由定義可得：

\[
T_1(E_{11})=E_{11},\quad
T_1(E_{12})=E_{12},\quad
T_1(E_{21})=E_{21},\quad
T_1(E_{22})=E_{22}.
\]

把這些向量寫成 \(\alpha\)-座標：

\[
[E_{11}]_{\alpha}=
\begin{pmatrix}1\\0\\0\\0\end{pmatrix},\quad
[E_{12}]_{\alpha}=
\begin{pmatrix}0\\0\\1\\0\end{pmatrix},\quad
[E_{21}]_{\alpha}=
\begin{pmatrix}0\\0\\0\\1\end{pmatrix},\quad
[E_{22}]_{\alpha}=
\begin{pmatrix}0\\1\\0\\0\end{pmatrix}.
\]

所以
\[
[T_1]_{\varepsilon}^{\alpha}
=
\begin{pmatrix}
1&0&0&0\\
0&0&0&1\\
0&1&0&0\\
0&0&1&0
\end{pmatrix}.
\]

再來，
\[
T_2(E_{11})=E_{11},\quad
T_2(E_{12})=E_{21},\quad
T_2(E_{21})=E_{12},\quad
T_2(E_{22})=E_{22}.
\]

寫成 \(\alpha\)-座標：

\[
[E_{11}]_{\alpha}=
\begin{pmatrix}1\\0\\0\\0\end{pmatrix},\quad
[E_{21}]_{\alpha}=
\begin{pmatrix}0\\0\\0\\1\end{pmatrix},\quad
[E_{12}]_{\alpha}=
\begin{pmatrix}0\\0\\1\\0\end{pmatrix},\quad
[E_{22}]_{\alpha}=
\begin{pmatrix}0\\1\\0\\0\end{pmatrix}.
\]

因此
\[
[T_2]_{\varepsilon}^{\alpha}
=
\begin{pmatrix}
1&0&0&0\\
0&0&0&1\\
0&0&1&0\\
0&1&0&0
\end{pmatrix}.
\]

\[
\varepsilon=(E_{11},E_{12},E_{21},E_{22}),\qquad
\alpha=(E_{11},E_{22},E_{12},E_{21}).
\]

由於
\[
T_1(A)=A,
\]
故
\[
T_1(E_{11})=E_{11},\quad
T_1(E_{22})=E_{22},\quad
T_1(E_{12})=E_{12},\quad
T_1(E_{21})=E_{21}.
\]

把它們寫成 \(\varepsilon\)-座標：

\[
[E_{11}]_{\varepsilon}=
\begin{pmatrix}
1\\0\\0\\0
\end{pmatrix},\quad
[E_{22}]_{\varepsilon}=
\begin{pmatrix}
0\\0\\0\\1
\end{pmatrix},\quad
[E_{12}]_{\varepsilon}=
\begin{pmatrix}
0\\1\\0\\0
\end{pmatrix},\quad
[E_{21}]_{\varepsilon}=
\begin{pmatrix}
0\\0\\1\\0
\end{pmatrix}.
\]

因此
\[
[T_1]_{\alpha}^{\varepsilon}
=
\begin{pmatrix}
1&0&0&0\\
0&0&1&0\\
0&0&0&1\\
0&1&0&0
\end{pmatrix}.
\]

又因為
\[
T_2(A)=A^t,
\]
故
\[
T_2(E_{11})=E_{11},\quad
T_2(E_{22})=E_{22},\quad
T_2(E_{12})=E_{21},\quad
T_2(E_{21})=E_{12}.
\]

把它們寫成 \(\varepsilon\)-座標：

\[
[E_{11}]_{\varepsilon}=
\begin{pmatrix}
1\\0\\0\\0
\end{pmatrix},\quad
[E_{22}]_{\varepsilon}=
\begin{pmatrix}
0\\0\\0\\1
\end{pmatrix},\quad
[E_{21}]_{\varepsilon}=
\begin{pmatrix}
0\\0\\1\\0
\end{pmatrix},\quad
[E_{12}]_{\varepsilon}=
\begin{pmatrix}
0\\1\\0\\0
\end{pmatrix}.
\]

因此
\[
[T_2]_{\alpha}^{\varepsilon}
=
\begin{pmatrix}
1&0&0&0\\
0&0&0&1\\
0&0&1&0\\
0&1&0&0
\end{pmatrix}.
\]

第二小題可直接利用線性映射的定義，
將 \(\varepsilon=(E_{11},E_{12},E_{21},E_{22})\) 的基底向量逐一代入
\[
T_1+cT_2.
\]

因為
\[
T_1(A)=A,\qquad T_2(A)=A^t,
\]
所以
\[
(T_1+cT_2)(A)=A+cA^t.
\]

現在對 \(\varepsilon\) 的各基底向量計算：

\[
(T_1+cT_2)(E_{11})=E_{11}+cE_{11}=(1+c)E_{11}.
\]
寫成 \(\alpha\)-座標為
\[
[(T_1+cT_2)(E_{11})]_{\alpha}
=
\begin{pmatrix}
1+c\\0\\0\\0
\end{pmatrix}.
\]

\[
(T_1+cT_2)(E_{12})=E_{12}+cE_{21}.
\]
由於
\[
\alpha=(E_{11},E_{22},E_{12},E_{21}),
\]
故其 \(\alpha\)-座標為
\[
[(T_1+cT_2)(E_{12})]_{\alpha}
=
\begin{pmatrix}
0\\0\\1\\c
\end{pmatrix}.
\]

\[
(T_1+cT_2)(E_{21})=E_{21}+cE_{12}.
\]
故其 \(\alpha\)-座標為
\[
[(T_1+cT_2)(E_{21})]_{\alpha}
=
\begin{pmatrix}
0\\0\\c\\1
\end{pmatrix}.
\]

\[
(T_1+cT_2)(E_{22})=E_{22}+cE_{22}=(1+c)E_{22}.
\]
寫成 \(\alpha\)-座標為
\[
[(T_1+cT_2)(E_{22})]_{\alpha}
=
\begin{pmatrix}
0\\1+c\\0\\0
\end{pmatrix}.
\]

因此
\[
[T_1+cT_2]_{\varepsilon}^{\alpha}
=
\begin{pmatrix}
1+c&0&0&0\\
0&0&0&1+c\\
0&1&c&0\\
0&c&1&0
\end{pmatrix}.
\]

另一方面，
\[
[T_1]_{\varepsilon}^{\alpha}
=
\begin{pmatrix}
1&0&0&0\\
0&0&0&1\\
0&1&0&0\\
0&0&1&0
\end{pmatrix},
\qquad
[T_2]_{\varepsilon}^{\alpha}
=
\begin{pmatrix}
1&0&0&0\\
0&0&0&1\\
0&0&1&0\\
0&1&0&0
\end{pmatrix}.
\]

所以
\[
[T_1]_{\varepsilon}^{\alpha}+c[T_2]_{\varepsilon}^{\alpha}
=
\begin{pmatrix}
1&0&0&0\\
0&0&0&1\\
0&1&0&0\\
0&0&1&0
\end{pmatrix}
+
c
\begin{pmatrix}
1&0&0&0\\
0&0&0&1\\
0&0&1&0\\
0&1&0&0
\end{pmatrix}
\]

\[
=
\begin{pmatrix}
1+c&0&0&0\\
0&0&0&1+c\\
0&1&c&0\\
0&c&1&0
\end{pmatrix}.
\]

故
\[
[T_1+cT_2]_{\varepsilon}^{\alpha}
=
[T_1]_{\varepsilon}^{\alpha}+c[T_2]_{\varepsilon}^{\alpha}.
\]

由(2)得
\[
[T_1+cT_2]_{\varepsilon}^{\alpha}
=
\begin{pmatrix}
1+c&0&0&0\\
0&0&0&1+c\\
0&1&c&0\\
0&c&1&0
\end{pmatrix}.
\]

此線性映射為 isomorphism
\[
[T_1+cT_2]_{\varepsilon}^{\alpha} \text{ 可逆}
\iff \det([T_1+cT_2]_{\varepsilon}^{\alpha})\neq 0.
\]

降階計算\(\det([T_1+cT_2]_{\varepsilon}^{\alpha})=(1+c)(1+c)(1+c)(1-c)\neq 0\)相當於
\[
1+c \neq 0,\qquad
1-c \neq 0.
\]

故充要條件為
\[
c\neq 1,\qquad c\neq -1.
\]

也就是
\[
T_1+cT_2 \text{ 為 isomorphism } \iff c\neq \pm 1.
\]

取 c=-2，則
\[
[T_1-2T_2]_{\varepsilon}^{\alpha}
=
\begin{pmatrix}
-1&0&0&0\\
0&0&0&-1\\
0&1&-2&0\\
0&-2&1&0
\end{pmatrix}.
\]

設其反矩陣為
\[
([T_1-2T_2]_{\varepsilon}^{\alpha})^{-1}.
\]

由
\[
T_2^2=T_1
\]
可知
\[
(T_1-2T_2)(aT_1+bT_2)
=
(a-2b)T_1+(b-2a)T_2.
\]

要它等於 \(T_1\)，故需滿足
\[
a-2b=1,\qquad b-2a=0.
\]

由第二式得
\[
b=2a.
\]
代回第一式：
\[
a-4a=1
\Rightarrow -3a=1
\Rightarrow a=-\frac13.
\]
因此
\[
b=-\frac23.
\]

所以
\[
(T_1-2T_2)^{-1}
=
-\frac13 T_1-\frac23 T_2.
\]

因此反矩陣就是此線性映射在
「定義域基底為 \(\alpha\)，值域基底為 \(\varepsilon\)」下的矩陣表示，即
\[
([T_1-2T_2]_{\varepsilon}^{\alpha})^{-1}
=
[(T_1-2T_2)^{-1}]_{\alpha}^{\varepsilon}.
\]

接著算各個 \alpha 基底向量的像：

\[
(T_1-2T_2)^{-1}(E_{11})=-E_{11},
\qquad
(T_1-2T_2)^{-1}(E_{22})=-E_{22},
\]
\[
(T_1-2T_2)^{-1}(E_{12})
=
-\frac13E_{12}-\frac23E_{21},
\]
\[
(T_1-2T_2)^{-1}(E_{21})
=
-\frac23E_{12}-\frac13E_{21}.
\]

故
\[
([T_1-2T_2]_{\varepsilon}^{\alpha})^{-1}
=
\begin{pmatrix}
-1&0&0&0\\
0&0&-\frac13&-\frac23\\
0&0&-\frac23&-\frac13\\
0&-1&0&0
\end{pmatrix}.
\]