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\(\mathbb{R}^n\) 中的矩陣表示法

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This work by Jephian Lin is licensed under a Creative Commons Attribution 4.0 International License.

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from lingeo import random_int_list, random_good_matrix

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

  • \([f]{\bf b} =f({\bf b})\),
  • \([\operatorname{id}]_{\mathcal{E}_n}^\alpha {\bf b} = [{\bf b}]_\alpha\),
  • \([\operatorname{id}]_{\mathcal{E}_m}^\beta f({\bf b}) = [f({\bf b})]_\beta\).

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

  • projection

Experiments

Exercise 1

執行以下程式碼。
已知 \(f\)\(\mathbb{R}^3\)\(\mathbb{R}^2\) 的線性函數﹐
\(\alpha\)\(\beta\) 分別為 \(\mathbb{R}^3\)\(\mathbb{R}^2\) 的一組基底。

[程式碼有更新]

### code
set_random_seed(0)
print_ans = False
m,n = 2,3
alpha = random_good_matrix(n,n,n, bound=3)
beta = random_good_matrix(m,m,m, bound=3)
A = matrix(m, random_int_list(m*n))
v = vector(random_int_list(n, 3))
b = alpha * v

print("alpha contains %s vectors:"%n)
for j in range(n):
    print("v%s ="%(j+1), alpha.column(j))

print("beta contains %s vectors:"%m)
for i in range(m):
    print("u%s ="%(i+1), beta.column(i))

for j in range(n):
    print( "f(v%s) = "%(j+1) + " + ".join("%s u%s"%(A[i,j],i+1) for i in range(m)) )
    
print("b =", b)

if print_ans:
    print("[b]_alpha =", v)
    print("[f(b)]_beta =", A*v)
    print("f(b) =", beta * A * v)
    print("[f]_alpha^beta =")
    show(A)
    print("[f] =")
    show(beta * A * alpha.inverse())

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)

\([{\bf b}]_\alpha\)\([f({\bf b})]_\beta\) 、及 \(f({\bf b})\)

  • 標點
  • \begin{aligned} ... \end{aligned} 把一些等號對齊(參考 風格指引

答:

\(U\)\(V\) 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。


\({\bf e}_1=(1,0,0)\)
\({\bf e}_2=(0,1,0)\)
\({\bf e}_3=(0,0,1)\)

\[\left[\begin{array}{ccc|cccc} {\bf v}_1& {\bf v}_2 & {\bf v}_3 & {\bf b} & {\bf e}_1 & {\bf e}_2 & {\bf e}_3 \end{array}\right]\] \[\left[\begin{array}{ccc|cccc} 1 & 3 & 1 & 6 & 1 & 0 & 0\\ -3 & -8 & -1 & -19 & 0 & 1 & 0\\ 0 & -1 & -1 & -1 & 0 & 0 & 1 \end{array}\right]\] 化簡 \[\left[\begin{array}{ccc|cccc} 1 & 0 & 0 & -1 & 7 & 2 & 5\\ 0 & 1 & 0 & 3 & -3 & -1 & -2\\ 0 & 0 & 1 & -2 & 3 & 1 & 1 \end{array}\right] \]

  1. \([{\bf b}]_\alpha=\begin{bmatrix} -1 \\ 3 \\ -2 \end{bmatrix}.\)

\[\begin{aligned}[] [f({\bf b})]_\beta &=[f(-1{\bf v}_1+3{\bf v}_2-2{\bf v}_3)]_\beta \\ &=-[f({\bf v}_1)]_\beta+3[f({\bf v}_2)]_\beta-2[f({\bf v}_3)]_\beta \\ &=-\begin{bmatrix}4 \\2\end{bmatrix}+3\begin{bmatrix}-4 \\4\end{bmatrix}-2\begin{bmatrix}-3 \\-4\end{bmatrix}\\ &=\begin{bmatrix}-10 \\18\end{bmatrix}. \end{aligned} \]

\[\begin{aligned}f({\bf b}) &=-10{\bf u}_1+18{\bf u}_2 \\ &=-10\begin{bmatrix}1 \\2\end{bmatrix} +18\begin{bmatrix}1 \\3\end{bmatrix} \\ &=\begin{bmatrix}8 \\34\end{bmatrix}. \end{aligned} \]

Exercise 1(b)

\([f]_\alpha^\beta\)\([f]\)

  • \(f(\be_1)\) 那行之前加: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。
  • 標點

答:

  1. 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 \[[f]_\alpha^\beta =\begin{bmatrix} | & | & | \\ [f({\bf v}_1)]_\beta & [f({\bf v}_2)]_\beta & [f({\bf v}_3)]_\beta \\ | & | & | \\ \end{bmatrix} =\begin{bmatrix}4 & -4 & -3 \\ 2 & 4 & -4 \end{bmatrix}. \]

  2. 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

\(f : \mathbb{R}^n \rightarrow \mathbb{R}^m\) 為一線性函數、
\(\alpha = \{{\bf v}_1, \ldots, {\bf v}_n\}\)\(\mathbb{R}^n\) 的一組基底、
\(\beta = \{{\bf u}_1, \ldots, {\bf u}_m\}\)\(\mathbb{R}^m\) 的一組基底。

Exercise 2(a)

\(m = n = 3\)
\(\alpha = \beta\)
\[A = \begin{bmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & 0 & -\frac{2}{\sqrt{6}} \\ \end{bmatrix} \] 的各行向量。
已知
\(f({\bf v}_1) = {\bf u}_1\)
\(f({\bf v}_2) = {\bf u}_2\)
\(f({\bf v}_3) = {\bf 0}\)
\([f]_\alpha^\beta\)\([f]\) 並說明 \(f\) 的作用。

  • 有一個 \(\be_3\) 打錯
  • 基底不會是矩陣,可以把那兩個單位矩陣刪掉;後面幾題一樣
  • 最後一式用 aligned 將等號對齊;後面幾題一樣
  • \(f\) 作用為投影 > \(f\) 的作用為將向量投影到 \(\vspan\{\bu_1,\bu_2\}\) 這個平面。

答:
因為 \([f({\bf v}_1)]_\beta\)\(\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}\) \([f({\bf v}_2)]_\beta\)\(\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}\) \([f({\bf v}_3)]_\beta\)\(\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}\)

所以 \([f]_\alpha^\beta = \begin{bmatrix} | & | & | \\ [f({\bf v}_1)]_\beta & [f({\bf v}_2)]_\beta & [f({\bf v}_3)]_\beta \\ | & | & | \\ \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix}\)

\({\mathcal{E}_n} = \{{\bf e}_1, {\bf e}_2, {\bf e}_3\}\)\(\mathbb{R}^n\) 的標準基底,
\({\mathcal{E}_m} = \{{\bf e}_1, {\bf e}_2, {\bf e}_3\}\)\(\mathbb{R}^m\) 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)

\(m = n = 3\)
\(\alpha = \beta\)
\[A = \begin{bmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & 0 & -\frac{2}{\sqrt{6}} \\ \end{bmatrix} \] 的各行向量。
已知
\(f({\bf v}_1) = {\bf u}_1\)
\(f({\bf v}_2) = {\bf u}_2\)
\(f({\bf v}_3) = -{\bf u}_3\)
\([f]_\alpha^\beta\)\([f]\) 並說明 \(f\) 的作用。

  • 在這題中,f會使\({\bf u}_3\)變號但\({\bf u}_1\),\({\bf u}_2\)不變,故為鏡射。> 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。

答:
從題目中可以知道 \([f({\bf v}_1)]_\beta\)\(\begin{bmatrix} 1 \\ 0 \\ 0\\ \end{bmatrix}\) \([f({\bf v}_2)]_\beta\)\(\begin{bmatrix} 0 \\ 1 \\ 0\\ \end{bmatrix}\) \([f({\bf v}_3)]_\beta\)\(\begin{bmatrix} 0 \\ 0 \\ -1\\ \end{bmatrix}\)

因此能得出 \([f]_\alpha^\beta = \begin{bmatrix} | & | & | \\ [f({\bf v}_1)]_\beta & [f({\bf v}_2)]_\beta & [f({\bf v}_3)]_\beta \\ | & | & | \\ \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{bmatrix}\)

\({\mathcal{E}_n}=\{{\bf e}_1, {\bf e}_2, {\bf e}_3\}\) 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©

\(m = n = 3\)
\(\alpha = \beta\)
\[A = \begin{bmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & 0 & -\frac{2}{\sqrt{6}} \\ \end{bmatrix} \] 的各行向量。
已知
\(f({\bf v}_1) = {\bf u}_2\)
\(f({\bf v}_2) = -{\bf u}_1\)
\(f({\bf v}_3) = {\bf u}_3\)
\([f]_\alpha^\beta\)\([f]\) 並說明 \(f\) 的作用。

  • 在這題中,能從\([f]_\alpha^\beta\)得知f為旋轉矩陣。 > 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。

答:
從題目中可以知道 \([f({\bf v}_1)]_\beta\)\(\begin{bmatrix} 0 \\ 1 \\ 0\\ \end{bmatrix}\) \([f({\bf v}_2)]_\beta\)\(\begin{bmatrix} -1 \\ 0 \\ 0\\ \end{bmatrix}\) \([f({\bf v}_3)]_\beta\)\(\begin{bmatrix} 0 \\ 0 \\ 1\\ \end{bmatrix}\)

因此能得出\([f]_\alpha^\beta = \begin{bmatrix} | & | & | \\ [f({\bf v}_1)]_\beta & [f({\bf v}_2)]_\beta & [f({\bf v}_3)]_\beta \\ | & | & | \\ \end{bmatrix}=\begin{bmatrix} 0 & -1 & 0\\ 1 & 0 & 0\\ 0 & 0 & 1 \end{bmatrix}\)

\({\mathcal{E}_n}=\{{\bf e}_1, {\bf e}_2, {\bf e}_3\}\) 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)

\(m = 2\)\(n = 3\)
\(\alpha\)
\[A = \begin{bmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & 0 & -\frac{2}{\sqrt{6}} \\ \end{bmatrix} \] 的各行向量、
\(\beta\)
\[B = \begin{bmatrix} 1 & 1 \\ 0 & 1 \\ \end{bmatrix} \] 的各行向量。
已知
\(f({\bf v}_1) = 3{\bf u}_1\)
\(f({\bf v}_2) = 4{\bf u}_2\)
\(f({\bf v}_3) = {\bf 0}\)
\([f]_\alpha^\beta\)\([f]\)

  • 不要中英雜夾
  • \([f]\) 的部份要用 \([f] = ([\operatorname{id}]_{\mathcal{E}_m}^\beta)^{-1} [f]_\alpha^\beta) [\operatorname{id}]_{\mathcal{E}_n}^\alpha = [\operatorname{id}]^{\mathcal{E}_m}_\beta[f]_\alpha^\beta ([\operatorname{id}]^{\mathcal{E}_n}_\alpha)^{-1}\) 計算
  • \(\be^{'}\) > \(\be'\)

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 \([f({\bf v}_1)]_\beta\)\(\begin{bmatrix} 3 \\ 0 \\ \end{bmatrix}\) \([f({\bf v}_2)]_\beta\)\(\begin{bmatrix} 0 \\ 4 \\ \end{bmatrix}\) \([f({\bf v}_3)]_\beta\)\(\begin{bmatrix} 0 \\ 0 \\ \end{bmatrix}\),故得 \[[f]_\alpha^\beta = \begin{bmatrix} | & | & | \\ [f({\bf v}_1)]_\beta & [f({\bf v}_2)]_\beta & [f({\bf v}_3)]_\beta \\ | & | & | \\ \end{bmatrix} = \begin{bmatrix} 3 & 0 & 0 \\ 0 & 4 & 0 \\ \end{bmatrix} \]
\({\mathcal{E}_m}=\{{\bf e}_1, {\bf e}_2\}\)\(\mathbb{R}^m\) 的標準基底,\({\mathcal{E}_n}=\{{\bf e}_1', {\bf e}_2', {\bf e}_3'\}\)\(\mathbb{R}^n\) 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

\(f : \mathbb{R}^n \rightarrow \mathbb{R}^m\) 為一線性函數。
\(\mathcal{E}_n\)\(\mathcal{E}_m\) 分別為 \(\mathbb{R}^n\)\(\mathbb{R}^m\) 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

  • \(\mathcal{E}_n\)\(\mathcal{E}_m\) 分別為 \(\mathbb{R}^n\)\(\mathbb{R}^m\) 的一組基底,
    >
    \(\mathcal{E}_n\)\(\mathcal{E}_m\) 分別為 \(\mathbb{R}^n\)\(\mathbb{R}^m\) 的一組基底,
    且令 \(\mathcal{E}_n = \{\be_1,\ldots, \be_n\}\)
  • \(\alpha\) 那句拿掉
  • \(\mathcal{E}_i\) > \(\be_i\)
  • \([f]_{\mathcal{E}_n}^{\mathcal{E}^m}\) 的公式有錯,修正完應該答案就要差不多了;公式後面的東西我都不知道在幹麻

答:
\(\mathcal{E}_n\)\(\mathcal{E}_m\) 分別為 \(\mathbb{R}^n\)\(\mathbb{R}^m\) 的一組基底,
且令 \(\mathcal{E}_n = \{\be_1,\ldots, \be_n\}\)

已知 \[[f] = \begin{bmatrix} | & ~ & | \\ f({\bf e}_1) & \cdots & f({\bf e}_n) \\ | & ~ & | \\ \end{bmatrix}, \] \[[f]_{\mathcal{E}_n}^{\mathcal{E}_m} = \begin{bmatrix} | & ~ & | \\ [f({\bf e}_1)]_{\mathcal{E}_m} & \cdots & [f({\bf e}_n)]_{\mathcal{E}_m} \\ | & ~ & | \\ \end{bmatrix}, \]

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

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