if $\vec{w}\in W,$ there exists $\vec{v}\in V$ such that $T(\vec{v})=\vec{w}$.

• $T$ 是单射。即若 $T(\vec{x})=T(\vec{y}),$ 则 $\vec{x}=\vec{y}.$
• $T$ 是满射。即若 $\vec{w}\in W,$ 则存在 $\vec{v}\in V$ 使得 $T(\vec{v})=\vec{w}$。

Two such subspaces which have an isomorphism as described above are said to be isomorphic.

Consider the following example of an isomorphism.

Example 1: Isomorphism 同构

Let $T: \mathbb{R}^2 \mapsto \mathbb{R}^2$ be defined by $T \begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}x + y \\ x - y\end{pmatrix}$ Show that $T$ is an isomorphism.

Solution 解答

To prove that $T$ is an isomorphism we must show

1. $T$ is a linear transformation;
2. $T$ is one to one;
3. $T$ is onto.

1. $T$ 是线性变换；
2. $T$ 是单射；
3. $T$ 是满射。

We proceed as follows.

1. $T$ is a linear transformation:

$\begin{aligned}T \left( k \begin{pmatrix}x_1 \\ y_1\end{pmatrix} + p \begin{pmatrix}x_2 \\ y_2\end{pmatrix} \right) &= T \left( \begin{pmatrix}kx_1 \\ ky_1\end{pmatrix} + \begin{pmatrix}px_2 \\ py_2\end{pmatrix} \right) \\ &= T \left( \begin{pmatrix}kx_1 + px_2 \\ ky_1 + py_2\end{pmatrix} \right) \\ &= \begin{pmatrix}(kx_1 + px_2) + (ky_1 + py_2) \\ (kx_1 + px_2) - (ky_1 + py_2)\end{pmatrix} \\ &= \begin{pmatrix}(kx_1 + ky_1) + (px_2 + py_2) \\ (kx_1 - ky_1) + (px_2 - py_2)\end{pmatrix} \\ &= \begin{pmatrix}kx_1 + ky_1 \\ kx_1 - ky_1\end{pmatrix} + \begin{pmatrix}px_2 + py_2 \\ px_2 - py_2\end{pmatrix} \\ &= k \begin{pmatrix}x_1 + y_1 \\ x_1 - y_1\end{pmatrix} + p \begin{pmatrix}x_2 + y_2 \\ x_2 - y_2\end{pmatrix} \\ &= k T \left( \begin{pmatrix}x_1 \\ y_1\end{pmatrix} \right) + p T \left( \begin{pmatrix}x_2 \\ y_2\end{pmatrix} \right)\end{aligned}$

Therefore $T$ is linear.

2. $T$ is one to one:

We need to show that if $T (\vec{x}) = \vec{0}$ for a vector $\vec{x} \in \mathbb{R}^2$, then it follows that $\vec{x} = \vec{0}$. Let $\vec{x} = \begin{pmatrix}x \\ y\end{pmatrix}$.

$T \left( \begin{pmatrix}x \\ y\end{pmatrix} \right) = \begin{pmatrix}x + y\\ x - y\end{pmatrix} = \begin{pmatrix}0 \\ 0\end{pmatrix} $ This provides a system of equations given by $\begin{aligned}x + y &= 0\\ x - y &= 0\end{aligned}$ You can verify that the solution to this system if $x = y =0$. Therefore $\vec{x} = \begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}0 \\ 0\end{pmatrix} $ and $T$ is one to one.

3. $T$ is onto:

Let $a,b$ be scalars. We want to check if there is always a solution to $T \left( \begin{pmatrix}x \\ y\end{pmatrix} \right) = \begin{pmatrix}x + y\\ x - y\end{pmatrix} = \begin{pmatrix}a \\ b\end{pmatrix}$

This can be represented as the system of equations $\begin{aligned}x + y &= a\\ x - y &= b\end{aligned}$

Setting up the augmented matrix and row reducing gives $\left[\begin{array}{cc\|c}1 & 1 & a \\ 1 & -1 & b\end{array}\right] \rightarrow \cdots \rightarrow \left[\begin{array}{cc\|c}1 & 0 & \frac{a+b}{2} \\ 0 & 1 & \frac{a-b}{2}\end{array}\right]$ This has a solution for all $a,b$ and therefore $T$ is onto.

Therefore $T$ is an isomorphism.

An important property of isomorphisms is that its inverse is also an isomorphism.

Proposition 1: Inverse of an Isomorphism 同构的逆

Let $T:V\rightarrow W$ be an isomorphism and $V,W$ be subspaces of $\mathbb{R}^n$. Then $T^{-1}:W\rightarrow V$ is also an isomorphism.

Proof

Let $T$ be an isomorphism. Since $T$ is onto, a typical vector in $W$ is of the form $T(\vec{v})$ where $\vec{v} \in V$. Consider then for $a,b$ scalars, $T^{-1}\left( aT(\vec{v}\_{1})+bT(\vec{v}\_{2})\right)$ where $\vec{v}\_{1}, \vec{v}\_2 \in V$. Is this equal to $aT^{-1}\left( T (\vec{v}\_{1})\right) +bT^{-1}\left( T(\vec{v}\_{2})\right) =a\vec{v}\_{1}+b\vec{v}\_{2}?$ Since $T$ is one to one, this will be so if $T\left( a\vec{v}\_{1}+b\vec{v}\_{2}\right) =T\left( T^{-1}\left( aT(\vec{v}\_{1})+bT(\vec{v}\_{2})\right) \right) =aT(\vec{v}\_{1})+bT(\vec{v}\_{2}).$ However, the above statement is just the condition that $T$ is a linear map. Thus $T^{-1}$ is indeed a linear map. If $\vec{v} \in V$ is given, then $\vec{v}=T^{-1}\left( T(\vec{v})\right)$ and so $T^{-1}$ is onto. If $T^{-1} (\vec{v})=0,$ then $\vec{v}=T\left( T^{-1}(\vec{v})\right) =T(\vec{0})=\vec{0}$ and so $T^{-1}$ is one to one.

Another important result is that the composition of multiple isomorphisms is also an isomorphism.

Proposition 2: Composition of Isomorphisms 同构的复合

Let $T:V\rightarrow W$ and $S:W\rightarrow Z$ be isomorphisms where $V,W,Z$ are subspaces of $\mathbb{R}^n$. Then $S\circ T$ defined by $\left( S\circ T\right) \left( \vec{v} \right) = S\left( T\left( \vec{v} \right) \right)$ is also an isomorphism.

Proof

Suppose $T:V\rightarrow W$ and $S:W\rightarrow Z$ are isomorphisms. Why is $S\circ T$ a linear map? For $a,b$ scalars,

$$\begin{aligned}S\circ T\left( a\vec{v}\_{1}+b(\vec{v}\_{2})\right) &= S\left( T\left(a\vec{v}\_{1}+b\vec{v}\_{2}\right) \right) =S\left( aT\vec{v}\_{1}+bT\vec{v}\_{2}\right) \\ &=aS\left( T\vec{v}\_{1}\right) +bS\left( T\vec{v}\_{2}\right) = a\left( S\circ T\right) \left( \vec{v}\_{1}\right) +b\left( S\circ T\right) \left( \vec{v}\_{2}\right)\end{aligned}$$

Hence $S\circ T$ is a linear map. If $\left( S\circ T\right) \left( \vec{v} \right) =0,$ then $S\left( T\left( \vec{v} \right) \right) =0$ and it follows that $T(\vec{v})=\vec{0}$ and hence by this lemma again, $\vec{v}=\vec{0}$. Thus $S\circ T$ is one to one. It remains to verify that it is onto. Let $\vec{z} \in Z$. Then since $S$ is onto, there exists $\vec{w} \in W$ such that $S(\vec{w})=\vec{z}.$ Also, since $T$ is onto, there exists $\vec{v}\in V$ such that $T(\vec{v})=\vec{w}.$ It follows that $S\left( T\left( \vec{v}\right) \right) =\vec{z}$ and so $S\circ T$ is also onto.

Consider two subspaces $V$ and $W$, and suppose there exists an isomorphism mapping one to the other. In this way the two subspaces are related, which we can write as $V \sim W$. Then the previous two propositions together claim that $\sim$ is an equivalence relation. That is: $\sim$ satisfies the following conditions:

• $V\sim V$
• If $V\sim W,$ it follows that $W\sim V$
• If $V\sim W$ and $W\sim Z,$ then $V\sim Z$

• $V\sim V$
• 若 $V\sim W,$ 则 $W\sim V$
• 若 $V\sim W$ 且 $W\sim Z,$ 则 $V\sim Z$

We leave the verification of these conditions as an exercise.

Consider the following example.

Example 2: Matrix Isomorphism 矩阵同构

Let $T:\mathbb{R}^{n}\rightarrow \mathbb{R}^{n}$ be defined by $T(\vec{x}) = A(\vec{x})$ where $A$ is an invertible $n\times n$ matrix. Then $T$ is an isomorphism.

Solution 解答

The reason for this is that, since $A$ is invertible, the only vector it sends to $\vec{0}$ is the zero vector. Hence if $A(\vec{x})=A(\vec{y}),$ then $A\left( \vec{x}-\vec{y}\right) =\vec{0}$ and so $\vec{x}=\vec{y}$. It is onto because if

$$\vec{y}\in \mathbb{R}^{n},A\left( A^{-1} (\vec{y})\right) =\left( AA^{-1}\right) (\vec{y}) =\vec{y}.$$

In fact, all isomorphisms from $\mathbb{R}^{n}$ to $\mathbb{R}^{n}$ can be expressed as $T(\vec{x}) = A(\vec{x})$ where $A$ is an invertible $n \times n$ matrix. One simply considers the matrix whose $i^{th}$ column is $T\vec{e}\_{i}$.

Recall that a basis of a subspace $V$ is a set of linearly independent vectors which span $V$. The following fundamental lemma describes the relation between bases and isomorphisms.

Lemma 1: Mapping Bases 映射基

Let $T:V\rightarrow W$ be a linear transformation where $V,W$ are subspaces of $\mathbb{R}^n$. If $T$ is one to one, then it has the property that if $\left\{ \vec{u}\_{1},\cdots ,\vec{u}\_{k}\right\}$ is linearly independent, so is $\left\{ T(\vec{u}\_{1}),\cdots ,T(\vec{u}\_{k})\right\}$.

More generally, $T$ is an isomorphism if and only if whenever $\left\{ \vec{v}\_{1},\cdots ,\vec{v}\_{n}\right\}$ is a basis for $V,$ it follows that $\left\{ T (\vec{v}\_{1}),\cdots ,T(\vec{v}\_{n})\right\}$ is a basis for $W$.

Proof

First suppose that $T$ is a linear transformation and is one to one and $\left\{ \vec{u}\_{1},\cdots ,\vec{u}\_{k}\right\}$ is linearly independent. It is required to show that $\left\{ T(\vec{u}\_{1}),\cdots ,T(\vec{ u}\_{k})\right\}$ is also linearly independent. Suppose then that $\sum\_{i=1}^{k}c\_{i}T(\vec{u}\_{i})=\vec{0}$ Then, since $T$ is linear, $T\left( \sum\_{i=1}^{n}c\_{i}\vec{u}\_{i}\right) =\vec{0}$ Since $T$ is one to one, it follows that $\sum\_{i=1}^{n}c\_{i}\vec{u}\_{i}=0$ Now the fact that $\left\{ \vec{u}\_{1},\cdots ,\vec{u}\_{n}\right\}$ is linearly independent implies that each $c\_{i}=0$. Hence $\left\{ T(\vec{u} \_{1}),\cdots ,T(\vec{u}\_{n})\right\}$ is linearly independent.

Now suppose that $T$ is an isomorphism and $\left\{ \vec{v}\_{1},\cdots ,\vec{ v}\_{n}\right\}$ is a basis for $V$. It was just shown that $\left\{ T(\vec{v} \_{1}),\cdots ,T(\vec{v}\_{n})\right\}$ is linearly independent. It remains to verify that span$\left\{ T(\vec{v}\_{1}),\cdots ,T(\vec{v}\_{n})\right\}=W$. If $\vec{w}\in W,$ then since $T$ is onto there exists $\vec{v}\in V$ such that $T(\vec{v})=\vec{w}$. Since $\left\{ \vec{v} \_{1},\cdots ,\vec{v}\_{n}\right\}$ is a basis, it follows that there exists scalars $\left\{ c\_{i}\right\} \_{i=1}^{n}$ such that $\sum\_{i=1}^{n}c\_{i}\vec{v}\_{i}=\vec{v}. $ Hence, $\vec{w}=T(\vec{v})=T\left( \sum\_{i=1}^{n}c\_{i}\vec{v}\_{i}\right) =\sum\_{i=1}^{n}c\_{i}T(\vec{v}\_{i}) $ It follows that span$\left\{ T(\vec{v}\_{1}),\cdots ,T(\vec{v}\_{n})\right\} =W$ showing that this set of vectors is a basis for $W$.

Next suppose that $T$ is a linear transformation which takes a basis to a basis. This means that if $\left\{ \vec{v}\_{1},\cdots ,\vec{v}\_{n}\right\}$ is a basis for $V,$ it follows $\left\{ T(\vec{v}\_{1}),\cdots ,T(\vec{v}\_{n})\right\}$ is a basis for $W.$ Then if $w\in W,$ there exist scalars $c\_{i}$ such that $w=\sum\_{i=1}^{n}c\_{i}T(\vec{v}\_{i})=T\left( \sum\_{i=1}^{n}c\_{i}\vec{v}\_{i}\right)$ showing that $T$ is onto. If $T\left( \sum\_{i=1}^{n}c\_{i}\vec{v}\_{i}\right) =\vec{0}$ then $\sum\_{i=1}^{n}c\_{i}T(\vec{v}\_{i})=\vec{0}$ and since the vectors $\left\{ T(\vec{v} \_{1}),\cdots ,T(\vec{v}\_{n})\right\}$ are linearly independent, it follows that each $c\_{i}=0.$ Since $\sum\_{i=1}^{n}c\_{i}\vec{v}\_{i}$ is a typical vector in $V$, this has shown that if $T(\vec{v})=\vec{0}$ then $\vec{v}=\vec{0}$ and so $T$ is also one to one. Thus $T$ is an isomorphism.$

The following theorem illustrates a very useful idea for defining an isomorphism. Basically, if you know what it does to a basis, then you can construct the isomorphism.

Theorem 1: Isomorphic Subspaces 同构的子空间

Suppose $V$ and $W$ are two subspaces of $\mathbb{R}^n$. Then the two subspaces are isomorphic if and only if they have the same dimension. In the case that the two subspaces have the same dimension, then for a linear map $T:V\rightarrow W$, the following are equivalent.

1. $T$ is one to one.
2. $T$ is onto.
3. $T$ is an isomorphism.

1. $T$ 是单射。
2. $T$ 是满射。
3. $T$ 是同构。

Proof

Suppose first that these two subspaces have the same dimension. Let a basis for $V$ be $\left\{ \vec{v}\_{1},\cdots ,\vec{v}\_{n}\right\}$ and let a basis for $W$ be $\left\{ \vec{w}\_{1},\cdots ,\vec{w}\_{n}\right\}$. Now define $T$ as follows. $T(\vec{v}\_{i})=\vec{w}\_{i}$ for $\sum\_{i=1}^{n}c\_{i}\vec{v}\_{i}$ an arbitrary vector of $V,$ $T\left( \sum\_{i=1}^{n}c\_{i}\vec{v}\_{i}\right) = \sum\_{i=1}^{n}c\_{i}T \vec{v}\_{i}=\sum\_{i=1}^{n}c\_{i}\vec{w}\_{i}.$ It is necessary to verify that this is well defined. Suppose then that $\sum\_{i=1}^{n}c\_{i}\vec{v}\_{i}=\sum\_{i=1}^{n}\hat{c}\_{i}\vec{v}\_{i}$ Then $\sum\_{i=1}^{n}\left( c\_{i}-\hat{c}\_{i}\right) \vec{v}\_{i}=\vec{0}$ and since $\left\{ \vec{v}\_{1},\cdots ,\vec{v}\_{n}\right\}$ is a basis, $c\_{i}=\hat{c}\_{i}$ for each $i$. Hence $\sum\_{i=1}^{n}c\_{i}\vec{w}\_{i}=\sum\_{i=1}^{n}\hat{c}\_{i}\vec{w}\_{i}$ and so the mapping is well defined. Also if $a,b$ are scalars, $\begin{aligned}T\left( a\sum\_{i=1}^{n}c\_{i}\vec{v}\_{i}+b\sum\_{i=1}^{n}\hat{c}\_{i}\vec{v}\_{i}\right) &=T\left( \sum\_{i=1}^{n}\left( ac\_{i}+b\hat{c}\_{i}\right) \vec{v}\_{i}\right) =\sum\_{i=1}^{n}\left( ac\_{i}+b\hat{c}\_{i}\right) \vec{w}\_{i} \\ &=a\sum\_{i=1}^{n}c\_{i}\vec{w}\_{i}+b\sum\_{i=1}^{n}\hat{c}\_{i}\vec{w}\_{i} \\ &=aT\left( \sum\_{i=1}^{n}c\_{i}\vec{v}\_{i}\right) +bT\left( \sum\_{i=1}^{n} \hat{c}\_{i}\vec{v}\_{i}\right)\end{aligned}$ Thus $T$ is a linear transformation.$

Now if $T\left( \sum\_{i=1}^{n}c\_{i}\vec{v}\_{i}\right) =\sum\_{i=1}^{n}c\_{i}\vec{w}\_{i}=\vec{0}, $ then since the $\left\{ \vec{w}\_{1},\cdots ,\vec{w}\_{n}\right\}$ are independent, each $c\_{i}=0$ and so $\sum\_{i=1}^{n}c\_{i}\vec{v}\_{i}=\vec{0}$ also. Hence $T$ is one to one. If $\sum\_{i=1}^{n}c\_{i}\vec{w}\_{i}$ is a vector in $W,$ then it equals $\sum\_{i=1}^{n}c\_{i}T(\vec{v}\_{i})=T\left( \sum\_{i=1}^{n}c\_{i}\vec{v}\_{i}\right) $ showing that $T$ is also onto. Hence $T$ is an isomorphism and so $V$ and $W$ are isomorphic.$

Next suppose $T:V \mapsto W$ is an isomorphism, so these two subspaces are isomorphic. Then for $\left\{ \vec{v}\_{1},\cdots ,\vec{v} \_{n}\right\}$ a basis for $V$, it follows that a basis for $W$ is $\left\{ T(\vec{v}\_{1}),\cdots ,T(\vec{v}\_{n})\right\}$ showing that the two subspaces have the same dimension.

Now suppose the two subspaces have the same dimension. Consider the three claimed equivalences.

First consider the claim that $1.)\Rightarrow 2.).$ If $T$ is one to one and if $\left\{ \vec{v}\_{1},\cdots ,\vec{v} \_{n}\right\}$ is a basis for $V,$ then $\left\{ T(\vec{v}\_{1}),\cdots ,T(\vec{v }\_{n})\right\}$ is linearly independent. If it is not a basis, then it must fail to span $W$. But then there would exist $\vec{w}\notin \mathrm{span} \left\{ T(\vec{v}\_{1}),\cdots ,T(\vec{v}\_{n})\right\}$ and it follows that $\left\{ T(\vec{v}\_{1}),\cdots ,T(\vec{v}\_{n}),\vec{w} \right\}$ would be linearly independent which is impossible because there exists a basis for $W$ of $n$ vectors.

Hence $\mathrm{span}\left\{ T(\vec{v}\_{1}),\cdots ,T(\vec{v}\_{n})\right\} =W$ and so $\left\{ T(\vec{v}\_{1}),\cdots ,T(\vec{v}\_{n})\right\}$ is a basis. If $\vec{w}\in W,$ there exist scalars $c\_{i}$ such that $\vec{w}=\sum\_{i=1}^{n}c\_{i}T(\vec{v}\_{i})=T\left( \sum\_{i=1}^{n}c\_{i}\vec{v} \_{i}\right) $ showing that $T$ is onto. This shows that $1.)\Rightarrow 2.).$

Next consider the claim that $2.)\Rightarrow 3.).$ Since $2.)$ holds, it follows that $T$ is onto. It remains to verify that $T$ is one to one. Since $T$ is onto, there exists a basis of the form $\left\{ T(\vec{v}\_{i}),\cdots ,T(\vec{v}\_{n})\right\} .$ Then it follows that $\left\{ \vec{v}\_{1},\cdots , \vec{v}\_{n}\right\}$ is linearly independent. Suppose $\sum\_{i=1}^{n}c\_{i}\vec{v}\_{i}=\vec{0}$ Then $\sum\_{i=1}^{n}c\_{i}T(\vec{v}\_{i})=\vec{0}$ Hence each $c\_{i}=0$ and so, $\left\{ \vec{v}\_{1},\cdots ,\vec{v} \_{n}\right\}$ is a basis for $V$. Now it follows that a typical vector in $V$ is of the form $\sum\_{i=1}^{n}c\_{i}\vec{v}\_{i}$. If $T\left( \sum\_{i=1}^{n}c\_{i}\vec{v}\_{i}\right) =\vec{0},$ it follows that $\sum\_{i=1}^{n}c\_{i}T(\vec{v}\_{i})=\vec{0}$ and so, since $\left\{ T(\vec{v}\_{i}),\cdots ,T(\vec{v}\_{n})\right\}$ is independent, it follows each $c\_{i}=0$ and hence $\sum\_{i=1}^{n}c\_{i}\vec{v} \_{i}=\vec{0}$. Thus $T$ is one to one as well as onto and so it is an isomorphism.$

If $T$ is an isomorphism, it is both one to one and onto by definition so $3.)$ implies both $1.)$ and $2.)$.

Note the interesting way of defining a linear transformation in the first part of the argument by describing what it does to a basis and then “extending it linearly” to the entire subspace.

Example 4: Isomorphic Subspaces 同构的子空间

Let $V=\mathbb{R}^{3}$ and let $W$ denote $\mathrm{span}\left\{ \begin{pmatrix}1 \\ 2 \\ 1 \\ 1\end{pmatrix} ,\begin{pmatrix}0 \\ 1 \\ 0 \\ 1\end{pmatrix} ,\begin{pmatrix}1 \\ 1 \\ 2 \\ 0\end{pmatrix} \right\}nonumber $ Show that $V$ and $W$ are isomorphic.$

Solution 解答

First observe that these subspaces are both of dimension 3 and so they are isomorphic by Theorem 1. The three vectors which span $W$ are easily seen to be linearly independent by making them the columns of a matrix and row reducing to the reduced row-echelon form.

You can exhibit an isomorphism of these two spaces as follows. $T(\vec{e}\_{1})=\begin{pmatrix}1 \\ 2 \\ 1 \\ 1\end{pmatrix}, T(\vec{e}\_{2})=\begin{pmatrix}0 \\ 1 \\ 0 \\ 1\end{pmatrix}, T(\vec{e}\_{3})=\begin{pmatrix}1 \\ 1 \\ 2 \\ 0\end{pmatrix} $ and extend linearly. Recall that the matrix of this linear transformation is just the matrix having these vectors as columns. Thus the matrix of this isomorphism is $\begin{pmatrix}1 & 0 & 1 \\ 2 & 1 & 1 \\ 1 & 0 & 2 \\ 1 & 1 & 0\end{pmatrix} $ You should check that multiplication on the left by this matrix does reproduce the claimed effect resulting from an application by $T$.

Consider the following example.

Example 5: Finding the Matrix of an Isomorphism 求同构的矩阵

Let $V=\mathbb{R}^{3}$ and let $W$ denote

$$\mathrm{span}\left\{ \begin{pmatrix}1 \\ 2 \\ 1 \\ 1\end{pmatrix} ,\begin{pmatrix}0 \\ 1 \\ 0 \\ 1\end{pmatrix} ,\begin{pmatrix}1 \\ 1 \\ 2 \\ 0\end{pmatrix} \right\}nonumber$$

Let $T: V \mapsto W$ be defined as follows. $T\begin{pmatrix}1 \\ 1 \\ 0\end{pmatrix} =\begin{pmatrix}1 \\ 2 \\ 1 \\ 1\end{pmatrix} ,T\begin{pmatrix}0 \\ 1 \\ 1\end{pmatrix} =\begin{pmatrix}0 \\ 1 \\ 0 \\ 1\end{pmatrix} ,T\begin{pmatrix}1 \\ 1 \\ 1\end{pmatrix} =\begin{pmatrix}1 \\ 1 \\ 2 \\ 0\end{pmatrix} $ Find the matrix of this isomorphism $T$.$

Solution 解答

First note that the vectors $\begin{pmatrix}1 \\ 1 \\ 0\end{pmatrix} ,\begin{pmatrix}0 \\ 1 \\ 1\end{pmatrix} ,\begin{pmatrix}1 \\ 1 \\ 1\end{pmatrix} $ are indeed a basis for $\mathbb{R}^{3}$ as can be seen by making them the columns of a matrix and using the reduced row-echelon form.

Now recall the matrix of $T$ is a $4\times 3$ matrix $A$ which gives the same effect as $T.$ Thus, from the way we multiply matrices, $A\begin{pmatrix}1 & 0 & 1 \\ 1 & 1 & 1 \\ 0 & 1 & 1\end{pmatrix} =\begin{pmatrix}1 & 0 & 1 \\ 2 & 1 & 1 \\ 1 & 0 & 2 \\ 1 & 1 & 0\end{pmatrix} $ Hence, $A=\begin{pmatrix}1 & 0 & 1 \\ 2 & 1 & 1 \\ 1 & 0 & 2 \\ 1 & 1 & 0\end{pmatrix} \begin{pmatrix}1 & 0 & 1 \\ 1 & 1 & 1 \\ 0 & 1 & 1\end{pmatrix} ^{-1}=\begin{pmatrix}1 & 0 & 0 \\ 0 & 2 & -1 \\ 2 & -1 & 1 \\ -1 & 2 & -1\end{pmatrix} $

Note how the span of the columns of this new matrix must be the same as the span of the vectors defining $W$.

This idea of defining a linear transformation by what it does on a basis works for linear maps which are not necessarily isomorphisms.

Example 6: Finding the Matrix of an Isomorphism 求同构的矩阵

Let $V=\mathbb{R}^{3}$ and let $W$ denote $\mathrm{span}\left\{ \begin{pmatrix}1 \\ 0 \\ 1 \\ 1\end{pmatrix} ,\begin{pmatrix}0 \\ 1 \\ 0 \\ 1\end{pmatrix} ,\begin{pmatrix}1 \\ 1 \\ 1 \\ 2\end{pmatrix} \right\}nonumber $ Let $T: V \mapsto W$ be defined as follows. $T\begin{pmatrix}1 \\ 1 \\ 0\end{pmatrix} = \begin{pmatrix}1 \\ 0 \\ 1 \\ 1\end{pmatrix} ,T\begin{pmatrix}0 \\ 1 \\ 1\end{pmatrix} =\begin{pmatrix}0 \\ 1 \\ 0 \\ 1\end{pmatrix} ,T\begin{pmatrix}1 \\ 1 \\ 1\end{pmatrix} =\begin{pmatrix}1 \\ 1 \\ 1 \\ 2\end{pmatrix} $ Find the matrix of this linear transformation.

Solution 解答

Note that in this case, the three vectors which span $W$ are not linearly independent. Nevertheless the above procedure will still work. The reasoning is the same as before. If $A$ is this matrix, then $A\begin{pmatrix}1 & 0 & 1 \\ 1 & 1 & 1 \\ 0 & 1 & 1\end{pmatrix} =\begin{pmatrix}1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 2\end{pmatrix} $ and so $A=\begin{pmatrix}1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 2\end{pmatrix} \begin{pmatrix}1 & 0 & 1 \\ 1 & 1 & 1 \\ 0 & 1 & 1\end{pmatrix} ^{-1}=\begin{pmatrix}1 & 0 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \\ 1 & 0 & 1\end{pmatrix} $

The columns of this last matrix are obviously not linearly independent.

Outcomes 学习目标

1. Describe the kernel and image of a linear transformation, and find a basis for each.

1. 描述线性变换的核与像，并求二者的一个基。

In this section we will consider the case where the linear transformation is not necessarily an isomorphism. First consider the following important definition.

Definition 1: Kernel and Image 定义 1：核与像

Let $V$ and $W$ be subspaces of $\mathbb{R}^n$ and let $T:V\mapsto W$ be a linear transformation. Then the image of $T$ denoted as $\mathrm{im}\left( T\right)$ is defined to be the set $\mathrm{im}\left( T\right) = \left\{T (\vec{v}):\vec{v}\in V\right\}nonumber $ In words, it consists of all vectors in $W$ which equal $T(\vec{v})$ for some $\vec{v}\in V$.

The kernel of $T$, written $\ker \left( T\right)$, consists of all $\vec{v}\in V$ such that $T(\vec{v})=\vec{0}$. That is, $\ker \left( T\right) =\left\{ \vec{v}\in V:T(\vec{v})=\vec{0}\right\}nonumber $

It follows that $\mathrm{im}\left( T\right)$ and $\ker \left( T\right)$ are subspaces of $W$ and $V$ respectively.

Proposition 1: Kernel and Image as Subspaces 命题 1：作为子空间的核与像

Let $V, W$ be subspaces of $\mathbb{R}^n$ and let $T:V\rightarrow W$ be a linear transformation. Then $\ker \left( T\right)$ is a subspace of $V$ and $\mathrm{im}\left( T\right)$ is a subspace of $W$.

Proof

First consider $\ker \left( T\right) .$ It is necessary to show that if $\vec{v}\_{1},\vec{v}\_{2}$ are vectors in $\ker \left( T\right)$ and if $a,b$ are scalars, then $a\vec{v}\_{1}+b\vec{v}\_{2}$ is also in $\ker \left( T\right) .$ But $T\left( a\vec{v}\_{1}+b\vec{v}\_{2}\right) =aT(\vec{v}\_{1})+bT(\vec{v}\_{2})=a\vec{0}+b\vec{0}=\vec{0} $

Thus $\ker \left( T\right)$ is a subspace of $V$.

Next suppose $T(\vec{v}\_{1}),T(\vec{v}\_{2})$ are two vectors in $\mathrm{im}\left( T\right) .$ Then if $a,b$ are scalars, $aT(\vec{v}\_{2})+bT(\vec{v}\_{2})=T\left( a\vec{v}\_{1}+b\vec{v}\_{2}\right)$ and this last vector is in $\mathrm{im}\left( T\right)$ by definition.

We will now examine how to find the kernel and image of a linear transformation and describe the basis of each.

Example 1: Kernel and Image of a Linear Transformation 示例 1：一个线性变换的核与像

Let $T: \mathbb{R}^4 \mapsto \mathbb{R}^2$ be defined by

$$T \begin{pmatrix}a \\ b \\ c \\ d\end{pmatrix} = \begin{pmatrix}a - b \\ c + d\end{pmatrix}$$

Then $T$ is a linear transformation. Find a basis for $\mathrm{ker}(T)$ and $\mathrm{im}(T)$.

Solution

You can verify that $T$ is a linear transformation.

First we will find a basis for $\mathrm{ker}(T)$. To do so, we want to find a way to describe all vectors $\vec{x} \in \mathbb{R}^4$ such that $T(\vec{x}) = \vec{0}$. Let $\vec{x} = \begin{pmatrix}a \\ b \\ c \\ d\end{pmatrix}$ be such a vector. Then$

$$T \begin{pmatrix}a \\ b \\ c \\ d\end{pmatrix} = \begin{pmatrix}a - b \\ c + d\end{pmatrix} = \left ( \begin{array}{c} 0 \\ 0 \end{array} \right )$$

The values of $a, b, c, d$ that make this true are given by solutions to the system

$$\begin{aligned}a - b &= 0 \\ c + d &= 0\end{aligned}$$

The solution to this system is $a = s, b = s, c = t, d = -t$ where $s, t$ are scalars. We can describe $\mathrm{ker}(T)$ as follows.

$$\mathrm{ker}(T) = \left\{ \begin{pmatrix}s \\ s \\ t \\ -t\end{pmatrix} \right\} = \mathrm{span} \left\{ \begin{pmatrix}1 \\ 1 \\ 0 \\ 0\end{pmatrix}, \begin{pmatrix}0 \\ 0 \\ 1 \\ -1\end{pmatrix} \right\}$$

Notice that this set is linearly independent and therefore forms a basis for $\mathrm{ker}(T)$.

We move on to finding a basis for $\mathrm{im}(T)$. We can write the image of $T$ as $\mathrm{im}(T) = \left\{ \begin{pmatrix}a - b \\ c + d\end{pmatrix} \right\}$$

We can write this in the form $\mathrm{span} = \left\{ \begin{pmatrix}1 \\ 0\end{pmatrix}, \begin{pmatrix}-1 \\ 0\end{pmatrix}, \begin{pmatrix}0 \\ 1\end{pmatrix}, \begin{pmatrix}0 \\ 1\end{pmatrix} \right\}nonumber$

This set is clearly not linearly independent. By removing unnecessary vectors from the set we can create a linearly independent set with the same span. This gives a basis for $\mathrm{im}(T)$ as $\mathrm{im}(T) = \mathrm{span} \left\{ \begin{pmatrix}1 \\ 0\end{pmatrix}, \begin{pmatrix}0 \\ 1\end{pmatrix} \right\}nonumber$

Recall that a linear transformation $T$ is called one to one if and only if $T(\vec{x}) = \vec{0}$ implies $\vec{x} = \vec{0}$. Using the concept of kernel, we can state this theorem in another way.

Theorem 1: One to One and Kernel 定理 1：单射与核

Let $T$ be a linear transformation where $\mathrm{ker}(T)$ is the kernel of $T$. Then $T$ is one to one if and only if $\mathrm{ker}(T)$ consists of only the zero vector.

A major result is the relation between the dimension of the kernel and dimension of the image of a linear transformation. In the previous example $\mathrm{ker}(T)$ had dimension $2$, and $\mathrm{im}(T)$ also had dimension of $2$. Is it a coincidence that the dimension of $\mathbb{M}\_{22}$ is $4 = 2 + 2$? Consider the following theorem.

Theorem 2: Dimension of Kernel and Image 定理 2：核与像的维数

Let $T:V\rightarrow W$ be a linear transformation where $V,W$ are subspaces of $\mathbb{R}^n$. Suppose the dimension of $V$ is $m$. Then $m=\dim \left( \ker \left( T\right) \right) +\dim \left( \mathrm{im}\left( T\right) \right) $

Proof

From Proposition 1, $\mathrm{im}\left( T\right)$ is a subspace of $W.$ We know that there exists a basis for $\mathrm{im}\left( T\right)$, $\left\{ T(\vec{v} \_{1}),\cdots ,T(\vec{v}\_{r})\right\} .$ Similarly, there is a basis for $\ker \left( T\right) ,\left\{ \vec{u}\_{1},\cdots ,\vec{u}\_{s}\right\}. Then if $\vec{v}\in V,$ there exist scalars $c\_{i}$ such that $T(\vec{v})=\sum\_{i=1}^{r}c\_{i}T(\vec{v}\_{i}) $ Hence $T\left( \vec{v}-\sum\_{i=1}^{r}c\_{i}\vec{v}\_{i}\right) =0.$ It follows that $\vec{v}-\sum\_{i=1}^{r}c\_{i}\vec{v}\_{i}$ is in $\ker \left( T\right)$. Hence there are scalars $a\_{i}$ such that $\vec{v}-\sum\_{i=1}^{r}c\_{i}\vec{v}\_{i}=\sum\_{j=1}^{s}a\_{j}\vec{u}\_{j} $ Hence $\vec{v}=\sum\_{i=1}^{r}c\_{i}\vec{v}\_{i}+\sum\_{j=1}^{s}a\_{j}\vec{u} \_{j}.$ Since $\vec{v}$ is arbitrary, it follows that $V=\mathrm{span}\left\{ \vec{u}\_{1},\cdots ,\vec{u}\_{s},\vec{v}\_{1},\cdots , \vec{v}\_{r}\right\}nonumber$$

If the vectors $\left\{ \vec{u}\_{1},\cdots ,\vec{u}\_{s},\vec{v}\_{1},\cdots , \vec{v}\_{r}\right\} are linearly independent, then it will follow that this set is a basis. Suppose then that $\sum\_{i=1}^{r}c\_{i}\vec{v}\_{i}+\sum\_{j=1}^{s}a\_{j}\vec{u}\_{j}=0 $ Apply $T$ to both sides to obtain $\sum\_{i=1}^{r}c\_{i}T(\vec{v}\_{i})+\sum\_{j=1}^{s}a\_{j}T(\vec{u}) \_{j}=\sum\_{i=1}^{r}c\_{i}T(\vec{v}\_{i})=0 $ Since $\left\{ T(\vec{v}\_{1}),\cdots ,T(\vec{v}\_{r})\right\} is linearly independent, it follows that each $c\_{i}=0.$ Hence $\sum\_{j=1}^{s}a\_{j}\vec{u }\_{j}=0$ and so, since the $\left\{ \vec{u}\_{1},\cdots ,\vec{u}\_{s}\right\} are linearly independent, it follows that each $a\_{j}=0$ also. Therefore $\left\{ \vec{u}\_{1},\cdots ,\vec{u}\_{s},\vec{v}\_{1},\cdots ,\vec{v} \_{r}\right\} is a basis for $V$ and so $n=s+r=\dim \left( \ker \left( T\right) \right) +\dim \left( \mathrm{im}\left( T\right) \right) $

The above theorem leads to the next corollary.

Corollary 1 推论 1

Let $T:V\rightarrow W$ be a linear transformation where $V,W$ are subspaces of $\mathbb{R}^n$. Suppose the dimension of $V$ is $m$. Then $\dim \left( \ker \left( T\right) \right) \leq m $ $\dim \left( \mathrm{im}\left( T \right) \right) \leq m $

This follows directly from the fact that $n=\dim \left( \ker \left( T\right) \right) +\dim \left( \mathrm{im}\left( T\right) \right)$.

Consider the following example.

Example 2 示例 2

Let $T:\mathbb{R}^{2}\rightarrow \mathbb{R}^{3}$ be defined by $T(\vec{x})=\begin{pmatrix}1 & 0 \\ 1 & 0 \\ 0 & 1\end{pmatrix} \vec{x} $ Then $\mathrm{im}\left( T\right) =V$ is a subspace of $\mathbb{R}^{3}$ and $T$ is an isomorphism of $\mathbb{R}^{2}$ and $V$. Find a $2\times 3$ matrix $A$ such that the restriction of multiplication by $A$ to $V=\mathrm{im}\left( T\right)$ equals $T^{-1}$.

Solution

Since the two columns of the above matrix are linearly independent, we conclude that $\mathrm{dim}(\mathrm{im}(T)) = 2$ and therefore $\mathrm{dim}(\mathrm{ker}(T)) = 2 - \mathrm{dim}(\mathrm{im}(T)) = 2-2 = 0$ by Theorem 2. Then by Theorem 1 it follows that $T$ is one to one.

Thus $T$ is an isomorphism of $\mathbb{R }^{2}$ and the two dimensional subspace of $\mathbb{R}^{3}$ which is the span of the columns of the given matrix. Now in particular, $T(\vec{e}\_{1})=\begin{pmatrix}1 \\ 1 \\ 0\end{pmatrix} ,$ T(\vec{e}\_{2})=\begin{pmatrix}0 \\ 0 \\ 1\end{pmatrix} $$

Thus $T^{-1}\begin{pmatrix}1 \\ 1 \\ 0\end{pmatrix} =\vec{e}\_{1},$ T^{-1}\begin{pmatrix}0 \\ 0 \\ 1\end{pmatrix} =\vec{e}\_{2} $$

Extend $T^{-1}$ to all of $\mathbb{R}^{3}$ by defining $T^{-1}\begin{pmatrix}0 \\ 1 \\ 0\end{pmatrix} =\vec{e}\_{1} $ Notice that the vectors $\left\{ \begin{pmatrix}1 \\ 1 \\ 0\end{pmatrix} ,\begin{pmatrix}0 \\ 0 \\ 1\end{pmatrix} ,\begin{pmatrix}0 \\ 1 \\ 0\end{pmatrix} \right\} $ are linearly independent so $T^{-1}$ can be extended linearly to yield a linear transformation defined on $\mathbb{R}^{3}$. The matrix of $T^{-1}$ denoted as $A$ needs to satisfy $A\begin{pmatrix}1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0\end{pmatrix} =\begin{pmatrix}1 & 0 & 1 \\ 0 & 1 & 0\end{pmatrix} $ and so $A=\begin{pmatrix}1 & 0 & 1 \\ 0 & 1 & 0\end{pmatrix} \begin{pmatrix}1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0\end{pmatrix}^{-1}=\begin{pmatrix}0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix} $$

Note that $\begin{pmatrix}0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix} \begin{pmatrix}1 \\ 1 \\ 0\end{pmatrix} =\begin{pmatrix}1 \\ 0\end{pmatrix} $ $\begin{pmatrix}0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix} \begin{pmatrix}0 \\ 0 \\ 1\end{pmatrix} =\begin{pmatrix}0 \\ 1\end{pmatrix} $ so the restriction to $V$ of matrix multiplication by this matrix yields $T^{-1}.$

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Original page: https://math.libretexts.org/Bookshelves/Linear_Algebra/A_First_Course_in_Linear_Algebra_(Kuttler)/05%3A_Linear_Transformations/5.08%3A_The_Matrix_of_a_Linear_Transformation_II

Outcomes 学习目标

1. Find the matrix of a linear transformation with respect to general bases.

1. 求线性变换关于一般基的矩阵。

We begin this section with an important lemma.

Lemma 1: Mapping of a Basis 引理 1：基的映射

Let $T: \mathbb{R}^n \mapsto \mathbb{R}^n$ be an isomorphism. Then $T$ maps any basis of $\mathbb{R}^n$ to another basis for $\mathbb{R}^n$.

Conversely, if $T: \mathbb{R}^n \mapsto \mathbb{R}^n$ is a linear transformation which maps a basis of $\mathbb{R}^n$ to another basis of $\mathbb{R}^n$, then it is an isomorphism.

Proof

First, suppose $T:\mathbb{R}^n \mapsto \mathbb{R}^n$ is a linear transformation which is one to one and onto. Let $\left\{ \vec{v}\_{1},\cdots ,\vec{v}\_{n}\right\}$ be a basis for $\mathbb{R}^n$. We wish to show that $\left\{ T(\vec{v}\_{1}),\cdots , T(\vec{v}\_{n})\right\}$ is also a basis for $\mathbb{R}^n$.

First consider why it is linearly independent. Suppose $\sum\_{k=1}^{n}a\_{k}T(\vec{v}\_{k})=\vec{0}$. Then by linearity we have $T\left( \sum\_{k=1}^{n}a\_{k}\vec{v}\_{k}\right) =\vec{0}$ and since $T$ is one to one, it follows that $\sum\_{k=1}^{n}a\_{k}\vec{v}\_{k}=\vec{0}$. This requires that each $a\_{k}=0$ because $\left\{ \vec{v}\_{1},\cdots, \vec{v}\_{n}\right\}$ is independent, and it follows that $\left\{ T(\vec{v}\_{1}),\cdots , T(\vec{v}\_{n})\right\}$ is linearly independent.

Next take $\vec{w}\in \mathbb{R}^n.$ Since $T$ is onto, there exists $\vec{v}\in \mathbb{R}^n$ such that $T(\vec{v})=\vec{w}$. Since $\left\{ \vec{v}\_{1},\cdots ,\vec{v}\_{n}\right\}$ is a basis, in particular it is a spanning set and there are scalars $b\_{k}$ such that $T\left( \sum\_{k=1}^{n}b\_{k}\vec{v} \_{k}\right) =T\left( \vec{v}\right) =\vec{w}$. Therefore $\vec{w} =\sum\_{k=1}^{n}b\_{k}T(\vec{v}\_{k})$ which is in the $\mathrm{span}\left\{ T(\vec{v}\_{1}),\cdots , T(\vec{v}\_{n})\right\} .$ Therefore, $\left\{ T(\vec{v}\_{1}),\cdots ,T(\vec{v}\_{n}) \right\}$ is a basis as claimed.

Suppose now that $T: \mathbb{R}^n \mapsto \mathbb{R}^n$ is a linear transformation such that $T(\vec{v}\_{i})=\vec{w}\_{i}$ where $\left\{vec{v} \_{1},\cdots ,\vec{v}\_{n}\right\}$ and $\left\{ \vec{w}\_{1},\cdots , \vec{w}\_{n}\right\}$ are two bases for $\mathbb{R}^n$.

To show that $T$ is one to one, let $T\left( \sum\_{k=1}^{n}c\_{k}\vec{v}\_{k}\right) =\vec{0}$. Then $\sum\_{k=1}^{n}c\_{k}T(\vec{v}\_{k})=\sum\_{k=1}^{n}c\_{k}\vec{w}\_{k}=\vec{ 0}$. It follows that each $c\_{k} = 0$ because it is given that $\left\{ \vec{w} \_{1},\cdots,\vec{w}\_{n}\right\}$ is linearly independent. Hence $T\left( \sum\_{k=1}^{n}c\_{k}\vec{v}\_{k}\right) =\vec{0}$ implies that $\sum\_{k=1}^{n}c\_{k}\vec{v}\_{k}=\vec{0}$ and so $T$ is one to one.

To show that $T$ is onto, let $\vec{w}$ be an arbitrary vector in $\mathbb{R}^n$. This vector can be written as $\vec{w} = \sum\_{k=1}^{n}d_k\vec{w}\_k = \sum\_{k=1}^{n}d\_{k}T(\vec{v}\_{k})=T\left( \sum\_{k=1}^{n}d\_{k} \vec{v}\_{k}\right) .$ Therefore, $T$ is also onto.

Consider now an important definition.

Definition 1: Coordinate Vector 定义 1：坐标向量

Let $B = \left\{ \vec{v}\_1, \vec{v}\_2, \cdots, \vec{v}\_n \right\}$ be a basis for $\mathbb{R}^n$ and let $\vec{x}$ be an arbitrary vector in $\mathbb{R}^n$. Then $\vec{x}$ is uniquely represented as $\vec{x} = a_1\vec{v}\_1 + a_2\vec{v}\_2 + \cdots + a_n\vec{v}\_n$ for scalars $a_1, \cdots, a_n.$

The coordinate vector of $\vec{x}$ with respect to the basis $B$, written $C_B(\vec{x})$ or $\vec{x}\_B$, is given by $C_B(\vec{x}) = C_B \left( a_1\vec{v}\_1 + a_2\vec{v}\_2 + \cdots + a_n\vec{v}\_n \right) = \begin{pmatrix}a_1 \\ a_2 \\ \vdots \\ a_n\end{pmatrix} $

Consider the following example.

Example 1: Coordinate Vector 示例 1：坐标向量

Let $B = \left\{ \begin{pmatrix}1 \\ 0\end{pmatrix}, \begin{pmatrix}-1 \\ 1\end{pmatrix} \right\}$ be a basis of $\mathbb{R}^2$ and let $\vec{x} = \begin{pmatrix}3 \\ -1\end{pmatrix}$ be a vector in $\mathbb{R}^2$. Find $C_B(\vec{x})$.

Solution

First, note the order of the basis is important so label the vectors in the basis $B$ as $B = \left\{ \begin{pmatrix}1 \\ 0\end{pmatrix}, \begin{pmatrix}-1 \\ 1\end{pmatrix} \right\} = \left\{ \vec{v}\_1, \vec{v}\_2 \right\}nonumber $ Now we need to find $a_1, a_2$ such that $\vec{x} = a_1 \vec{v}\_1 + a_2 \vec{v}\_2$, that is: $\begin{pmatrix}3 \\ -1\end{pmatrix} = a_1 \begin{pmatrix}1 \\ 0\end{pmatrix} + a_2 \begin{pmatrix}-1 \\ 1\end{pmatrix} $ Solving this system gives $a_1 = 2, a_2 = -1$. Therefore the coordinate vector of $\vec{x}$ with respect to the basis $B$ is $C_B(\vec{x}) = \begin{pmatrix}a_1 \\ a_2\end{pmatrix} = \begin{pmatrix}2 \\ -1\end{pmatrix} $

Given any basis $B$, one can easily verify that the coordinate function is actually an isomorphism.

Theorem 1: $C_B$Transformation is a Linear 定理 1：坐标函数 $C_B$ 是线性变换

For any basis $B$ of $\mathbb{R}^n$, the coordinate function $C_B: \mathbb{R}^n \rightarrow \mathbb{R}^n $ is a linear transformation, and moreover an isomorphism.

We now discuss the main result of this section, that is how to represent a linear transformation with respect to different bases.

Theorem 2: The Matrix of a Linear 定理 2：线性变换的矩阵

Let $T: \mathbb{R}^n \mapsto \mathbb{R}^m$ be a linear transformation, and let $B_1$ and $B_2$ be bases of $\mathbb{R}^{n}$ and $\mathbb{R}^{m}$ respectively.

Then the following holds $C\_{B_2} T = M\_{B\_{2} B\_{1}} C\_{B_1} \label{matrixequation}$ where $M\_{B\_{2} B\_{1}}$ is a unique $m \times n$ matrix.

If the basis $B_1$ is given by $B_1 = \left\{ \vec{v}\_1, \cdots, \vec{v}\_n \right\}$ in this order, then $M\_{B\_{2} B\_{1}} = \left[ C\_{B_2}(T(\vec{v}\_1)) \quad C\_{B_2}(T(\vec{v}\_2)) \quad \cdots \quad C\_{B_2}(T(\vec{v}\_n)) \right]$

Proof

The above equation $\eqref{matrixequation}$ can be represented by the following diagram. $\begin{array}{rcl} & T & \\ \mathbb{R}^n & \rightarrow & \mathbb{R}^m \\ C_{B_{1}}\downarrow & & \downarrow C_{B_{2}} \\ \mathbb{R}^{n} & \rightarrow & \mathbb{R}^{m} \\ & M_{B_{2} B_{1}} & \end{array}$

Since $C\_{B_1}$ is an isomorphism, then the matrix we are looking for is the matrix of the linear transformation $C\_{B_2} T C^{-1}\_{B_1} : \mathbb{R}^n \mapsto \mathbb{R}^m. $ By Theorem 5.2.2, the columns are given by the image of the standard basis $\left\{ \vec{e}\_1, \vec{e}\_2, \cdots, \vec{e}\_n \right\}$ . But since $C^{-1}\_{B_1}( \vec{e}\_i) = \vec{v}\_i$, we readily obtain that $M\_{B\_{2} B\_{1}} = \left[ C\_{B_2}T C^{-1}\_{B_1} (\vec{e}\_1) \quad C\_{B_2}T C^{-1}\_{B_1} (\vec{e}\_2) \quad \cdots \quad C\_{B_2}T C^{-1}\_{B_1} (\vec{e}\_n) \right] = \left[ C\_{B_2}(T(\vec{v}\_1)) \quad C\_{B_2}(T(\vec{v}\_2)) \quad \cdots \quad C\_{B_2}(T(\vec{v}\_n)) \right]$ and this completes the proof.

Consider the following example.

Example 2: Matrix of a Linear 示例 2：线性变换的矩阵

Let $T: \mathbb{R}^2 \mapsto \mathbb{R}^2$ be a linear transformation defined by $T \left( \begin{pmatrix}a \\ b\end{pmatrix} \right) = \begin{pmatrix}b \\ a\end{pmatrix}$.

Consider the two bases $B_1 = \left\{ \vec{v}\_{1}, \vec{v}\_{2} \right\} = \left\{ \begin{pmatrix}1 \\ 0\end{pmatrix}, \begin{pmatrix}-1 \\ 1\end{pmatrix} \right\}nonumber $ and $B_2 = \left\{ \begin{pmatrix}1 \\ 1\end{pmatrix}, \begin{pmatrix}1 \\ -1\end{pmatrix} \right\}nonumber $

Find the matrix $M\_{B_2,B_1}$ of $T$ with respect to the bases $B_1$ and $B_2$.

Solution

By Theorem 2, the columns of $M\_{B\_{2} B\_{1}}$ are the coordinate vectors of $T(\vec{v}\_{1}), T(\vec{v}\_{2})$ with respect to $B_2$.

Since $T \left( \begin{pmatrix}1 \\ 0\end{pmatrix} \right) = \begin{pmatrix}0 \\ 1\end{pmatrix} , $ a standard calculation yields $\begin{pmatrix}0 \\ 1\end{pmatrix} = \left(\frac{1}{2} \right)\begin{pmatrix}1 \\ 1\end{pmatrix} + \left(-\frac{1}{2} \right) \begin{pmatrix}1 \\ -1\end{pmatrix}, $ the first column of $M\_{B\_{2} B\_{1}}$ is $\begin{pmatrix}\frac{1}{2}\\ -\frac{1}{2}\end{pmatrix}$.

The second column is found in a similar way. We have $T \left( \begin{pmatrix}-1 \\ 1\end{pmatrix} \right) = \begin{pmatrix}1 \\ -1\end{pmatrix} , $ and with respect to $B_2$ calculate: $\begin{pmatrix}1 \\ -1\end{pmatrix} = 0 \begin{pmatrix}1 \\ 1\end{pmatrix} + 1 \begin{pmatrix}1 \\ -1\end{pmatrix} $ Hence the second column of $M\_{B\_{2} B\_{1}}$ is given by $\begin{pmatrix}0 \\ 1\end{pmatrix}$. We thus obtain $M\_{B\_{2} B\_{1}} = \begin{pmatrix}\frac{1}{2} & 0 \\ -\frac{1}{2} & 1\end{pmatrix} $

We can verify that this is the correct matrix $M\_{B\_{2} B\_{1}}$ on the specific example $\vec{v} = \begin{pmatrix}3 \\ -1\end{pmatrix} $ First applying $T$ gives $T( \vec{v} ) = T \left( \begin{pmatrix}3 \\ -1\end{pmatrix} \right) = \begin{pmatrix}-1\\ 3\end{pmatrix} $ and one can compute that $C\_{B_2} \left( \begin{pmatrix}-1 \\ 3\end{pmatrix} \right) = \begin{pmatrix}1\\ -2\end{pmatrix} $

On the other hand, one compute $C\_{B_1}( \vec{v})$ as $C\_{B_1} \left( \begin{pmatrix}3 \\ -1\end{pmatrix} \right) = \begin{pmatrix}2\\ -1\end{pmatrix} , $ and finally applying $M\_{B_1 B_2}$ gives $\begin{pmatrix}\frac{1}{2} & 0 \\ -\frac{1}{2} & 1\end{pmatrix} \begin{pmatrix}2 \\ -1\end{pmatrix} = \begin{pmatrix}1 \\ -2\end{pmatrix} $ as above.

We see that the same vector results from either method, as suggested by Theorem 2.

If the bases $B_1$ and $B_2$ are equal, say $B$, then we write $M\_{B}$ instead of $M\_{B B}$. The following example illustrates how to compute such a matrix. Note that this is what we did earlier when we considered only $B_1=B_2$ to be the standard basis.

Example 3: Matrix of a Linear Transformation with respect to an Arbitrary 示例 3：关于任意基的线性变换矩阵

Consider the basis $B$ of $\mathbb{R}^3$ given by $B = \left\{ \vec{v}\_1 , \vec{v}\_2, \vec{v}\_3 \right\} = \left\{ \begin{pmatrix}1 \\ 0 \\ 1\end{pmatrix} ,\begin{pmatrix}1 \\ 1 \\ 1\end{pmatrix} ,\begin{pmatrix}-1 \\ 1 \\ 0\end{pmatrix} \right\}nonumber $ And let $T :\mathbb{R}^{3}\mapsto \mathbb{R}^{3}$ be the linear transformation defined on $B$ as: $T\begin{pmatrix}1 \\ 0 \\ 1\end{pmatrix} =\begin{pmatrix}1 \\ -1 \\ 1\end{pmatrix} ,T \begin{pmatrix}1 \\ 1 \\ 1\end{pmatrix} =\begin{pmatrix}1 \\ 2 \\ -1\end{pmatrix} ,T\begin{pmatrix}-1 \\ 1 \\ 0\end{pmatrix} =\begin{pmatrix}0 \\ 1 \\ 1\end{pmatrix} $

1. Find the matrix $M\_B$ of $T$ relative to the basis $B$.
2. Then find the usual matrix of $T$ with respect to the standard basis of $\mathbb{R}^{3}$.

1. 求 $T$ 关于基 $B$ 的矩阵 $M\_B$。
2. 再求 $T$ 关于 $\mathbb{R}^{3}$ 标准基的通常矩阵。

Solution

Equation $\eqref{matrixequation}$ gives $C_BT=M\_{B}C_B$, and thus $M\_{B} = C_BTC^{-1}\_B$.

Now $C_B(\vec{v}\_i) = \vec{e}\_i$, so the matrix of $C_B^{-1}$ (with respect to the standard basis) is given by $\left[ C_B^{-1}(\vec{e}\_1) \quad C_B^{-1}(\vec{e}\_2) \quad C_B^{-1}(\vec{e}\_2) \right] = \begin{pmatrix}1 & 1 & -1 \\ 0 & 1 & 1 \\ 1 & 1 & 0\end{pmatrix} $ Moreover the matrix of $T C_B^{-1}$ is given by $\left[ TC_B^{-1}(\vec{e}\_1) \quad TC_B^{-1}(\vec{e}\_2) \quad TC_B^{-1}(\vec{e}\_2) \right] = \begin{pmatrix}1 & 1 & 0 \\ -1 & 2 & 1 \\ 1 & -1 & 1\end{pmatrix} $ Thus $\begin{array}{rcl} M_{B} & = & C_B T C^{-1}_B = (C_B^{-1})^{-1} (T C^{-1}_B) \\ & = & \begin{pmatrix}1 & 1 & -1 \\ 0 & 1 & 1 \\ 1 & 1 & 0\end{pmatrix}^{-1}\begin{pmatrix}1 & 1 & 0 \\ -1 & 2 & 1 \\ 1 & -1 & 1\end{pmatrix} \\ & = & \begin{pmatrix}2 & -5 & 1 \\ -1 & 4 & 0 \\ 0 & -2 & 1\end{pmatrix} \end{array}$

Consider how this works. Let $\vec{b} = \begin{pmatrix}b_1 \\ b_2 \\ b_3\end{pmatrix}$ be an arbitrary vector in $\mathbb{R}^3$.

Apply $C^{-1}\_{B}$ to $\vec{b}$ to get $b_1\begin{pmatrix}1 \\ 0 \\ 1\end{pmatrix} + b_2\begin{pmatrix}1 \\ 1 \\ 1\end{pmatrix} + b_3\begin{pmatrix}-1 \\ 1 \\ 0\end{pmatrix} $ Apply $T$ to this linear combination to obtain $b_1\begin{pmatrix}1 \\ -1 \\ 1\end{pmatrix} + b_2\begin{pmatrix}1 \\ 2 \\ -1\end{pmatrix} + b_3\begin{pmatrix}0 \\ 1 \\ 1\end{pmatrix} =\begin{pmatrix}b_1+b_2 \\ -b_1 + 2b_2+ b_3 \\ b_1-b_2+b_3\end{pmatrix} $ Now take the matrix $M\_{B}$ of the transformation (as found above) and multiply it by $\vec{b}$. $\begin{pmatrix}2 & -5 & 1 \\ -1 & 4 & 0 \\ 0 & -2 & 1\end{pmatrix} \begin{pmatrix}b_1 \\ b_2 \\ b_3\end{pmatrix} =\begin{pmatrix}2b_1-5b_2+b_3 \\ -b_1 + 4b_2 \\ -2b_2 + b_3\end{pmatrix} $ Is this the coordinate vector of the above relative to the given basis? We check as follows. $\left( 2b_1-5b_2+b_3\right) \begin{pmatrix}1 \\ 0 \\ 1\end{pmatrix} +\left( -b_1 + 4b_2\right) \begin{pmatrix}1 \\ 1 \\ 1\end{pmatrix} +\left( -2b_2+b_3\right) \begin{pmatrix}-1 \\ 1 \\ 0\end{pmatrix} $ $= \begin{pmatrix}b_1+b_2 \\ -b_1 + 2b_2+b_3 \\ b_1-b_2+b_3\end{pmatrix} $ You see it is the same thing.

Now lets find the matrix of $T$ with respect to the standard basis. Let $A$ be this matrix. That is, multiplication by $A$ is the same as doing $T$. Thus $A\begin{pmatrix}1 & 1 & -1 \\ 0 & 1 & 1 \\ 1 & 1 & 0\end{pmatrix} =\begin{pmatrix}1 & 1 & 0 \\ -1 & 2 & 1 \\ 1 & -1 & 1\end{pmatrix} $ Hence $A=\begin{pmatrix}1 & 1 & 0 \\ -1 & 2 & 1 \\ 1 & -1 & 1\end{pmatrix} \begin{pmatrix}1 & 1 & -1 \\ 0 & 1 & 1 \\ 1 & 1 & 0\end{pmatrix} ^{-1}=\begin{pmatrix}0 & 0 & 1 \\ 2 & 3 & -3 \\ -3 & -2 & 4\end{pmatrix} $ Of course this is a very different matrix than the matrix of the linear transformation with respect to the non standard basis.

Outcomes 学习目标

1. Use linear transformations to determine the particular solution and general solution to a system of equations.
2. Find the kernel of a linear transformation.

1. 利用线性变换确定方程组的特解与通解。
2. 求一个线性变换的核。

Recall the definition of a linear transformation discussed above. $T$ is a linear transformation if whenever $\vec{x}, \vec{y}$ are vectors and $k,p$ are scalars, $T\left( k\vec{x}+p\vec{y}\right) =k T \left( \vec{x} \right) +p T\left(\vec{y} \right) $ Thus linear transformations distribute across addition and pass scalars to the outside.

It turns out that we can use linear transformations to solve linear systems of equations. Indeed given a system of linear equations of the form $A\vec{x}=\vec{b}$, one may rephrase this as $T(\vec{x})=\vec{b}$ where $T$ is the linear transformation $T_A$ induced by the coefficient matrix $A$. With this in mind consider the following definition.

Definition 1: Particular Solution of a System of Equations 方程组的特解

Suppose a linear system of equations can be written in the form $T\left(\vec{x}\right)=\vec{b} $ If $T\left(\vec{x}\_{p}\right)=\vec{b},$ then $\vec{x}\_{p}$ is called a particular solution of the linear system.

Recall that a system is called homogeneous if every equation in the system is equal to $0$. Suppose we represent a homogeneous system of equations by $T\left(\vec{x}\right)=0$. It turns out that the $\vec{x}$ for which $T \left(\vec{x}\right) = 0$ are part of a special set called the null space of $T$. We may also refer to the null space as the kernel of $T$, and we write $ker\left(T\right)$.

Consider the following definition.

Definition 2: Null Space or Kernel of a Linear Transformation 线性变换的零空间或核

Let $T$ be a linear transformation. Define $\ker \left( T\right) = \left\{ \vec{x}:T \left(\vec{x} \right)= \vec{0} \right\}nonumber $ The kernel, $\ker \left( T\right)$ consists of the set of all vectors $\vec{x}$ for which $T (\vec{x}) = \vec{0}$. This is also called the null space of $T$.

We may also refer to the kernel of $T$ as the solution space of the equation $T \left(\vec{x}\right) = \vec{0}$.

Consider the following example.

Example 1: The Kernel of the Derivative 导数的核

Let $\frac{d}{dx}$ denote the linear transformation defined on $f,$ the functions which are defined on $\mathbb{R}$ and have a continuous derivative. Find $\ker \left( \frac{d}{dx}\right) .$

Solution

The example asks for functions $f$ which the property that $\frac{df}{dx} =0.$ As you may know from calculus, these functions are the constant functions. Thus $\ker \left( \frac{d}{dx}\right)$ is the set of constant functions.

Definition 2 states that $\ker \left( T\right)$ is the set of solutions to the equation, $T\left( \vec{x} \right) = \vec{0}$ Since we can write $T\left( \vec{x} \right)$ as $A\vec{x}$, you have been solving such equations for quite some time.

We have spent a lot of time finding solutions to systems of equations in general, as well as homogeneous systems. Suppose we look at a system given by $A\vec{x}=\vec{b}$, and consider the related homogeneous system. By this, we mean that we replace $\vec{b}$ by $\vec{0}$ and look at $A\vec{x}=\vec{0}$. It turns out that there is a very important relationship between the solutions of the original system and the solutions of the associated homogeneous system. In the following theorem, we use linear transformations to denote a system of equations. Remember that $T\left(\vec{x}\right) = A\vec{x}$.

Theorem 1: Particular Solution and General Solution 特解与通解

Suppose $\vec{x}\_{p}$ is a solution to the linear system given by , $T\left( \vec{x} \right) = \vec{b} $ Then if $\vec{y}$ is any other solution to $T\left(\vec{x}\right)=\vec{b}$, there exists $\vec{x}\_0 \in \ker \left( T\right)$ such that $\vec{y} = \vec{x}\_{p}+ \vec{x}\_0 $ Hence, every solution to the linear system can be written as a sum of a particular solution, $\vec{x}\_p$, and a solution $\vec{x}\_0$ to the associated homogeneous system given by $T\left(\vec{x}\right)=\vec{0}$.

Proof

Consider $\vec{y} - \vec{x}\_{p}= \vec{y} + \left( -1\right) \vec{x}\_{p}.$ Then $T\left( \vec{y} - \vec{x}\_{p}\right) =T\left(\vec{y}\right) -T\left( \vec{x}\_{p} \right)$. Since $\vec{y}$ and $\vec{x}\_{p}$ are both solutions to the system, it follows that $T\left(\vec{y}\right)= \vec{b}$ and $T\left(\vec{x}\_p\right) = \vec{b}$.

Hence, $T\left(\vec{y}\right)-T\left( \vec{x}\_{p} \right) =\vec{b} - \vec{b} = \vec{0}$. Let $\vec{x}\_0 = \vec{y} - \vec{x}\_{p}$. Then, $T\left(\vec{x}\_0\right)= \vec{0}$ so $\vec{x}\_0$ is a solution to the associated homogeneous system and so is in $\ker \left(T\right)$.

Sometimes people remember the above theorem in the following form. The solutions to the system $T\left(\vec{x}\right)=\vec{b}$ are given by $\vec{x}\_{p}+\ker \left( T\right)$ where $\vec{x}\_{p}$ is a particular solution to $T\left(\vec{x}\right)=\vec{b}$.

For now, we have been speaking about the kernel or null space of a linear transformation $T$. However, we know that every linear transformation $T$ is determined by some matrix $A$. Therefore, we can also speak about the null space of a matrix. Consider the following example.

Example 2: The Null Space of a Matrix 矩阵的零空间

Let $A=\begin{pmatrix}1 & 2 & 3 & 0 \\ 2 & 1 & 1 & 2 \\ 4 & 5 & 7 & 2\end{pmatrix} $ Find $\mathrm{null} \left( A\right)$. Equivalently, find the solutions to the system of equations $A\vec{x}=\vec{0}$.

Solution

We are asked to find $\left\{ \vec{x} : A\vec{x} = \vec{0}\right\} .$ In other words we want to solve the system, $A\vec{x}=\vec{0}$. Let $\vec{x} = \begin{pmatrix}x \\ y \\ z \\ w\end{pmatrix}.$ Then this amounts to solving $\begin{pmatrix}1 & 2 & 3 & 0 \\ 2 & 1 & 1 & 2 \\ 4 & 5 & 7 & 2\end{pmatrix} \begin{pmatrix}x \\ y \\ z \\ w\end{pmatrix} =\begin{pmatrix}0 \\ 0 \\ 0\end{pmatrix} $

This is the linear system $\begin{array}{c} x+2y+3z=0 $ 2x+y+z+2w=0 $ 4x+5y+7z+2w=0 \end{array} $ To solve, set up the augmented matrix and row reduce to find the reduced row-echelon form.

$$\left[\begin{array}{rrrr\|r}1 & 2 & 3 & 0 & 0 \\ 2 & 1 & 1 & 2 & 0 \\ 4 & 5 & 7 & 2 & 0\end{array}\right] \rightarrow \cdots \rightarrow \left[\begin{array}{rrrr\|r}1 & 0 & - \frac{1}{3} & \frac{4}{3} & 0 \\ 0 & 1 & \frac{5}{3} & - \frac{2}{3} & 0 \\ 0 & 0 & 0 & 0 & 0\end{array}\right]$$

This yields $x= \frac{1}{3}z- \frac{4}{3}w$ and $y= \frac{2}{3}w- \frac{5}{3}z.$ Since $\mathrm{null} \left( A\right)$ consists of the solutions to this system, it consists vectors of the form, $\begin{pmatrix}\frac{1}{3}z- \frac{4}{3}w \\ \frac{2}{3}w- \frac{5}{3}z \\ z \\ w\end{pmatrix} =z \begin{pmatrix}\frac{1}{3} \\ - \frac{5}{3} \\ 1 \\ 0\end{pmatrix} +w \begin{pmatrix}- \frac{4}{3} \\ \frac{2}{3} \\ 0 \\ 1\end{pmatrix} $

Consider the following example.

Example 3: A General Solution 通解

The general solution of a linear system of equations is the set of all possible solutions. Find the general solution to the linear system, $\begin{pmatrix}1 & 2 & 3 & 0 \\ 2 & 1 & 1 & 2 \\ 4 & 5 & 7 & 2\end{pmatrix} \begin{pmatrix}x \\ y \\ z \\ w\end{pmatrix} =\begin{pmatrix}9 \\ 7 \\ 25\end{pmatrix} $

given that $\begin{pmatrix}x \\ y \\ z \\ w\end{pmatrix}=\begin{pmatrix}1 \\ 1 \\ 2 \\ 1\end{pmatrix}$ is one solution.

Solution

Note the matrix of this system is the same as the matrix in Example 2. Therefore, from Theorem 1, you will obtain all solutions to the above linear system by adding a particular solution $\vec{x}\_p$ to the solutions of the associated homogeneous system, $\vec{x}$. One particular solution is given above by $\vec{x}\_p = \begin{pmatrix}x \\ y \\ z \\ w\end{pmatrix}=\begin{pmatrix}1 \\ 1 \\ 2 \\ 1\end{pmatrix} $

Using this particular solution along with the solutions found in Example 2, we obtain the following solutions, $z\begin{pmatrix}\frac{1}{3} \\ - \frac{5}{3} \\ 1 \\ 0\end{pmatrix} +w\begin{pmatrix}- \frac{4}{3} \\ \frac{2}{3} \\ 0 \\ 1\end{pmatrix} +\begin{pmatrix}1 \\ 1 \\ 2 \\ 1\end{pmatrix} $

Hence, any solution to the above linear system is of this form.

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Exercise 1 习题 1

Show the map $T$: $\mathbb{R}^n \to \mathbb{R}^m$ defined by $T (\vec{x}) = A\vec{x}$ where $A$ is an $m\times n$ matrix and $\vec{x}$ is an $m\times 1$ column vector is a linear transformation.

Answer This result follows from the properties of matrix multiplication.

Exercise 2 习题 2

Show that the function $T_{\vec{u}}$ defined by $T_{\vec{u}} (\vec{v}) =\vec{v}−proj_{\vec{u}} (\vec{v})$ is also a linear transformation.

Answer $$\begin{aligned}T_{\vec{u}}(a\vec{v}+b\vec{w})&=a\vec{v}+b\vec{w}-\frac{(a\vec{v}+b\vec{w}\bullet\vec{u})}{\|\vec{u}\|^2}\vec{u} \\ &=a\vec{v}-a\frac{(\vec{v}\bullet\vec{u})}{\|\vec{u}\|^2}\vec{u}+b\vec{w}-b\frac{(\vec{w}\bullet\vec{u})}{\|\vec{u}\|^2}\vec{u} \\ &=aT_{\vec{u}}(\vec{v})+bT_{\vec{u}}(\vec{w})\end{aligned}$$

Exercise 3 习题 3

Let $\vec{u}$ be a fixed vector. The function $T_{\vec{u}}$ defined by $T_{\vec{u}}\vec{v} =\vec{u}+\vec{v}$ has the effect of translating all vectors by adding $\vec{u}\neq\vec{0}$. Show this is not a linear transformation. Explain why it is not possible to represent $T_{\vec{u}}$ in $\mathbb{R}^3$ by multiplying by a $3×3$ matrix.

Answer Linear transformations take $\vec{0}$ to $\vec{0}$ which $T$ does not. Also $T_{\vec{a}} (\vec{u}+\vec{v})\neq T_{\vec{a}}\vec{u}+T_{\vec{a}}\vec{v}$.

Exercise 4 习题 4

Consider the following functions which map $\mathbb{R}^n$ to $\mathbb{R}^n$.

• $T$ multiplies the $j$th component of $\vec{x}$ by a nonzero number $b$.
• $T$ replaces the $i$th component of $\vec{x}$ with $b$ times the $j$th component added to the $i$th component.
• $T$ switches the $i$th and $j$th components.

• $T$ 将 $\vec{x}$ 的第 $j$ 个分量乘以非零数 $b$。
• $T$ 将 $\vec{x}$ 的第 $i$ 个分量替换为第 $j$ 个分量的 $b$ 倍与该第 $i$ 个分量之和。
• $T$ 交换第 $i$ 个与第 $j$ 个分量。

Show these functions are linear transformations and describe their matrices $A$ such that $T (\vec{x}) = A\vec{x}$.

Answer

• The matrix of $T$ is the elementary matrix which multiplies the $j$th diagonal entry of the identity matrix by $b$.
• The matrix of $T$ is the elementary matrix which takes $b$ times the $j$th row and adds to the $i$th row.
• The matrix of $T$ is the elementary matrix which switches the $i$th and the $j$th rows where the two components are in the $i$th and $j$th positions.

• $T$ 的矩阵是将单位矩阵第 $j$ 个对角元乘以 $b$ 所得的初等矩阵。
• $T$ 的矩阵是将第 $j$ 行的 $b$ 倍加到第 $i$ 行所得的初等矩阵。
• $T$ 的矩阵是交换第 $i$ 行与第 $j$ 行（这两个分量分别位于第 $i$、第 $j$ 个位置）的初等矩阵。

Exercise 5 习题 5

You are given a linear transformation $T$: $\mathbb{R}^n \to \mathbb{R}^m$ and you know that $T(A_i)=B_i$ where $\left(A_1\cdots a_n\right)^{-1}$ exists. Show that the matrix of $T$ is of the form $B_1\cdots B_n\left(A_1\cdots A_n\right)^{-1}$.

Answer Suppose $\begin{pmatrix}\vec{c}_1^{T} \\ \vdots \\ \vec{c}_n^T\end{pmatrix}=\begin{pmatrix}\vec{a}_1&\cdots &\vec{a}_n\end{pmatrix}^{-1}$ Thus $\vec{c}_i^T\vec{a}_j=\delta_{ij}$. Therefore $\begin{aligned}\begin{pmatrix}\vec{b}_1&\cdots&\vec{b}_n\end{pmatrix}\begin{pmatrix}\vec{a}_1&\cdots&\vec{a}_n\end{pmatrix}^{-1}\vec{a}_i&=\begin{pmatrix}\vec{b}_1&\cdots&\vec{b}_n\end{pmatrix}\begin{pmatrix}\vec{c}_1^T \\ \vdots \\ \vec{c}_n^T\end{pmatrix}\vec{a}_i\\ &=\begin{pmatrix}\vec{b}_1&\cdots&\vec{b}_n\end{pmatrix}\vec{e}_i \\ &=\vec{b}_i\end{aligned}$ Thus $T\vec{a}_i=\begin{pmatrix}\vec{b}_1&\cdots&\vec{b}_n\end{pmatrix}\begin{pmatrix}\vec{a}_1&\cdots&\vec{a}_n\end{pmatrix}^{-1}\vec{a}_i=A\vec{a}_i$. If $\vec{x}$ is arbitrary, then since the matrix $\begin{pmatrix}\vec{a}_1&\cdots&\vec{a}_n\end{pmatrix}$ is invertible, there exists a unique $\vec{y}$ such that $\begin{pmatrix}\vec{a}_1&\cdots&\vec{a}_n\end{pmatrix}\vec{y}=\vec{x}$ Hence $T\vec{x}=T\left(\sum\limits_{i=1}^n y_i\vec{a}_i\right)=\sum\limits_{i=1}^n y_iT\vec{a}_i=\sum\limits_{i=1}^n y_iA\vec{a}_i=A\left(\sum\limits_{i=1}^ny_i\vec{a}_i\right)=A\vec{x}$

Exercise 6 习题 6

Suppose $T$ is a linear transformation such that $\begin{aligned}T\begin{pmatrix}1\\2\\-6\end{pmatrix}&=\begin{pmatrix}5\\1\\3\end{pmatrix} \\ T\begin{pmatrix}-1\\-1\\5\end{pmatrix}&=\begin{pmatrix}1\\1\\5\end{pmatrix} \\ T\begin{pmatrix}0\\-1\\2\end{pmatrix}&=\begin{pmatrix}5\\3\\-2\end{pmatrix}\end{aligned}$ Find the matrix of $T$. That is find $A$ such that $T(\vec{x}) = A\vec{x}$.

Answer $$\begin{pmatrix}5&1&5\\1&1&3\\3&5&-2\end{pmatrix}\begin{pmatrix}3&2&1\\2&2&1\\4&1&1\end{pmatrix}=\begin{pmatrix}37&17&11 \\ 17&7&5\\11&14&6\end{pmatrix}$$

Exercise 7 习题 7

Suppose $T$ is a linear transformation such that $\begin{aligned}T\begin{pmatrix}1\\1\\-8\end{pmatrix}&=\begin{pmatrix}1\\3\\1\end{pmatrix} \\ T\begin{pmatrix}-1\\0\\6\end{pmatrix}&=\begin{pmatrix}2\\4\\1\end{pmatrix} \\ T\begin{pmatrix}0\\-1\\3\end{pmatrix}&=\begin{pmatrix}6\\1\\-1\end{pmatrix}\end{aligned}$ Find the matrix of $T$. That is find $A$ such that $T(\vec{x})=A\vec{x}$.

Answer $$\begin{pmatrix}1&2&6\\3&4&1\\1&1&-1\end{pmatrix}\begin{pmatrix}6&3&1\\5&3&1\\6&2&1\end{pmatrix}=\begin{pmatrix}52&21&9\\44&23&8\\5&4&1\end{pmatrix}$$

Exercise 8 习题 8

Suppose $T$ is a linear transformation such that $\begin{aligned}T\begin{pmatrix}1\\3\\-7\end{pmatrix}&=\begin{pmatrix}-3\\1\\3\end{pmatrix} \\ T\begin{pmatrix}-1\\-2\\6\end{pmatrix}&=\begin{pmatrix}1\\3\\-3\end{pmatrix} \\ T\begin{pmatrix}0\\-1\\2\end{pmatrix}&=\begin{pmatrix}5\\3\\-3\end{pmatrix}\end{aligned}$ Find the matrix of $T$. That is find $A$ such that $T(\vec{x})=A\vec{x}$.

Answer $$\begin{pmatrix}-3&1&5\\1&3&3\\3&-3&-3\end{pmatrix}\begin{pmatrix}2&2&1\\1&2&1\\4&1&1\end{pmatrix}=\begin{pmatrix}15&1&3\\17&11&7\\-9&-3&-3\end{pmatrix}$$

Exercise 9 习题 9

Suppose $T$ is a linear transformation such that $\begin{aligned}T\begin{pmatrix}1\\1\\-7\end{pmatrix}&=\begin{pmatrix}3\\3\\3\end{pmatrix} \\ T\begin{pmatrix}-1\\0\\6\end{pmatrix}&=\begin{pmatrix}1\\2\\3\end{pmatrix} \\ T\begin{pmatrix}0\\-1\\2\end{pmatrix}&=\begin{pmatrix}1\\3\\-1\end{pmatrix}\end{aligned}$ Find the matrix of $T$. That is find $A$ such that $T(\vec{x})=A\vec{x}$.

Answer $$\begin{pmatrix}3&1&1\\3&2&3\\3&3&-1\end{pmatrix}\begin{pmatrix}6&2&1\\5&2&1\\6&1&1\end{pmatrix}=\begin{pmatrix}29&9&5\\46&13&8\\27&11&5\end{pmatrix}$$

Exercise 10 习题 10

Suppose $T$ is a linear transformation such that $\begin{aligned}T\begin{pmatrix}1\\2\\-18\end{pmatrix}&=\begin{pmatrix}5\\2\\5\end{pmatrix} \\ T\begin{pmatrix}-1\\-1\\15\end{pmatrix}&=\begin{pmatrix}3\\3\\5\end{pmatrix} \\ T\begin{pmatrix}0\\-1\\4\end{pmatrix}&=\begin{pmatrix}2\\5\\-2\end{pmatrix}\end{aligned}$ Find the matrix of $T$. That is find $A$ such that $T(\vec{x})=A\vec{x}$.

Answer $$\begin{pmatrix}5&3&2\\2&3&5\\5&5&-2\end{pmatrix}\begin{pmatrix}11&4&1\\10&4&1\\12&3&1\end{pmatrix}=\begin{pmatrix}109&38&10\\112&35&10\\81&34&8\end{pmatrix}$$

Exercise 11 习题 11

Consider the following functions $T : \mathbb{R}^3 \to \mathbb{R}^2$. Show that each is a linear transformation and determine for each the matrix $A$ such that $T(\vec{x}) = A\vec{x}$.

• $T\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}x+2y+3z \\ 2y-3x+z\end{pmatrix}$
• $T\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}7x+2y+z \\ 3x-11y+2z\end{pmatrix}$
• $T\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}3x+2y+z \\ x+2y+6z\end{pmatrix}$
• $T\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}2y-5x+z \\ x+y+z\end{pmatrix}$

• $T\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}x+2y+3z \\ 2y-3x+z\end{pmatrix}$
• $T\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}7x+2y+z \\ 3x-11y+2z\end{pmatrix}$
• $T\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}3x+2y+z \\ x+2y+6z\end{pmatrix}$
• $T\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}2y-5x+z \\ x+y+z\end{pmatrix}$

Exercise 12 习题 12

Consider the following functions $T : \mathbb{R}^3 \to \mathbb{R}^2$. Explain why each of these functions $T$ is not linear.

• $T\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}x+2y+3z+1 \\ 2y-3x+z\end{pmatrix}$
• $T\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}x+2y^2+3z \\ 2y+3x+z\end{pmatrix}$
• $T\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}\sin x+2y+3z \\ 2y+3x+z\end{pmatrix}$
• $T\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}x+2y+3z \\ 2y+3x-\ln z\end{pmatrix}$

• $T\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}x+2y+3z+1 \\ 2y-3x+z\end{pmatrix}$
• $T\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}x+2y^2+3z \\ 2y+3x+z\end{pmatrix}$
• $T\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}\sin x+2y+3z \\ 2y+3x+z\end{pmatrix}$
• $T\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}x+2y+3z \\ 2y+3x-\ln z\end{pmatrix}$

Exercise 13 习题 13

Suppose $\left(A_1\cdots A_n\right)^{-1}$ exists where each $A_j ∈ \mathbb{R}^n$ and let vectors $B_1,\cdots ,B_n$ in $\mathbb{R}^m$ be given. Show that there always exists a linear transformation $T$ such that $T(A_i) = B_i$.

Exercise 14 习题 14

Find the matrix for $T (\vec{w}) = proj_{\vec{v}}(\vec{w})$ where $\vec{v}=\begin{pmatrix}1&-2&3\end{pmatrix}^T$.

Answer Recall that $proj_{\vec{u}}(\vec{v})=\frac{\vec{v}\bullet\vec{u}}{\|\vec{u}\|^2}\vec{u}$ and so the desired matrix has $i$th column equal to $proj_{\vec{u}}(\vec{e}_i)$. Therefore, the matrix desired is $\frac{1}{14}\begin{pmatrix}1&-2&3\\-2&4&-6\\3&-6&9\end{pmatrix}$

Exercise 15 习题 15

Find the matrix for $T (\vec{w}) = proj_{\vec{v}}(\vec{w})$ where $\vec{v}=\begin{pmatrix}1&5&3\end{pmatrix}^T$.

Answer $$\frac{1}{35}\begin{pmatrix}1&5&3\\5&25&15\\3&15&9\end{pmatrix}$$

Exercise 16 习题 16

Find the matrix for $T (\vec{w}) = proj_{\vec{v}}(\vec{w})$ where $\vec{v}=\begin{pmatrix}1&0&3\end{pmatrix}^T$.

Answer $$\frac{1}{10}\begin{pmatrix}1&0&3\\0&0&0\\3&0&9\end{pmatrix}$$

Exercise 17 习题 17

Show that if a function $T :\mathbb{R}^n \to \mathbb{R}^m$ is linear, then it is always the case that $T(\vec{0})=\vec{0}$.

Exercise 18 习题 18

Let $T$ be a linear transformation induced by the matrix $A=\begin{pmatrix}3&1\\-1&2\end{pmatrix}$ and $S$ a linear transformation induced by $B=\begin{pmatrix}0&-2\\4&2\end{pmatrix}$. Find matrix of $S\circ T$ and find $(S\circ T)(\vec{x})$ for $\vec{x}=\begin{pmatrix}2\\-1\end{pmatrix}$.

Answer The matrix of $S\circ T$ is given by $BA$. $\begin{pmatrix}0&-2\\4&2\end{pmatrix}\begin{pmatrix}3&1\\-1&2\end{pmatrix}=\begin{pmatrix}2&-4\\10&8\end{pmatrix}$ Now, $(S\circ T)(\vec{x})=(BA)\vec{x}$. $\begin{pmatrix}2&-4\\10&8\end{pmatrix}\begin{pmatrix}2\\-1\end{pmatrix}=\begin{pmatrix}8\\12\end{pmatrix}$

Exercise 19 习题 19

Let $T$ be a linear transformation and suppose $T\left(\begin{pmatrix}1\\-4\end{pmatrix}\right)=\begin{pmatrix}2\\-3\end{pmatrix}$. Suppose $S$ is a linear transformation induced by the matrix $B=\begin{pmatrix}1&2\\-1&3\end{pmatrix}$. Find $(S\circ T)(\vec{x})$ for $\vec{x}=\begin{pmatrix}1\\-4\end{pmatrix}$.

Answer To find $(S\circ T)(\vec{x})$ we compute $S(T(\vec{x}))$. $\begin{pmatrix}1&2\\-1&3\end{pmatrix}\begin{pmatrix}2\\-3\end{pmatrix}=\begin{pmatrix}-4\\-11\end{pmatrix}$

Exercise 20 习题 20

Let $T$ be a linear transformation induced by the matrix $A=\begin{pmatrix}2&3\\1&1\end{pmatrix}$ and $S$ a linear transformation induced by $B=\begin{pmatrix}-1&3\\1&-2\end{pmatrix}$. Find matrix of $S\circ T$ and find $(S\circ T)(\vec{x})$ for $\vec{x}=\begin{pmatrix}5\\6\end{pmatrix}$.

Exercise 21 习题 21

Let $T$ be a linear transformation induced by the matrix $A=\begin{pmatrix}2&1\\5&2\end{pmatrix}$. Find the matrix of $T^{-1}$.

Answer The matrix of $T^{-1}$ is $A^{-1}$. $\begin{pmatrix}2&1\\5&2\end{pmatrix}^{-1}=\begin{pmatrix}-2&1\\5&-2\end{pmatrix}$

Exercise 22 习题 22

Let $T$ be a linear transformation induced by the matrix $A=\begin{pmatrix}4&-3\\2&-2\end{pmatrix}$. Find the matrix of $T^{-1}$.

Exercise 23 习题 23

Let $T$ be a linear transformation and suppose $T\left(\begin{pmatrix}1\\2\end{pmatrix}\right)=\begin{pmatrix}9\\8\end{pmatrix},T\left(\begin{pmatrix}0\\-1\end{pmatrix}\right)=\begin{pmatrix}-4\\-3\end{pmatrix}$. Find the matrix of $T^{-1}$.

Exercise 24 习题 24

Find the matrix for the linear transformation which rotates every vector in $\mathbb{R}^2$ through an angle of $\pi/3$.

Answer $$\begin{pmatrix}\cos\left(\frac{\pi}{3}\right)&-\sin\left(\frac{\pi}{3}\right) \\ \sin\left(\frac{\pi}{3}\right)&\cos\left(\frac{\pi}{3}\right)\end{pmatrix}=\begin{pmatrix}\frac{1}{2}&-\frac{1}{2}\sqrt{3} \\ \frac{1}{2}\sqrt{3}&\frac{1}{2}\end{pmatrix}$$

Exercise 25 习题 25

Find the matrix for the linear transformation which rotates every vector in $\mathbb{R}^2$ through an angle of $\pi/4$.

Answer $$\begin{pmatrix}\cos\left(\frac{\pi}{4}\right)&-\sin\left(\frac{\pi}{4}\right) \\ \sin\left(\frac{\pi}{4}\right)&\cos\left(\frac{\pi}{4}\right)\end{pmatrix}=\begin{pmatrix}\frac{1}{2}\sqrt{2}&-\frac{1}{2}\sqrt{2} \\ \frac{1}{2}\sqrt{2}&\frac{1}{2}\sqrt{2}\end{pmatrix}$$

Exercise 26 习题 26

Find the matrix for the linear transformation which rotates every vector in $\mathbb{R}^2$ through an angle of $-\pi/3$.

Answer $$\begin{pmatrix}\cos\left(-\frac{\pi}{3}\right)&-\sin\left(-\frac{\pi}{3}\right) \\ \sin\left(-\frac{\pi}{3}\right)&\cos\left(-\frac{\pi}{3}\right)\end{pmatrix}=\begin{pmatrix}\frac{1}{2}&\frac{1}{2}\sqrt{3} \\ -\frac{1}{2}\sqrt{3}&\frac{1}{2}\end{pmatrix}$$

Exercise 27 习题 27

Find the matrix for the linear transformation which rotates every vector in $\mathbb{R}^2$ through an angle of $2\pi/3$.

Answer $$\begin{pmatrix}\cos\left(\frac{2\pi}{3}\right)&-\sin\left(\frac{2\pi}{3}\right) \\ \sin\left(\frac{2\pi}{3}\right)&\cos\left(\frac{2\pi}{3}\right)\end{pmatrix}=\begin{pmatrix}-\frac{1}{2}&-\frac{1}{2}\sqrt{3} \\ \frac{1}{2}\sqrt{3}&-\frac{1}{2}\end{pmatrix}$$

Exercise 28 习题 28

Find the matrix for the linear transformation which rotates every vector in $\mathbb{R}^2$ through an angle of $\pi/12$. Hint: Note that $\pi /12=\pi /3-\pi /4$.

Answer $$\begin{aligned}&\begin{pmatrix}\cos\left(\frac{\pi}{3}\right)&-\sin\left(\frac{\pi}{3}\right) \\ \sin\left(\frac{\pi}{3}\right)&\cos\left(\frac{\pi}{3}\right)\end{pmatrix}\begin{pmatrix}\cos\left(-\frac{\pi}{4}\right)&-\sin\left(-\frac{\pi}{4}\right) \\ \sin\left(-\frac{\pi}{4}\right)&\cos\left(-\frac{\pi}{4}\right)\end{pmatrix} \\ =&\begin{pmatrix}\frac{1}{4}\sqrt{2}\sqrt{3}+\frac{1}{4}\sqrt{2}&\frac{1}{4}\sqrt{2}-\frac{1}{4}\sqrt{2}\sqrt{3} \\ \frac{1}{4}\sqrt{2}\sqrt{3}-\frac{1}{4}\sqrt{2}&\frac{1}{4}\sqrt{2}\sqrt{3}+\frac{1}{4}\sqrt{2}\end{pmatrix}\end{aligned}$$

Exercise 29 习题 29

Find the matrix for the linear transformation which rotates every vector in $\mathbb{R}^2$ through an angle of $2\pi/3$ and then reflects across the $x$ axis.

Answer $$\begin{pmatrix}1&0\\0&-1\end{pmatrix}\begin{pmatrix}\cos\left(\frac{2\pi}{3}\right)&-\sin\left(\frac{2\pi}{3}\right) \\ \sin\left(\frac{2\pi}{3}\right)&\cos\left(\frac{2\pi}{3}\right)\end{pmatrix}=\begin{pmatrix}-\frac{1}{2}&-\frac{1}{2}\sqrt{3}\\-\frac{1}{2}\sqrt{3}&\frac{1}{2}\end{pmatrix}$$

Exercise 30 习题 30

Find the matrix for the linear transformation which rotates every vector in $\mathbb{R}^2$ through an angle of $\pi/3$ and then reflects across the $x$ axis.

Answer $$\begin{pmatrix}1&0\\0&-1\end{pmatrix}\begin{pmatrix}\cos\left(\frac{\pi}{3}\right)&-\sin\left(\frac{\pi}{3}\right) \\ \sin\left(\frac{\pi}{3}\right)&\cos\left(\frac{\pi}{3}\right)\end{pmatrix}=\begin{pmatrix}\frac{1}{2}&-\frac{1}{2}\sqrt{3}\\-\frac{1}{2}\sqrt{3}&-\frac{1}{2}\end{pmatrix}$$

Exercise 31 习题 31

Find the matrix for the linear transformation which rotates every vector in $\mathbb{R}^2$ through an angle of $\pi/4$ and then reflects across the $x$ axis.

Answer $$\begin{pmatrix}1&0\\0&-1\end{pmatrix}\begin{pmatrix}\cos\left(\frac{\pi}{4}\right)&-\sin\left(\frac{\pi}{4}\right) \\ \sin\left(\frac{\pi}{4}\right)&\cos\left(\frac{\pi}{4}\right)\end{pmatrix}=\begin{pmatrix}\frac{1}{2}\sqrt{2}&-\frac{1}{2}\sqrt{2}\\-\frac{1}{2}\sqrt{2}&-\frac{1}{2}\sqrt{2}\end{pmatrix}$$

Exercise 32 习题 32

Find the matrix for the linear transformation which rotates every vector in $\mathbb{R}^2$ through an angle of $\pi/6$ and then reflects across the $x$ axis followed by a reflection across the $y$ axis.

Answer $$\begin{pmatrix}-1&0\\0&1\end{pmatrix}\begin{pmatrix}\cos\left(\frac{\pi}{6}\right)&-\sin\left(\frac{\pi}{6}\right) \\ \sin\left(\frac{\pi}{6}\right)&\cos\left(\frac{\pi}{6}\right)\end{pmatrix}=\begin{pmatrix}-\frac{1}{2}\sqrt{3}&\frac{1}{2}\\ \frac{1}{2}&\frac{1}{2}\sqrt{3}\end{pmatrix}$$

Exercise 33 习题 33

Find the matrix for the linear transformation which rotates every vector in $\mathbb{R}^2$ across the $x$ axis and then rotates every vector through an angle of $\pi /4$.

Answer $$\begin{pmatrix}\cos\left(\frac{\pi}{4}\right)&-\sin\left(\frac{\pi}{4}\right) \\ \sin\left(\frac{\pi}{4}\right)&\cos\left(\frac{\pi}{4}\right)\end{pmatrix}\begin{pmatrix}1&0\\0&-1\end{pmatrix}=\begin{pmatrix}\frac{1}{2}\sqrt{2}&\frac{1}{2}\sqrt{2} \\ \frac{1}{2}\sqrt{2}&-\frac{1}{2}\sqrt{2}\end{pmatrix}$$

Exercise 34 习题 34

Find the matrix for the linear transformation which rotates every vector in $\mathbb{R}^2$ across the $y$ axis and then rotates every vector through an angle of $\pi /4$.

Answer $$\begin{pmatrix}\cos\left(\frac{\pi}{4}\right)&-\sin\left(\frac{\pi}{4}\right) \\ \sin\left(\frac{\pi}{4}\right)&\cos\left(\frac{\pi}{4}\right)\end{pmatrix}\begin{pmatrix}-1&0\\0&1\end{pmatrix}=\begin{pmatrix}-\frac{1}{2}\sqrt{2}&-\frac{1}{2}\sqrt{2} \\ -\frac{1}{2}\sqrt{2}&\frac{1}{2}\sqrt{2}\end{pmatrix}$$

Exercise 35 习题 35

Find the matrix for the linear transformation which rotates every vector in $\mathbb{R}^2$ across the $x$ axis and then rotates every vector through an angle of $\pi /6$.

Answer $$\begin{pmatrix}\cos\left(\frac{\pi}{6}\right)&-\sin\left(\frac{\pi}{6}\right) \\ \sin\left(\frac{\pi}{6}\right)&\cos\left(\frac{\pi}{6}\right)\end{pmatrix}\begin{pmatrix}1&0\\0&-1\end{pmatrix}=\begin{pmatrix}\frac{1}{2}\sqrt{3}&\frac{1}{2} \\ \frac{1}{2}&-\frac{1}{2}\sqrt{3}\end{pmatrix}$$

Exercise 36 习题 36

Find the matrix for the linear transformation which rotates every vector in $\mathbb{R}^2$ across the $y$ axis and then rotates every vector through an angle of $\pi /6$.

Answer $$\begin{pmatrix}\cos\left(\frac{\pi}{6}\right)&-\sin\left(\frac{\pi}{6}\right) \\ \sin\left(\frac{\pi}{6}\right)&\cos\left(\frac{\pi}{6}\right)\end{pmatrix}\begin{pmatrix}-1&0\\0&1\end{pmatrix}=\begin{pmatrix}-\frac{1}{2}\sqrt{3}&-\frac{1}{2} \\ -\frac{1}{2}&\frac{1}{2}\sqrt{3}\end{pmatrix}$$

Exercise 37 习题 37

Find the matrix for the linear transformation which rotates every vector in $\mathbb{R}^2$ through an angle of $5\pi /12$. Hint: Note that $5\pi /12=2\pi /3-\pi /4$.

Answer $\begin{pmatrix}\cos\left(\frac{2\pi}{3}\right)&-\sin\left(\frac{2\pi}{3}\right) \\ \sin\left(\frac{2\pi}{3}\right)&\cos\left(\frac{2\pi}{3}\right)\end{pmatrix}\begin{pmatrix}\cos\left(-\frac{\pi}{4}\right)&-\sin\left(-\frac{\pi}{4}\right) \\ \sin\left(-\frac{\pi}{4}\right)&\cos\left(-\frac{\pi}{4}\right)\end{pmatrix}=$ $\begin{pmatrix}\frac{1}{4}\sqrt{2}\sqrt{3}-\frac{1}{4}\sqrt{2}&-\frac{1}{4}\sqrt{2}\sqrt{3}-\frac{1}{4}\sqrt{2} \\ \frac{1}{4}\sqrt{2}\sqrt{3}+\frac{1}{4}\sqrt{2}&\frac{1}{4}\sqrt{2}\sqrt{3}-\frac{1}{4}\sqrt{2}\end{pmatrix}$ Note that it doesn't matter about the order in this case.

Exercise 38 习题 38

Find the matrix of the linear transformation which rotates every vector in $\mathbb{R}^3$ counter clockwise about the $z$ axis when viewed from the positive $z$ axis through an angle of $30^\circ$ and then reflects through the $xy$ plane.

Answer $$\begin{pmatrix}1&0&0\\0&1&0\\0&0&-1\end{pmatrix}\begin{pmatrix}\cos\left(\frac{\pi}{6}\right)&-\sin\left(\frac{\pi}{6}\right)&0 \\ \sin\left(\frac{\pi}{6}\right)&\cos\left(\frac{\pi}{6}\right)&0 \\ 0&0&1\end{pmatrix}=\begin{pmatrix}\frac{1}{2}\sqrt{3}&-\frac{1}{2}&0\\ \frac{1}{2}&\frac{1}{2}\sqrt{3}&0\\0&0&-1\end{pmatrix}$$

Exercise 39 习题 39

Let $\vec{u}=\begin{pmatrix}a\\b\end{pmatrix}$ be a unit vector in $\mathbb{R}^2$. Find the matrix which reflects all vectors across this vector, as shown in the following picture.

[figure]

Figure 1. Hint: Notice that $\begin{pmatrix}a\\b\end{pmatrix}=\begin{pmatrix}\cos\theta \\ \sin\theta\end{pmatrix}$ for some $\theta$. First rotate through $-\theta$. Next reflect through the $x$ axis. Finally rotate through $\theta$.

Answer $$\begin{aligned}&\begin{pmatrix}\cos (\theta )&-\sin(\theta) \\ \sin(\theta)&\cos(\theta)\end{pmatrix}\begin{pmatrix}1&0\\0&-1\end{pmatrix}\begin{pmatrix}\cos(-\theta)&-\sin(\theta) \\ \sin(-\theta)&\cos(-\theta)\end{pmatrix} \\ =&\begin{pmatrix}\cos^2\theta-\sin^2\theta &2\cos\theta\sin\theta \\ 2\cos\theta\sin\theta&\sin^2\theta-\cos^2\theta\end{pmatrix}\end{aligned}$$ Now to write in terms of $a,b$, note that $a/\sqrt{a^2+b^2}=\cos\theta$, $b/\sqrt{a^2+b^2}=\sin\theta$. Now plug this in to the above. The result is $$\begin{pmatrix}\frac{a^2-b^2}{a^2+b^2}&2\frac{ab}{a^2+b^2} \\ 2\frac{ab}{a^2+b^2}&\frac{b^2-a^2}{a^2+b^2}\end{pmatrix}=\frac{1}{a^2+b^2}\begin{pmatrix}a^2-b^2&2ab \\ 2ab&b^2-a^2\end{pmatrix}$$ Since this is a unit vector, $a^2+b^2=1$ and so you get $$\begin{pmatrix}a^2-b^2&2ab \\ 2ab&b^2-a^2\end{pmatrix}$$

Exercise 40 习题 40

Let $T$ be a linear transformation given by $T\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}2&1\\0&1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}$ Is $T$ one to one? Is $T$ onto?

Exercise 41 习题 41

Let $T$ be a linear transformation given by $T\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}-1&2\\2&1\\1&4\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}$ Is $T$ one to one? Is $T$ onto?

Exercise 42 习题 42

Let $T$ be a linear transformation given by $T\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}2&0&1\\1&2&-1\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}$ Is $T$ one to one? Is $T$ onto?

Exercise 43 习题 43

Let $T$ be a linear transformation given by $T\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}1&3&-5\\2&0&2\\2&4&-6\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}$ Is $T$ one to one? Is $T$ onto?

Exercise 44 习题 44

Give an example of a $3\times 2$ matrix with the property that the linear transformation determined by this matrix is one to one but not onto.

Answer $$\begin{pmatrix}1&0\\0&1\\0&0\end{pmatrix}$$

Exercise 45 习题 45

Suppose $A$ is an $m\times n$ matrix in which $m ≤ n$. Suppose also that the rank of $A$ equals $m$. Show that the transformation $T$ determined by $A$ maps $\mathbb{R}^n$ onto $\mathbb{R}^m$. Hint: The vectors $\vec{e}_1,\cdots ,\vec{e}_m$ occur as columns in the reduced row-echelon form for $A$.

Answer This says that the columns of $A$ have a subset of $m$ vectors which are linearly independent. Therefore, this set of vectors is a basis for $\mathbb{R}^m$. It follows that the span of the columns is all of $\mathbb{R}^m$. Thus $A$ is onto.

Exercise 46 习题 46

Suppose $A$ is an $m\times n$ matrix in which $m ≥ n$. Suppose also that the rank of $A$ equals $n$. Show that $A$ is one to one. Hint: If not, there exists a vector, $\vec{x}$ such that $A\vec{x} = 0$, and this implies at least one column of $A$ is a linear combination of the others. Show this would require the rank to be less than $n$.

Answer The columns are independent. Therefore, $A$ is one to one.

Exercise 47 习题 47

Explain why an $n\times n$ matrix $A$ is both one to one and onto if and only if its rank is $n$.

Answer The rank is $n$ is the same as saying the columns are independent which is the same as saying $A$ is one to one which is the same as saying the columns are a basis. Thus the span of the columns of $A$ is all of $\mathbb{R}^n$ and so $A$ is onto. If $A$ is onto, then the columns must be linearly independent since otherwise the span of these columns would have dimension less than $n$ and so the dimension of $\mathbb{R}^n$ would be less than $n$.

Exercise 48 习题 48

Let $V$ and $W$ be subspaces of $\mathbb{R}^n$ and $\mathbb{R}^m$ respectively and let $T : V \to W$ be a linear transformation. Suppose that $\vec{v}_1,\cdots ,\vec{v}_r$ is linearly independent. Show that it must be the case that $T\vec{v}_1,\cdots ,T\vec{v}_r$ is also linearly independent.

Answer If $\sum_i^r a_i\vec{v}_r=0$, then using linearity properties of $T$ we get $0=T(0)=T\left(\sum\limits_i^ra_i\vec{v}_r\right)=\sum\limits_i^ra_iT(\vec{v}_r)$. Since we assume that $T\vec{v}_a,\cdots ,T\vec{v}_r$ is linearly independent, we must have all $a_i = 0$, and therefore we conclude that $\vec{v}_1,\cdots ,\vec{v}_r$ is also linearly independent.

Exercise 49 习题 49

Let $V=span\left\{\begin{array}{c}1\\1\\2\\0\end{array}\right],\begin{pmatrix}0\\1\\1\\1\end{pmatrix},\begin{pmatrix}1\\1\\0\\1\end{pmatrix}\right\}$. Let $T\vec{x}=A\vec{x}$ where $A$ is the matrix $\begin{pmatrix}1&1&1&1\\0&1&1&0\\0&1&2&1\\1&1&1&2\end{pmatrix}$ Give a basis for $im(T)$.

Exercise 50 习题 50

Let $V=span\left\{\begin{array}{c}1\\0\\0\\1\end{array}\right],\begin{pmatrix}1\\1\\1\\1\end{pmatrix},\begin{pmatrix}1\\4\\4\\1\end{pmatrix}\right\}$. Let $T\vec{x}=A\vec{x}$ where $A$ is the matrix $\begin{pmatrix}1&1&1&1\\0&1&1&0\\0&1&2&1\\1&1&1&2\end{pmatrix}$ Find a basis for $im(T)$. In this case, the original vectors do not form an independent set.

Answer Since the third vector is a linear combinations of the first two, then the image of the third vector will also be a linear combinations of the image of the first two. However the image of the first two vectors are linearly independent (check!), and hence form a basis of the image. Thus a basis for $im(T)$ is: $V=span\left\{\begin{array}{c}2\\0\\1\\3\end{array}\right],\begin{pmatrix}4\\2\\4\\5\end{pmatrix}\right\}$.

Exercise 51 习题 51

If $\vec{v}_1,\cdots ,\vec{v}_r$ is linearly independent and $T$ is a one to one linear transformation, show that $T\vec{v}_1,\cdots ,T\vec{v}_r$ is also linearly independent. Give an example which shows that if $T$ is only linear, it can happen that, although $\vec{v}_1,\cdots ,\vec{v}_r$ is linearly independent, $T\vec{v}_1,\cdots ,T\vec{v}_r$ is not. In fact, show that it can happen that each of the $T\vec{v}_j$ equals $0$.

Exercise 52 习题 52

Let $V$ and $W$ be subspaces of $\mathbb{R}^n$ and $\mathbb{R}^m$ respectively and let $T : V \to W$ be a linear transformation. Show that if $T$ is onto $W$ and if $\vec{v}_1,\cdots ,\vec{v}_r$ is a basis for $V$, then $\operatorname{span}\left\{T\vec{v}_1,\cdots ,T\vec{v}_r\right\} = W$.

Exercise 53 习题 53

Define $T$: $\mathbb{R}^4\to\mathbb{R}^3$ as follows. $T\vec{x}=\begin{pmatrix}3&2&1&8\\2&2&-2&6\\1&1&-1&3\end{pmatrix}\vec{x}$ Find a basis for $im(T)$. Also find a basis for $\text{ker}(T)$.

Exercise 54 习题 54

Define $T$: $\mathbb{R}^3\to\mathbb{R}^3$ as follows. $T\vec{x}=\begin{pmatrix}1&2&0\\1&1&1\\0&1&1\end{pmatrix}\vec{x}$ where on the right, it is just matrix multiplication of the vector $\vec{x}$ which is meant. Explain why $T$ is an isomorphism of $\mathbb{R}^3$ to $\mathbb{R}^3$.

Exercise 55 习题 55

Suppose $T$: $\mathbb{R}^3\to\mathbb{R}^3$ is a linear transformation given by $T\vec{x}=A\vec{x}$ where $A$ is a $3\times 3$ matrix. Show that $T$ is an isomorphism if and only if $A$ is invertible.

Exercise 56 习题 56

Suppose $T$: $\mathbb{R}^n\to\mathbb{R}^m$ is a linear transformation given by $T\vec{x}=A\vec{x}$ where $A$ is an $m\times n$ matrix. Show that $T$ is never an ismorphism if $m\neq n$. In particular, show that if $m>n$, $T$ cannot be onto and if $m