(1) \(T_1: \mathbb{R}^5 \to \mathbb{R}^4\)

\[
T_1 \begin{bmatrix} a \\ b \\ c \\ d \\ e \end{bmatrix}
= \begin{bmatrix} a - c + 3d - e \\ a + 2d - e \\ 2a - c + 5d - e \\ -c + d \end{bmatrix}
\]

計算在標準 basis 的 images：
\[
\begin{aligned}
T_1\left(\begin{bmatrix}1\\0\\0\\0\\0\end{bmatrix}\right) &= \mathbf{v}_1 = \begin{bmatrix}1\\1\\2\\0\end{bmatrix},\quad
T_1\left(\begin{bmatrix}0\\1\\0\\0\\0\end{bmatrix}\right) = \mathbf{v}_2 = \begin{bmatrix}0\\0\\0\\0\end{bmatrix},\quad
T_1\left(\begin{bmatrix}0\\0\\1\\0\\0\end{bmatrix}\right) = \mathbf{v}_3 = \begin{bmatrix}-1\\0\\-1\\-1\end{bmatrix},\\[4pt]
T_1\left(\begin{bmatrix}0\\0\\0\\1\\0\end{bmatrix}\right) &= \mathbf{v}_4 = \begin{bmatrix}3\\2\\5\\1\end{bmatrix},\quad
T_1\left(\begin{bmatrix}0\\0\\0\\0\\1\end{bmatrix}\right) = \mathbf{v}_5 = \begin{bmatrix}-1\\-1\\-1\\0\end{bmatrix}
\end{aligned}
\]

令
\[
A = \begin{bmatrix}
1 & 0 & -1 & 3 & -1 \\
1 & 0 & 0 & 2 & -1 \\
2 & 0 & -1 & 5 & -1 \\
0 & 0 & -1 & 1 & 0
\end{bmatrix}
\]
則 \(T_1 = T_A : \mathbb{R}^5 \to \mathbb{R}^4,\ \mathbf{v} \mapsto A\mathbf{v}\)

因 \(A\) 由 Elementary Row Operation 變成reduce echelon form \(A'\) 為：（其實echelon form就可以）
\[
A' = \begin{bmatrix}
1 & 0 & 0 & 2 & -1 \\
0 & 0 & 1 & -1 & 0 \\
0 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 & 0
\end{bmatrix}
\]
\[
\begin{bmatrix}
1 & 0 & 0 & 2 & -1 \\
0 & 0 & 1 & -1 & 0 \\
0 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 & 0
\end{bmatrix}
\begin{bmatrix} a \\ b \\ c \\ d \\ e \end{bmatrix}
= \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}
\]

由 \(A'\) 可得：
- pivot variables: \(a, c, e\)
- free variables: \(b, d\)

解方程組：
\[
\begin{cases}
a + 2d - e = 0 \\
c - d = 0 \\
e = 0
\end{cases}
\Rightarrow
\begin{cases}
a = -2d \\
c = d \\
e = 0
\end{cases}
\]

- \(b=1,\ d=0\) → \(\mathbf{u}_1 = \begin{bmatrix}0\\1\\0\\0\\0\end{bmatrix}\)
- \(b=0,\ d=1\) → \(\mathbf{u}_2 = \begin{bmatrix}-2\\0\\1\\1\\0\end{bmatrix}\)

\[
\text{basis of } \ker(T_1) = N(T_1) = N(A) = \left\{
\mathbf{u}_1,\ \mathbf{u}_2
\right\}
\]

取原矩陣 \(A\) 對應 \(A'\) 中 pivot 所在的 column：
\[
\text{basis of } \operatorname{im}(T_1) = R(T_1) = \operatorname{Col}(A) = \left\{
\begin{bmatrix}1\\1\\2\\0\end{bmatrix},\ 
\begin{bmatrix}-1\\0\\-1\\-1\end{bmatrix},\ 
\begin{bmatrix}-1\\-1\\-1\\0\end{bmatrix}
\right\}
\]

(2) \(T_2: P_2(\mathbb{R}) \to P_3(\mathbb{R})\)

\[
f(x) \mapsto x^2 f'(x)
\]

\(\ker(T_2) = \{ f(x) \in P_2(\mathbb{R}) \mid T_2(f(x)) = 0 \}\)
\[
x^2 f'(x) = 0
\]
當 \(x \neq 0\) 時，\(f'(x) = 0\)，故 \(f(x)\) 為常數  
常數多項式的空間由 \(\{1\}\) 所擴展：
\[
\text{basis of } \ker(T_2) = \{ \mathbf{1} \}
\]

\(\operatorname{im}(T_2) = \{ T_2(f(x)) \mid f(x) \in P_2(\mathbb{R}) \}\) 
令 \(f(x) = a_0 + a_1 x + a_2 x^2 \in P_2(\mathbb{R})\)：
\[
T_2(f(x)) = x^2 (a_1 + 2a_2 x) = a_1 x^2 + 2a_2 x^3 \in \operatorname{Span}\{\mathbf{x}^2, \mathbf{x}^3\}
\]
\[
\text{basis of } \operatorname{im}(T_2) = \{ \mathbf{x}^2, \mathbf{x}^3 \}
\]