林宥呈
29 March 2026
討論區

第二題
\[
\text{(2)}\quad
\operatorname{Tr}(1)
=[ \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}]_r
= \begin{bmatrix} -1 \\ 1 \\ 0 \\ 0 \end{bmatrix},
\]
\[
\operatorname{Tr}(x)
= [\begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix}]_r
= \begin{bmatrix} 1 \\ -1 \\ 1 \\ 0 \end{bmatrix},
\]
\[
\operatorname{Tr}(1 + x^2)
= [\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}]_r
= \begin{bmatrix} 0 \\ 1 \\ -1 \\ 1 \end{bmatrix},
\]
\[
[T]^{\beta}_r
=
\begin{bmatrix}
-1 & 1 & 0 \\
1 & -1 & 1 \\
0 & 1 & -1 \\
0 & 0 & 1
\end{bmatrix}.
\]
參考解答 (1)
\(\begin{align*}
\left[T(1)\right]_{\varepsilon^\prime}&=\left[\begin{bmatrix} 0+0 & 1\\ 0 & 0\end{bmatrix}\right]_{\varepsilon^\prime} \\
&=\begin{bmatrix} 0\\1\\0\\0\end{bmatrix}
\end{align*}\)
\(\begin{align*}
\left[T(x)\right]_{\varepsilon^\prime}&=\left[\begin{bmatrix} 1+0 & 1\\ 1 & 0\end{bmatrix}\right]_{\varepsilon^\prime} \\
&=\begin{bmatrix} 1\\0\\1\\0\end{bmatrix}
\end{align*}\)
\(\begin{align*}
\left[T(x^2)\right]_{\varepsilon^\prime}&=\left[\begin{bmatrix} 0+0 & 0\\ 0 & 1\end{bmatrix}\right]_{\varepsilon^\prime} \\
&=\begin{bmatrix} 1\\0\\0\\1\end{bmatrix}
\end{align*}\)
\(\begin{align*}
\Rightarrow [T]_{\varepsilon}^{\varepsilon ^\prime}&=\begin{bmatrix}
| & | & |\\
\left[T(1)\right]_{\varepsilon^\prime} & \left[T(x)\right]_{\varepsilon^\prime} & \left[T(x^2)\right]_{\varepsilon^\prime}\\
| & | & |
\end{bmatrix}\\
&=\begin{bmatrix}0&1&1\\1&0&0\\0&1&0\\0&0&1\end{bmatrix}
\end{align*}\)