(1) \(\beta = (\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{v}_4)\) 為 \(V\) 之一組 ordered basis。  

因：

\[
\begin{aligned}
\mathbf{w}_1 &= \mathbf{v}_1 - \mathbf{v}_2 + \mathbf{v}_3 + 3\mathbf{v}_4 \\
\mathbf{w}_2 &= -\mathbf{v}_1 - \mathbf{v}_2 + \mathbf{v}_3 + 2\mathbf{v}_4 \\
\mathbf{w}_3 &= 5\mathbf{v}_1 - \mathbf{v}_2 + \mathbf{v}_3 + 5\mathbf{v}_4 \\
\mathbf{w}_4 &= \mathbf{v}_2 - \mathbf{v}_3 - 3\mathbf{v}_4 \\
\mathbf{w}_5 &= \mathbf{v}_1 + \mathbf{v}_2
\end{aligned}
\]

令 \(T_\beta: V \to \mathbb{R}^4\) 為用 \(\beta\) 將 \(V\) 坐標化之函數，

故：

\[
T_\beta(\mathbf{w}_1) = \begin{bmatrix} 1 \\ -1 \\ 1 \\ 3 \end{bmatrix},\quad
T_\beta(\mathbf{w}_2) = \begin{bmatrix} -1 \\ -1 \\ 1 \\ 2 \end{bmatrix},\quad
T_\beta(\mathbf{w}_3) = \begin{bmatrix} 5 \\ -1 \\ 1 \\ 5 \end{bmatrix},\quad
T_\beta(\mathbf{w}_4) = \begin{bmatrix} 0 \\ 1 \\ -1 \\ -3 \end{bmatrix},\quad
T_\beta(\mathbf{w}_5) = \begin{bmatrix} 1 \\ 1 \\ 0 \\ 0 \end{bmatrix}
\]

將 \(T_\beta(\mathbf{w}_1), \dots, T_\beta(\mathbf{w}_5)\) 按 column 排成矩陣的形式，

即

\[
A = \begin{bmatrix}
T_\beta(\mathbf{w}_1) & \dots & T_\beta(\mathbf{w}_5)
\end{bmatrix}
= \begin{bmatrix}
1 & -1 & 5 & 0 & 1 \\
-1 & -1 & -1 & 1 & 1 \\
1 & 1 & 1 & -1 & 0 \\
3 & 2 & 5 & -3 & 0
\end{bmatrix}
\]

\[
\xrightarrow{\text{利用Elementory row operation 化為 echelon form}}
\begin{bmatrix}
1 & -1 & 5 & 0 & 1 \\
0 & -2 & 4 & 1 & 2 \\
0 & 0 & 0 & -\frac{1}{2} & 2 \\
0 & 0 & 0 & 0 & 1
\end{bmatrix}
\]

故由 \(A\) 之 echelon form 知 pivot variables 為 \(x_1, x_2, x_4, x_5\)，

因此，\(T_\beta(\mathbf{w}_1), T_\beta(\mathbf{w}_2), T_\beta(\mathbf{w}_4), T_\beta(\mathbf{w}_5)\) 為 linearly independent.

可知 \(\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_4, \mathbf{w}_5\) 也為 linearly independent (By Ex. 6.6 (1))

故 \(\beta' = (\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_4, \mathbf{w}_5)\) 也是 \(V\) 之一組 ordered basis.

(2) 因：

\[
\begin{aligned}
\mathbf{v}_1 &= \mathbf{w}_1 + \mathbf{w}_4 \\[2pt]
\mathbf{v}_2 &= -\mathbf{w}_1 - \mathbf{w}_4 + \mathbf{w}_5 \\[2pt]
\mathbf{v}_3 &= 2\mathbf{w}_1 + 3\mathbf{w}_2 + 4\mathbf{w}_4 + \mathbf{w}_5 \\[2pt]
\mathbf{v}_4 &= -\mathbf{w}_1 - \mathbf{w}_2 - 2\mathbf{w}_4
\end{aligned}
\]

故經座標化後得：

\[
T_{\beta'}(\mathbf{v}_1) = \begin{bmatrix} 1 \\ 0 \\ 1 \\ 0 \end{bmatrix},\quad
T_{\beta'}(\mathbf{v}_2) = \begin{bmatrix} -1 \\ 0 \\ -1 \\ 1 \end{bmatrix},\quad
T_{\beta'}(\mathbf{v}_3) = \begin{bmatrix} 2 \\ 3 \\ 4 \\ 1 \end{bmatrix},\quad
T_{\beta'}(\mathbf{v}_4) = \begin{bmatrix} -1 \\ -1 \\ -2 \\ 0 \end{bmatrix}
\]

(3) 令  

\[
B = \big[\, T_\beta(\mathbf{w}_1) \;\; T_\beta(\mathbf{w}_2) \;\; T_\beta(\mathbf{w}_4) \;\; T_\beta(\mathbf{w}_5) \,\big]
\]  

\[
B' = \big[\, T_{\beta'}(\mathbf{v}_1) \;\; T_{\beta'}(\mathbf{v}_2) \;\; T_{\beta'}(\mathbf{v}_3) \;\; T_{\beta'}(\mathbf{v}_4) \,\big]
\]

Noted that：

\[
BB' = 
\begin{bmatrix}
1 & -1 & 0 & 1 \\
-1 & -1 & 1 & 1 \\
1 & 1 & -1 & 0 \\
3 & 2 & -3 & 0
\end{bmatrix}
\begin{bmatrix}
1 & -1 & 2 & -1 \\
0 & 0 & 3 & -1 \\
1 & -1 & 4 & -2 \\
0 & 1 & 1 & 0
\end{bmatrix}
=
\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{bmatrix}
= I_4
\]

因此，\( BB' = I_4 \)，即 \( B \) 與 \( B' \) 互為反矩陣。

\[
\text{令 }
A=
\begin{bmatrix}
1 & -1 & 5 & 0 & 1\\
-1 & -1 & -1 & 1 & 1\\
1 & 1 & 1 & -1 & 0\\
3 & 2 & 5 & -3 & 0
\end{bmatrix},
\]
其中五個 column 分別是
\[
[w_1]_\beta,\ [w_2]_\beta,\ [w_3]_\beta,\ [w_4]_\beta,\ [w_5]_\beta.
\]

\[
\text{若取 }\beta'=(w_1,w_2,w_4,w_5),
\text{ 則考慮方陣 }
P=
\begin{bmatrix}
1 & -1 & 0 & 1\\
-1 & -1 & 1 & 1\\
1 & 1 & -1 & 0\\
3 & 2 & -3 & 0
\end{bmatrix}.
\]

\[
\text{考慮增廣矩陣 }[P\mid I]:
\]

\[
\left[
\begin{array}{cccc|cccc}
1 & -1 & 0 & 1 & 1 & 0 & 0 & 0\\
-1 & -1 & 1 & 1 & 0 & 1 & 0 & 0\\
1 & 1 & -1 & 0 & 0 & 0 & 1 & 0\\
3 & 2 & -3 & 0 & 0 & 0 & 0 & 1
\end{array}
\right]
\]

經列運算可得

\[
\left[
\begin{array}{cccc|cccc}
1 & 0 & 0 & 0 & 1 & -1 & 2 & -1\\
0 & 1 & 0 & 0 & 0 & 0 & 3 & -1\\
0 & 0 & 1 & 0 & 1 & -1 & 4 & -2\\
0 & 0 & 0 & 1 & 0 & 1 & 1 & 0
\end{array}
\right].
\]

因此

\[
Q=
\begin{bmatrix}
1 & -1 & 2 & -1\\
0 & 0 & 3 & -1\\
1 & -1 & 4 & -2\\
0 & 1 & 1 & 0
\end{bmatrix}.
\]

\[
\text{這個矩陣是從 }\beta\text{ 座標到 }\beta'\text{ 座標的轉換矩陣，}
\]
也就是說

\[
\begin{bmatrix}
\mid & \mid & \mid & \mid\\
T_{\beta'}(v_1) & T_{\beta'}(v_2) & T_{\beta'}(v_3) & T_{\beta'}(v_4)\\
\mid & \mid & \mid & \mid
\end{bmatrix}
=
Q.
\]

故

\[
T_{\beta'}(v_1)=
\begin{bmatrix}
1\\
0\\
1\\
0
\end{bmatrix},
\qquad
T_{\beta'}(v_2)=
\begin{bmatrix}
-1\\
0\\
-1\\
1
\end{bmatrix},
\qquad
T_{\beta'}(v_3)=
\begin{bmatrix}
2\\
3\\
4\\
1
\end{bmatrix},
\qquad
T_{\beta'}(v_4)=
\begin{bmatrix}
-1\\
-1\\
-2\\
0
\end{bmatrix}.
\]

\[
\text{另外，}
P=
\begin{bmatrix}
1 & -1 & 0 & 1\\
-1 & -1 & 1 & 1\\
1 & 1 & -1 & 0\\
3 & 2 & -3 & 0
\end{bmatrix}
\text{ 與 }
Q=
\begin{bmatrix}
1 & -1 & 2 & -1\\
0 & 0 & 3 & -1\\
1 & -1 & 4 & -2\\
0 & 1 & 1 & 0
\end{bmatrix}
\text{ 互為反方陣。}
\]