例题1:求一个 2 × 2 2\times 2 2×2的线性变换矩阵 T T T,使其可以将单位向量 e 1 = [ 1 0 ] e_1=\begin {bmatrix} 1\\0\end {bmatrix} e1=[10]变换到 e 1 + e 2 e_1+e_2 e1+e2,且将单位向量 e 2 = [ 0 1 ] e_2=\begin {bmatrix} 0\\1\end {bmatrix} e2=[01]变换到 e 1 − e 2 e_1-e_2 e1−e2。
解:
设此变换为:
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T=\begin {bmatrix} t_{11}&t_{12}\\t_{21}&t_{22}\end {bmatrix}
T=[t11t21t12t22]
把单位向量 e 1 = [ 1 0 ] e_1=\begin {bmatrix} 1\\0\end {bmatrix} e1=[10]变换到 e 1 + e 2 e_1+e_2 e1+e2,则有:
T e 1 = [ t 11 t 12 t 21 t 22 ] [ 1 0 ] = [ t 11 t 21 ] = [ 1 1 ] = e 1 + e 2 Te_1=\begin {bmatrix} t_{11}&t_{12}\\t_{21}&t_{22}\end {bmatrix}\begin {bmatrix} 1\\0\end {bmatrix}=\begin {bmatrix} t_{11}\\t_{21}\end {bmatrix}=\begin {bmatrix} 1\\1\end {bmatrix}=e_1+e_2 Te1=[t11t21t12t22][10]=[t11t21]=[11]=e1+e2
即: t 11 = 1 t_{11}=1 t11=1, t 21 = 1 t_{21}=1 t21=1
把单位向量 e 2 = [ 0 1 ] e_2=\begin {bmatrix} 0\\1\end {bmatrix} e2=[01]变换到 e 1 − e 2 e_1-e_2 e1−e2,则有:
T e 2 = [ t 11 t 12 t 21 t 22 ] [ 0 1 ] = [ t 12 t 22 ] = [ 1 − 1 ] = e 1 − e 2 Te_2=\begin {bmatrix} t_{11}&t_{12}\\t_{21}&t_{22}\end {bmatrix}\begin {bmatrix} 0\\1\end {bmatrix}=\begin {bmatrix} t_{12}\\t_{22}\end {bmatrix}=\begin {bmatrix} 1\\-1\end {bmatrix}=e_1-e_2 Te2=[t11t21t12t22][01]=[t12t22]=[1−1]=e1−e2
即: t 12 = 1 t_{12}=1 t12=1, t 22 = − 1 t_{22}=-1 t22=−1
所以此变换 T = [ 1 1 1 − 1 ] T=\begin {bmatrix} 1&1\\1&-1\end {bmatrix} T=[111−1]
例题2:如果一个 3 × 3 3\times 3 3×3的线性变换矩阵 T T T,使其可以将单位向量 e 1 = [ 1 0 0 ] e_1=\begin {bmatrix} 1\\0\\0\end {bmatrix} e1=⎣⎡100⎦⎤变换到 [ 3 − 2 1 ] \begin {bmatrix} 3\\-2\\1\end {bmatrix} ⎣⎡3−21⎦⎤,把单位向量 e 2 = [ 0 1 0 ] e_2=\begin {bmatrix} 0\\1\\0\end {bmatrix} e2=⎣⎡010⎦⎤变换到 [ 6 0 7 ] \begin {bmatrix} 6\\0\\7\end {bmatrix} ⎣⎡607⎦⎤,把单位向量 e 3 = [ 0 0 1 ] e_3=\begin {bmatrix} 0\\0\\1\end {bmatrix} e3=⎣⎡001⎦⎤变换到 [ 5 4 − 1 ] \begin {bmatrix} 5\\4\\-1\end {bmatrix} ⎣⎡54−1⎦⎤,求此变换矩阵 T T T。
解:
设此变换为:
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T=\begin {bmatrix} t_{11}&t_{12}&t_{13}\\t_{21}&t_{22}&t_{23}\\t_{31}&t_{32}&t_{33}\end {bmatrix}
T=⎣⎡t11t21t31t12t22t32t13t23t33⎦⎤
把单位向量 e 1 = [ 1 0 0 ] e_1=\begin {bmatrix} 1\\0\\0\end {bmatrix} e1=⎣⎡100⎦⎤变换到 [ 3 − 2 1 ] \begin {bmatrix} 3\\-2\\1\end {bmatrix} ⎣⎡3−21⎦⎤,即:
T e 1 = [ t 11 t 12 t 13 t 21 t 22 t 23 t 31 t 32 t 33 ] [ 1 0 0 ] = [ t 11 t 21 t 31 ] = [ 3 − 2 1 ] Te_1=\begin {bmatrix} t_{11}&t_{12}&t_{13}\\t_{21}&t_{22}&t_{23}\\t_{31}&t_{32}&t_{33}\end {bmatrix}\begin {bmatrix} 1\\0\\0\end {bmatrix}=\begin {bmatrix} t_{11}\\t_{21}\\t_{31}\end {bmatrix}=\begin {bmatrix} 3\\-2\\1\end {bmatrix} Te1=⎣⎡t11t21t31t12t22t32t13t23t33⎦⎤⎣⎡100⎦⎤=⎣⎡t11t21t31⎦⎤=⎣⎡3−21⎦⎤
把单位向量 e 2 = [ 0 1 0 ] e_2=\begin {bmatrix} 0\\1\\0\end {bmatrix} e2=⎣⎡010⎦⎤变换到 [ 6 0 7 ] \begin {bmatrix} 6\\0\\7\end {bmatrix} ⎣⎡607⎦⎤,即:
T e 2 = [ t 11 t 12 t 13 t 21 t 22 t 23 t 31 t 32 t 33 ] [ 0 1 0 ] = [ t 12 t 22 t 32 ] = [ 6 0 7 ] Te_2=\begin {bmatrix} t_{11}&t_{12}&t_{13}\\t_{21}&t_{22}&t_{23}\\t_{31}&t_{32}&t_{33}\end {bmatrix}\begin {bmatrix} 0\\1\\0\end {bmatrix}=\begin {bmatrix} t_{12}\\t_{22}\\t_{32}\end {bmatrix}=\begin {bmatrix} 6\\0\\7\end {bmatrix} Te2=⎣⎡t11t21t31t12t22t32t13t23t33⎦⎤⎣⎡010⎦⎤=⎣⎡t12t22t32⎦⎤=⎣⎡607⎦⎤
把单位向量 e 3 = [ 0 0 1 ] e_3=\begin {bmatrix} 0\\0\\1\end {bmatrix} e3=⎣⎡001⎦⎤变换到 [ 5 4 − 1 ] \begin {bmatrix} 5\\4\\-1\end {bmatrix} ⎣⎡54−1⎦⎤,即:
T e 3 = [ t 11 t 12 t 13 t 21 t 22 t 23 t 31 t 32 t 33 ] [ 0 0 1 ] = [ t 13 t 23 t 33 ] = [ 5 4 − 1 ] Te_3=\begin {bmatrix} t_{11}&t_{12}&t_{13}\\t_{21}&t_{22}&t_{23}\\t_{31}&t_{32}&t_{33}\end {bmatrix}\begin {bmatrix} 0\\0\\1\end {bmatrix}=\begin {bmatrix} t_{13}\\t_{23}\\t_{33}\end {bmatrix}=\begin {bmatrix} 5\\4\\-1\end {bmatrix} Te3=⎣⎡t11t21t31t12t22t32t13t23t33⎦⎤⎣⎡001⎦⎤=⎣⎡t13t23t33⎦⎤=⎣⎡54−1⎦⎤
所以此变换为:
T = [ 3 6 5 − 2 0 4 1 7 − 1 ] T=\begin {bmatrix} 3&6&5\\-2&0&4\\1&7&-1\end {bmatrix} T=⎣⎡3−2160754−1⎦⎤
例题3:求把二维空间 R 2 R^2 R2关于直线 y = x y=x y=x对称的线性变换矩阵 T T T。
解:
设此变换为:
T = [ t 11 t 12 t 21 t 22 ] T=\begin {bmatrix} t_{11}&t_{12}\\t_{21}&t_{22}\end {bmatrix} T=[t11t21t12t22]
把二维空间 R 2 R^2 R2关于直线 y = x y=x y=x对称,相当于把 e 1 = [ 1 0 ] e_1=\begin {bmatrix} 1\\0\end {bmatrix} e1=[10]变换到 [ 0 1 ] \begin {bmatrix} 0\\1\end {bmatrix} [01];
把 e 2 = [ 0 1 ] e_2=\begin {bmatrix} 0\\1\end {bmatrix} e2=[01]变换到 [ 1 0 ] \begin {bmatrix} 1\\0\end {bmatrix} [10],则有:
T e 1 = [ t 11 t 12 t 21 t 22 ] [ 1 0 ] = [ t 11 t 21 ] = [ 0 1 ] Te_1=\begin {bmatrix} t_{11}&t_{12}\\t_{21}&t_{22}\end {bmatrix}\begin {bmatrix} 1\\0\end {bmatrix}=\begin {bmatrix} t_{11}\\t_{21}\end {bmatrix}=\begin {bmatrix} 0\\1\end {bmatrix} Te1=[t11t21t12t22][10]=[t11t21]=[01]
T e 2 = [ t 11 t 12 t 21 t 22 ] [ 0 1 ] = [ t 12 t 22 ] = [ 1 0 ] Te_2=\begin {bmatrix} t_{11}&t_{12}\\t_{21}&t_{22}\end {bmatrix}\begin {bmatrix} 0\\1\end {bmatrix}=\begin {bmatrix} t_{12}\\t_{22}\end {bmatrix}=\begin {bmatrix} 1\\0\end {bmatrix} Te2=[t11t21t12t22][01]=[t12t22]=[10]
所以此变换 T = [ 0 1 1 0 ] T=\begin {bmatrix} 0&1\\1&0\end {bmatrix} T=[0110]
总结:
求出空间中所有单位向量的线性变化矩阵
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Te_1=T(e_1)
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Te_2=T(e_2), \cdots
Te2=T(e2),⋯,再拼合起来:
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T=[T(e_1) \qquad T(e_2) \qquad \cdots \qquad T(e_n)]
T=[T(e1)T(e2)⋯T(en)],这就是该线性变换的完整矩阵,对
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x=\begin {bmatrix} x_1\\x_2\\ \vdots \\x_n\end {bmatrix}
x=⎣⎢⎢⎢⎡x1x2⋮xn⎦⎥⎥⎥⎤,经过线性变换
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T(x)=[T(e_1) \qquad T(e_2) \qquad \cdots \qquad T(e_n)]\begin {bmatrix} x_1\\x_2\\ \vdots \\x_n\end {bmatrix}
T(x)=[T(e1)T(e2)⋯T(en)]⎣⎢⎢⎢⎡x1x2⋮xn⎦⎥⎥⎥⎤