## 答林容麟──關於座標變換矩陣的快捷算法

$\mathfrak{B}_1=\left\{\begin{bmatrix} 1\\ 1\\ 1 \end{bmatrix},\begin{bmatrix} 1\\ 0\\ 1 \end{bmatrix}\right\},~~\mathfrak{B}_2=\left\{\begin{bmatrix} 5\\ 4\\ 5 \end{bmatrix},\begin{bmatrix} -5\\ -6\\ -5 \end{bmatrix}\right\}$

\begin{aligned} \mathbf{x}&=c_1\mathbf{v}_1+\cdots+c_n\mathbf{v}_n,\\ \mathbf{x}&=d_1\mathbf{w}_1+\cdots+d_n\mathbf{w}_n.\end{aligned}

\begin{aligned} \begin{bmatrix} \mathbf{x} \end{bmatrix}_{\mathfrak{B}_2}&=\begin{bmatrix} c_1\mathbf{v}_1+\cdots+c_n\mathbf{v}_n \end{bmatrix}_{\mathfrak{B}_2}\\ &=c_1[\mathbf{v}_1]_{\mathfrak{B}_2}+\cdots+c_n[\mathbf{v}_n]_{\mathfrak{B}_2}\\ &=\begin{bmatrix} ~&~&~\\ [\mathbf{v}_1]_{\mathfrak{B}_2}&\cdots&[\mathbf{v}_n]_{\mathfrak{B}_2}\\ ~&~&~ \end{bmatrix}\begin{bmatrix} c_1\\ \vdots\\ c_n \end{bmatrix}\\ &=C_{\mathfrak{B}_1\rightarrow\mathfrak{B}_2}\cdot\begin{bmatrix} \mathbf{x} \end{bmatrix}_{\mathfrak{B}_1},\end{aligned}

$C_{\mathfrak{B}_1\rightarrow\mathfrak{B}_2}=\begin{bmatrix} ~&~&~\\ [\mathbf{v}_1]_{\mathfrak{B}_2}&\cdots&[\mathbf{v}_n]_{\mathfrak{B}_2}\\ ~&~&~ \end{bmatrix}$

$C_{\mathfrak{B}_2\rightarrow\mathfrak{B}_1}=\begin{bmatrix} ~&~&~\\ [\mathbf{w}_1]_{\mathfrak{B}_1}&\cdots&[\mathbf{w}_n]_{\mathfrak{B}_1}\\ ~&~&~ \end{bmatrix}$

$\mathbf{v}_1=\begin{bmatrix} 1\\ 1\\ 1 \end{bmatrix},~\mathbf{v}_2=\begin{bmatrix} 1\\ 0\\ 1 \end{bmatrix},~\mathbf{w}_1=\begin{bmatrix} 5\\ 4\\ 5 \end{bmatrix},~\mathbf{w}_2=\begin{bmatrix} -5\\ -6\\ -5 \end{bmatrix}$

$C_{\mathfrak{B}_1\to\mathfrak{B}_2}=\begin{bmatrix} c_{11}&c_{12}\\ c_{21}&c_{22} \end{bmatrix}$

$\begin{bmatrix} 5&-5\\ 4&-6\\ 5&-5 \end{bmatrix}\begin{bmatrix} c_{11}\\ c_{21} \end{bmatrix}=\begin{bmatrix} 1\\ 1\\ 1 \end{bmatrix},~\begin{bmatrix} 5&-5\\ 4&-6\\ 5&-5 \end{bmatrix}\begin{bmatrix} c_{12}\\ c_{22} \end{bmatrix}=\begin{bmatrix} 1\\ 0\\ 1 \end{bmatrix}$

$\begin{bmatrix} 5&-5\\ 4&-6\\ 5&-5 \end{bmatrix}\begin{bmatrix} c_{11}&c_{12}\\ c_{21}&c_{22} \end{bmatrix}=\begin{bmatrix} 1&1\\ 1&0\\ 1&1 \end{bmatrix}$

\begin{aligned}\begin{bmatrix} 5&-5&\vline&1&1\\ 4&-6&\vline&1&0\\ 5&-5&\vline&1&1 \end{bmatrix}&\to\left[\!\!\begin{array}{crccr} 5&-5&\vline&1&1\\ 0&-10&\vline&1&-4\\ 0&0&\vline&0&0 \end{array}\!\!\right]\to\left[\!\!\begin{array}{crcrc} 1&-1&\vline&\frac{1}{5}&\frac{1}{5}\\[0.3em] 0&1&\vline&-\frac{1}{10}&\frac{2}{5}\\ 0&0&\vline&0&0 \end{array}\!\!\right]\\ &\to\left[\!\!\begin{array}{cccrc} 1&0&\vline&\frac{1}{10}&\frac{3}{5}\\[0.3em] 0&1&\vline&-\frac{1}{10}&\frac{2}{5}\\ 0&0&\vline&0&0 \end{array}\!\!\right]\end{aligned}

$C_{\mathfrak{B}_1\to\mathfrak{B}_2}=\left[\!\!\begin{array}{rc} \frac{1}{10}&\frac{3}{5}\\[0.3em] -\frac{1}{10}&\frac{2}{5} \end{array}\!\!\right]$