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\(\mathbb{R}^n\) 中的矩陣表示法
This work by Jephian Lin is licensed under a Creative Commons Attribution 4.0 International License.
\(\newcommand{\trans}{^\top} \newcommand{\adj}{^{\rm adj}} \newcommand{\cof}{^{\rm cof}} \newcommand{\inp}[2]{\left\langle#1,#2\right\rangle} \newcommand{\dunion}{\mathbin{\dot\cup}} \newcommand{\bzero}{\mathbf{0}} \newcommand{\bone}{\mathbf{1}} \newcommand{\ba}{\mathbf{a}} \newcommand{\bb}{\mathbf{b}} \newcommand{\bc}{\mathbf{c}} \newcommand{\bd}{\mathbf{d}} \newcommand{\be}{\mathbf{e}} \newcommand{\bh}{\mathbf{h}} \newcommand{\bp}{\mathbf{p}} \newcommand{\bq}{\mathbf{q}} \newcommand{\br}{\mathbf{r}} \newcommand{\bx}{\mathbf{x}} \newcommand{\by}{\mathbf{y}} \newcommand{\bz}{\mathbf{z}} \newcommand{\bu}{\mathbf{u}} \newcommand{\bv}{\mathbf{v}} \newcommand{\bw}{\mathbf{w}} \newcommand{\tr}{\operatorname{tr}} \newcommand{\nul}{\operatorname{null}} \newcommand{\rank}{\operatorname{rank}} %\newcommand{\ker}{\operatorname{ker}} \newcommand{\range}{\operatorname{range}} \newcommand{\Col}{\operatorname{Col}} \newcommand{\Row}{\operatorname{Row}} \newcommand{\spec}{\operatorname{spec}} \newcommand{\vspan}{\operatorname{span}} \newcommand{\Vol}{\operatorname{Vol}} \newcommand{\sgn}{\operatorname{sgn}} \newcommand{\idmap}{\operatorname{id}} \newcommand{\am}{\operatorname{am}} \newcommand{\gm}{\operatorname{gm}} \newcommand{\mult}{\operatorname{mult}} \newcommand{\iner}{\operatorname{iner}}\)
Main idea
Recall that if \(f : \mathbb{R}^n \rightarrow \mathbb{R}^m\) is a linear function and \(\{ {\bf e}_1,\ldots, {\bf e}_n\}\) is the standard basis of \(\mathbb{R}^n\).
Then the matrix
\[[f] = \begin{bmatrix} | & ~ & | \\ f({\bf e}_1) & \cdots & f({\bf e}_n) \\ | & ~ & | \\ \end{bmatrix} \] has the property that \(f({\bf b}) = [f]{\bf b}\) for all \({\bf b}\in\mathbb{R}^n\).
Sometimes \(f({\bf e}_i)\) cannot be easily found, while the function \(f\) is described by the bases of \(\mathbb{R}^n\) and \(\mathbb{R}^m\) instead.
Let \(f : \mathbb{R}^n \rightarrow \mathbb{R}^m\) be a linear function,
\(\alpha = \{ {\bf v}_1, \ldots, {\bf v}_n \}\) be a basis of \(\mathbb{R}^n\), and
\(\beta\) a basis of \(\mathbb{R}^m\).
Then the matrix
\[[f]_\alpha^\beta = \begin{bmatrix} | & ~ & | \\ [f({\bf v}_1)]_\beta & \cdots & [f({\bf v}_n)]_\beta \\ | & ~ & | \\ \end{bmatrix} \] has the property that \([f({\bf b})]_\beta = [f]_\alpha^\beta [{\bf b}]_\alpha\).
Therefore, we call \([f]_\alpha^\beta\) the matrix representation of \(f\) with respect to \(\alpha\) and \(\beta\).
The equality can be visualized by the following diagram.
\[\begin{array}{ccc} {\bf b} & \xrightarrow{f} & f({\bf b}) \\ \downarrow & ~ & \downarrow \\ [{\bf b}]_\alpha & \xrightarrow{[f]_\alpha^\beta\cdot\square} & [f({\bf b})]_\beta \\ \end{array} \]
Let \(\mathcal{E}_n\) and \(\mathcal{E}_m\) be the standard basis of \(\mathbb{R}^n\) and \(\mathbb{R}^m\), respectively. Since we know
We know \([f] = ([\operatorname{id}]_{\mathcal{E}_m}^\beta)^{-1} [f]_\alpha^\beta [\operatorname{id}]_{\mathcal{E}_n}^\alpha = [\operatorname{id}]_\beta^{\mathcal{E}_m} [f]_\alpha^\beta ([\operatorname{id}]_\alpha^{\mathcal{E}_n})^{-1}\).
Side stories
Experiments
Exercise 1
執行以下程式碼。
已知 \(f\) 為 \(\mathbb{R}^3\) 到 \(\mathbb{R}^2\) 的線性函數﹐
而 \(\alpha\) 和 \(\beta\) 分別為 \(\mathbb{R}^3\) 和 \(\mathbb{R}^2\) 的一組基底。
[程式碼有更新]
藉由
seed=0
,得到\(\alpha\) 含有三個向量:
\({\bf v}_1 = (1, -3, 0)\)、
\({\bf v}_2 = (3, -8, -1)\)、
\({\bf v}_3 = (1, -1, -1)\)。
\(\beta\) 含有兩個向量:
\({\bf u}_1 = (1, 2)\)、
\({\bf u}_2 = (1, 3)\)。
\(f({\bf v}_1) = 4 {\bf u}_1 + 2 {\bf u}_2\)、
\(f({\bf v}_2) = -4 {\bf u}_1 + 4 {\bf u}_2\)、
\(f({\bf v}_3) = -3 {\bf u}_1 + -4 {\bf u}_2\)、
\({\bf b} = (6, -19, -1)\)。
Exercise 1(a)
求 \([{\bf b}]_\alpha\)、\([f({\bf b})]_\beta\) 、及 \(f({\bf b})\)。
\begin{aligned} ... \end{aligned}
把一些等號對齊(參考 風格指引)答:
令 \(U\) 和 \(V\) 為兩向量空間,
對任意向量 \({\bf u}, {\bf u}_1, {\bf u}_2\in U\) 及純量 \(k\in\mathbb{R}\).
如果 \[\begin{aligned} f({\bf u}_1 + {\bf u}_2) &= f({\bf u}_1) + f({\bf u}_2) \\ f(k{\bf u}) &= kf({\bf u}) \\ \end{aligned} \] 函數 \(f: U\rightarrow V\) 是線性。
設
\({\bf e}_1=(1,0,0)\)、
\({\bf e}_2=(0,1,0)\)、
\({\bf e}_3=(0,0,1)\)。
令 \[\left[\begin{array}{ccc|cccc} {\bf v}_1& {\bf v}_2 & {\bf v}_3 & {\bf b} & {\bf e}_1 & {\bf e}_2 & {\bf e}_3 \end{array}\right]\] 得 \[\left[\begin{array}{ccc|cccc} 1 & 3 & 1 & 6 & 1 & 0 & 0\\ -3 & -8 & -1 & -19 & 0 & 1 & 0\\ 0 & -1 & -1 & -1 & 0 & 0 & 1 \end{array}\right]\] 化簡 \[\left[\begin{array}{ccc|cccc} 1 & 0 & 0 & -1 & 7 & 2 & 5\\ 0 & 1 & 0 & 3 & -3 & -1 & -2\\ 0 & 0 & 1 & -2 & 3 & 1 & 1 \end{array}\right] \]
\([{\bf b}]_\alpha=\begin{bmatrix} -1 \\ 3 \\ -2 \end{bmatrix}.\)
\[\begin{aligned}[] [f({\bf b})]_\beta &=[f(-1{\bf v}_1+3{\bf v}_2-2{\bf v}_3)]_\beta \\ &=-[f({\bf v}_1)]_\beta+3[f({\bf v}_2)]_\beta-2[f({\bf v}_3)]_\beta \\ &=-\begin{bmatrix}4 \\2\end{bmatrix}+3\begin{bmatrix}-4 \\4\end{bmatrix}-2\begin{bmatrix}-3 \\-4\end{bmatrix}\\ &=\begin{bmatrix}-10 \\18\end{bmatrix}. \end{aligned} \]
\[\begin{aligned}f({\bf b}) &=-10{\bf u}_1+18{\bf u}_2 \\ &=-10\begin{bmatrix}1 \\2\end{bmatrix} +18\begin{bmatrix}1 \\3\end{bmatrix} \\ &=\begin{bmatrix}8 \\34\end{bmatrix}. \end{aligned} \]
Exercise 1(b)
求 \([f]_\alpha^\beta\) 及 \([f]\)。
答:
因為 \[\begin{aligned}[] [f({\bf v}_1)]_\beta &= [f]_\alpha^\beta \begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix} =\begin{bmatrix}4 \\ 2\end{bmatrix}, \\ [f({\bf v}_2)]_\beta &= [f]_\alpha^\beta \begin{bmatrix}0 \\ 1 \\ 0\end{bmatrix} =\begin{bmatrix}-4 \\ 4\end{bmatrix}, \\ [f({\bf v}_3)]_\beta &= [f]_\alpha^\beta \begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix} =\begin{bmatrix}-3 \\ -4\end{bmatrix}, \end{aligned} \] 所以 \[[f]_\alpha^\beta =\begin{bmatrix} | & | & | \\ [f({\bf v}_1)]_\beta & [f({\bf v}_2)]_\beta & [f({\bf v}_3)]_\beta \\ | & | & | \\ \end{bmatrix} =\begin{bmatrix}4 & -4 & -3 \\ 2 & 4 & -4 \end{bmatrix}. \]
由前一題的最簡階梯形可以將 \(\be_1,\be_2,\be_3\) 分別寫成 \(\{\bv_1,\bv_2,\bv_3\}\) 的線性組合。
\(f({\bf e}_1)=f(7{\bf v}_1-3{\bf v}_2+3{\bf v}_3) =7f({\bf v}_1)-3f({\bf v}_2)+3f({\bf v}_3) =\begin{bmatrix}21 \\ 32\end{bmatrix}\)、
\(f({\bf e}_2)=f(2{\bf v}_1-{\bf v}_2+{\bf v}_3) =2f({\bf v}_1)-f({\bf v}_2)+f({\bf v}_3) =\begin{bmatrix}5 \\ 6\end{bmatrix}\)、
\(f({\bf e}_3)=f(5{\bf v}_1-2{\bf v}_2+{\bf v}_3) =5f({\bf v}_1)-2f({\bf v}_2)+f({\bf v}_3) =\begin{bmatrix}23 \\ 44\end{bmatrix}\)。
\([f]= \begin{bmatrix} | & | & | \\ f({\bf e}_1) & f({\bf e}_2) & f({\bf e}_3)\\ | & | & | \\ \end{bmatrix} =\begin{bmatrix}21 & 5 & 23 \\ 32 & 6 & 44\end{bmatrix}.\)
Exercises
Exercise 2
令 \(f : \mathbb{R}^n \rightarrow \mathbb{R}^m\) 為一線性函數、
\(\alpha = \{{\bf v}_1, \ldots, {\bf v}_n\}\) 為 \(\mathbb{R}^n\) 的一組基底、
\(\beta = \{{\bf u}_1, \ldots, {\bf u}_m\}\) 為 \(\mathbb{R}^m\) 的一組基底。
Exercise 2(a)
令 \(m = n = 3\) 且
\(\alpha = \beta\) 為
\[A = \begin{bmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & 0 & -\frac{2}{\sqrt{6}} \\ \end{bmatrix} \] 的各行向量。
已知
\(f({\bf v}_1) = {\bf u}_1\)、
\(f({\bf v}_2) = {\bf u}_2\)、
\(f({\bf v}_3) = {\bf 0}\)。
求 \([f]_\alpha^\beta\) 及 \([f]\) 並說明 \(f\) 的作用。
aligned
將等號對齊;後面幾題一樣答:
因為 \([f({\bf v}_1)]_\beta\) 為 \(\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}\)、 \([f({\bf v}_2)]_\beta\) 為 \(\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}\)、 \([f({\bf v}_3)]_\beta\) 為 \(\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}\),
所以 \([f]_\alpha^\beta = \begin{bmatrix} | & | & | \\ [f({\bf v}_1)]_\beta & [f({\bf v}_2)]_\beta & [f({\bf v}_3)]_\beta \\ | & | & | \\ \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix}\)。
令 \({\mathcal{E}_n} = \{{\bf e}_1, {\bf e}_2, {\bf e}_3\}\) 為 \(\mathbb{R}^n\) 的標準基底,
\({\mathcal{E}_m} = \{{\bf e}_1, {\bf e}_2, {\bf e}_3\}\) 為 \(\mathbb{R}^m\) 的標準基底,
\(\begin{aligned} \\ [f] &= [\operatorname{id}]_\beta^{\mathcal{E}_m} [f]_\alpha^\beta ([\operatorname{id}]_\alpha^{\mathcal{E}_n})^{-1} \\ &= \begin{bmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & 0 & -\frac{2}{\sqrt{6}} \\ \end{bmatrix}\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix}\begin{bmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & 0 & -\frac{2}{\sqrt{6}} \\ \end{bmatrix}^{-1} \\ &= \begin{bmatrix} \frac{5}{6} & -\frac{1}{6} & \frac{1}{3} \\ -\frac{1}{6} & \frac{5}{6} & \frac{1}{3} \\ \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \end{bmatrix} \end{aligned}\)
觀察
\(f({\bf v}_1) = {\bf u}_1\)、
\(f({\bf v}_2) = {\bf u}_2\)、
\(f({\bf v}_3) = {\bf 0}\)。
\(f\) 的作用為將向量投影到 \(\vspan\{\bu_1,\bu_2\}\) 這個平面。
Exercise 2(b)
令 \(m = n = 3\) 且
\(\alpha = \beta\) 為
\[A = \begin{bmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & 0 & -\frac{2}{\sqrt{6}} \\ \end{bmatrix} \] 的各行向量。
已知
\(f({\bf v}_1) = {\bf u}_1\)、
\(f({\bf v}_2) = {\bf u}_2\)、
\(f({\bf v}_3) = -{\bf u}_3\)。
求 \([f]_\alpha^\beta\) 及 \([f]\) 並說明 \(f\) 的作用。
答:
從題目中可以知道 \([f({\bf v}_1)]_\beta\) 為 \(\begin{bmatrix} 1 \\ 0 \\ 0\\ \end{bmatrix}\)、 \([f({\bf v}_2)]_\beta\) 為 \(\begin{bmatrix} 0 \\ 1 \\ 0\\ \end{bmatrix}\)、 \([f({\bf v}_3)]_\beta\) 為 \(\begin{bmatrix} 0 \\ 0 \\ -1\\ \end{bmatrix}\),
因此能得出 \([f]_\alpha^\beta = \begin{bmatrix} | & | & | \\ [f({\bf v}_1)]_\beta & [f({\bf v}_2)]_\beta & [f({\bf v}_3)]_\beta \\ | & | & | \\ \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{bmatrix}\)。
令 \({\mathcal{E}_n}=\{{\bf e}_1, {\bf e}_2, {\bf e}_3\}\) 的行向量為 \(\mathbb{R}^n\) 的標準基底,
由此能計算出
\(\begin{aligned} \\ [f] &= [\operatorname{id}]_\beta^{\mathcal{E}_m} [f]_\alpha^\beta([\operatorname{id}]_\alpha^{\mathcal{E}_n})^{-1} \\ &= \begin{bmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & 0 & -\frac{2}{\sqrt{6}} \\ \end{bmatrix}\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{bmatrix}\begin{bmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & 0 & -\frac{2}{\sqrt{6}} \\ \end{bmatrix}^{-1} \\ &= \begin{bmatrix} \frac{2}{3} & -\frac{1}{3} & \frac{2}{3} \\ -\frac{1}{3} & \frac{2}{3} & \frac{2}{3} \\ \frac{2}{3} & \frac{2}{3} & -\frac{1}{3} \end{bmatrix} \end{aligned}\)
在這題中,\(f\) 會使 \({\bf u}_3\) 變號但 \({\bf u}_1\), \({\bf u}_2\) 不變,故為 \(f\) 的作用為對平面 \(\vspan\{\bu_1,\bu_2\}\) 鏡射。
Exercise 2©
令 \(m = n = 3\) 且
\(\alpha = \beta\) 為
\[A = \begin{bmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & 0 & -\frac{2}{\sqrt{6}} \\ \end{bmatrix} \] 的各行向量。
已知
\(f({\bf v}_1) = {\bf u}_2\)、
\(f({\bf v}_2) = -{\bf u}_1\)、
\(f({\bf v}_3) = {\bf u}_3\)。
求 \([f]_\alpha^\beta\) 及 \([f]\) 並說明 \(f\) 的作用。
答:
從題目中可以知道 \([f({\bf v}_1)]_\beta\) 為 \(\begin{bmatrix} 0 \\ 1 \\ 0\\ \end{bmatrix}\)、 \([f({\bf v}_2)]_\beta\) 為 \(\begin{bmatrix} -1 \\ 0 \\ 0\\ \end{bmatrix}\)、 \([f({\bf v}_3)]_\beta\) 為 \(\begin{bmatrix} 0 \\ 0 \\ 1\\ \end{bmatrix}\),
因此能得出\([f]_\alpha^\beta = \begin{bmatrix} | & | & | \\ [f({\bf v}_1)]_\beta & [f({\bf v}_2)]_\beta & [f({\bf v}_3)]_\beta \\ | & | & | \\ \end{bmatrix}=\begin{bmatrix} 0 & -1 & 0\\ 1 & 0 & 0\\ 0 & 0 & 1 \end{bmatrix}\)。
令 \({\mathcal{E}_n}=\{{\bf e}_1, {\bf e}_2, {\bf e}_3\}\) 的行向量為 \(\mathbb{R}^n\) 的標準基底,
由此能計算出
\(\begin{aligned} \\ [f] &= [\operatorname{id}]_\beta^{\mathcal{E}_m} [f]_\alpha^\beta ([\operatorname{id}]_\alpha^{\mathcal{E}_n})^{-1} \\ &= \begin{bmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & 0 & -\frac{2}{\sqrt{6}} \\ \end{bmatrix}\begin{bmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & 0 & -\frac{2}{\sqrt{6}} \\ \end{bmatrix}^{-1} \\ &= \begin{bmatrix} -\frac{1}{6} & -\frac{2}{\sqrt{6}}+\frac{1}{6} & -\frac{1}{\sqrt{6}}-\frac{1}{3} \\ \frac{2}{\sqrt{6}}+\frac{1}{6} & \frac{1}{6} & \frac{1}{\sqrt{6}}-\frac{1}{3} \\ \frac{1}{\sqrt{6}}-\frac{1}{3} & -\frac{1}{\sqrt{6}}-\frac{1}{3} & \frac{2}{3} \end{bmatrix} \end{aligned}\)
在這題中,能從\([f]_\alpha^\beta\)得知 \(f\) 的作用為將向量以 \(\bu_3\) 為軸,由 \(\bu_1\) 方向往 \(\bu_2\) 方向旋轉 \(90\) 度。
Exercise 2(d)
令 \(m = 2\)、\(n = 3\) 且
\(\alpha\) 為
\[A = \begin{bmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & 0 & -\frac{2}{\sqrt{6}} \\ \end{bmatrix} \] 的各行向量、
\(\beta\) 為
\[B = \begin{bmatrix} 1 & 1 \\ 0 & 1 \\ \end{bmatrix} \] 的各行向量。
已知
\(f({\bf v}_1) = 3{\bf u}_1\)、
\(f({\bf v}_2) = 4{\bf u}_2\)、
\(f({\bf v}_3) = {\bf 0}\)。
求 \([f]_\alpha^\beta\) 及 \([f]\)。
答:
\(\alpha\) 含有三個向量:
\({\bf v}_1 = (\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})\)、
\({\bf v}_2 = (\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}},0)\)、
\({\bf v}_3 = (\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}},-\frac{2}{\sqrt{6}})\)。
\(\beta\) 含有兩個向量:
\({\bf u}_1 = (1,0)\)、
\({\bf u}_2 = (1,1)\)。
經計算後得出
\(f({\bf v}_1) = 3{\bf u}_1 = 3{\bf u}_1 + 0{\bf u}_2\)、
\(f({\bf v}_2) = 4{\bf u}_2 = 0{\bf u}_1 + 4{\bf u}_2\)、
\(f({\bf v}_3) = {\bf 0} = 0{\bf u}_1 + 0{\bf u}_2\)。
由此可知 \([f({\bf v}_1)]_\beta\) 為 \(\begin{bmatrix} 3 \\ 0 \\ \end{bmatrix}\)、 \([f({\bf v}_2)]_\beta\) 為 \(\begin{bmatrix} 0 \\ 4 \\ \end{bmatrix}\)、 \([f({\bf v}_3)]_\beta\) 為 \(\begin{bmatrix} 0 \\ 0 \\ \end{bmatrix}\),故得 \[[f]_\alpha^\beta = \begin{bmatrix} | & | & | \\ [f({\bf v}_1)]_\beta & [f({\bf v}_2)]_\beta & [f({\bf v}_3)]_\beta \\ | & | & | \\ \end{bmatrix} = \begin{bmatrix} 3 & 0 & 0 \\ 0 & 4 & 0 \\ \end{bmatrix} \]。
令 \({\mathcal{E}_m}=\{{\bf e}_1, {\bf e}_2\}\) 為 \(\mathbb{R}^m\) 的標準基底,\({\mathcal{E}_n}=\{{\bf e}_1', {\bf e}_2', {\bf e}_3'\}\) 為 \(\mathbb{R}^n\) 的標準基底,得
\(\begin{aligned} \\ [f] &= [\operatorname{id}]^{\mathcal{E}_n}_\beta[f]_\alpha^\beta ([\operatorname{id}]^{\mathcal{E}_m}_\alpha)^{-1} \\ &= \begin{bmatrix} 1 & 1 \\ 0 & 1 \\ \end{bmatrix}\begin{bmatrix} 3 & 0 & 0 \\ 0 & 4 & 0 \\ \end{bmatrix}\begin{bmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & 0 & -\frac{2}{\sqrt{6}} \\ \end{bmatrix}^{-1} \\ &= \begin{bmatrix} \sqrt{3}+2\sqrt{2} & \sqrt{3}-2\sqrt{2} & \sqrt{3} \\ 2\sqrt{2} & -2\sqrt{2} & 0 \\ \end{bmatrix}. \end{aligned}\)
Exercise 3
若 \(f : \mathbb{R}^n \rightarrow \mathbb{R}^m\) 為一線性函數。
而 \(\mathcal{E}_n\) 和 \(\mathcal{E}_m\) 分別為 \(\mathbb{R}^n\) 和 \(\mathbb{R}^m\) 的一組基底。
說明 \([f]\) 就是 \([f]_{\mathcal{E}_n}^{\mathcal{E}_m}\)。
這題完全不知道你在寫什麼喔…
–>
設 \(\mathcal{E}_n\) 和 \(\mathcal{E}_m\) 分別為 \(\mathbb{R}^n\) 和 \(\mathbb{R}^m\) 的一組基底,
且令 \(\mathcal{E}_n = \{\be_1,\ldots, \be_n\}\)。
答:
設 \(\mathcal{E}_n\) 和 \(\mathcal{E}_m\) 分別為 \(\mathbb{R}^n\) 和 \(\mathbb{R}^m\) 的一組基底,
且令 \(\mathcal{E}_n = \{\be_1,\ldots, \be_n\}\)。
已知 \[[f] = \begin{bmatrix} | & ~ & | \\ f({\bf e}_1) & \cdots & f({\bf e}_n) \\ | & ~ & | \\ \end{bmatrix}, \] 且 \[[f]_{\mathcal{E}_n}^{\mathcal{E}_m} = \begin{bmatrix} | & ~ & | \\ [f({\bf e}_1)]_{\mathcal{E}_m} & \cdots & [f({\bf e}_n)]_{\mathcal{E}_m} \\ | & ~ & | \\ \end{bmatrix}, \]
因為 \(\mathcal{E}_m\) 為標準基底,所以任何向量都有 \([\bv]_{\mathcal{E}_m} = \bv\)。
由此可知,\([f]\) 就是 \([f]_{\mathcal{E}_n}^{\mathcal{E}_m}\)。
目前分數 6.5