## 線性映射與座標變換

$A=\left[\!\!\begin{array}{cr} 0&-1\\ 1&0 \end{array}\!\!\right],~~\mathbf{x}=\begin{bmatrix} 2\\ 4 \end{bmatrix}$

\begin{aligned} T(\mathbf{x}+\mathbf{z})&=T(\mathbf{x})+T(\mathbf{z})\\ T(c\mathbf{x})&=cT(\mathbf{x})\end{aligned}

$\mathbf{x}\xrightarrow[]{~A~}\mathbf{y}$

$\begin{bmatrix} 2\\ 4 \end{bmatrix}\xrightarrow[]{~A~}\left[\!\!\begin{array}{r} -4\\ 2 \end{array}\!\!\right]$

$X\xrightarrow[]{~A~}Y$

\begin{aligned} Y=AX&=\left[\!\!\begin{array}{cr} 0&-1\\ 1&0 \end{array}\!\!\right]\begin{bmatrix} 2&3&3&1&1\\ 4&3&1&1&3 \end{bmatrix}\\ &=\left[\!\!\begin{array}{rrrrr} -4&-3&-1&-1&-3\\ 2&3&3&1&1 \end{array}\!\!\right]\end{aligned}

\begin{aligned} \mathbf{x}&\xrightarrow[]{~B~}B\mathbf{x}\xrightarrow[]{~A~}A(B\mathbf{x})\end{aligned}

$\mathbf{x}\xrightarrow[]{~AB~}(AB)\mathbf{x}$

\begin{aligned} AB&=\left[\!\!\begin{array}{cr} 0&-1\\ 1&0 \end{array}\!\!\right]\left[\!\!\begin{array}{rr} -1&0\\ 0&-1 \end{array}\!\!\right]=\left[\!\!\begin{array}{rc} 0&1\\ -1&0 \end{array}\!\!\right]\end{aligned}

\begin{aligned} \mathbf{y}&=A\mathbf{x}=2\begin{bmatrix} 0\\ 1 \end{bmatrix}+4\left[\!\!\begin{array}{r} -1\\ 0 \end{array}\!\!\right]\end{aligned}

$\mathbf{v}_1=\begin{bmatrix} 0\\ 1 \end{bmatrix},~\mathbf{v}_2=\left[\!\!\begin{array}{r} -1\\ 0 \end{array}\!\!\right]$

$\mathbf{y}=A[\mathbf{y}]_{\mathfrak{B}}$

$A$ 即為有序基底構成的座標變換矩陣，$A=\begin{bmatrix}\mathbf{v}_1&\mathbf{v}_2\end{bmatrix}$。下圖說明了平面上的向量如何表示為基底向量的線性組合。

$?\xrightarrow[]{~A~}\mathbf{y}$

$A[?]_{\mathfrak{B}}=\mathbf{y}$

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