小结
- 线性变换的矩阵
- R2\mathbb{R}^{2}R2中的集合线性变换
- 满射与单射
线性变换的矩阵
I2=[1001]\boldsymbol{I_2}=\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}I2=[1001]的两列是e1=[10]\boldsymbol{e_1}=\begin{bmatrix} 1 \\ 0 \end{bmatrix}e1=[10]和e2=[01]\boldsymbol{e_2}=\begin{bmatrix} 0 \\ 1\end{bmatrix}e2=[01],设T\boldsymbol{T}T是R2\mathbb{R}^{2}R2到R3\mathbb{R}^{3}R3的线性变换,满足T(e1)=[5−72],T(e2)=[−380]\boldsymbol{T(e_1)}=\begin{bmatrix}5 \\ -7 \\ 2\end{bmatrix},\boldsymbol{T(e_2)}=\begin{bmatrix}-3 \\ 8 \\ 0\end{bmatrix}T(e1)=⎣⎡5−72⎦⎤,T(e2)=⎣⎡−380⎦⎤。求出R2\mathbb{R}^{2}R2中任意向量x\boldsymbol{x}x的像的公式。
解:
x=[x1x2]=x1[10]+x2[01]=x1e1+x2e2\boldsymbol{x}=\begin{bmatrix}x_1 \\ x_2\end{bmatrix}=x_1\begin{bmatrix}1 \\ 0\end{bmatrix} + x_2\begin{bmatrix}0 \\ 1\end{bmatrix}=x_1\boldsymbol{e_1} + x_2\boldsymbol{e_2}x=[x1x2]=x1[10]+x2[01]=x1e1+x2e2
因为T\boldsymbol{T}T是线性变换,所以:
T(x)=T(x1e1+x2e2)=x1T(e1)+x2T(e2)=x1[5−72]+x2[−380]=[5−3−7820]x\quad\boldsymbol{T(x)} = \boldsymbol{T(}x_1\boldsymbol{e_1} + x_2\boldsymbol{e_2)} \\ = x_1\boldsymbol{T(e_1)} + x_2\boldsymbol{T(e_2)} \\ = x_1\begin{bmatrix}5 \\ -7 \\ 2\end{bmatrix} + x_2\begin{bmatrix}-3 \\ 8 \\ 0\end{bmatrix} \\ = \begin{bmatrix} 5 & -3\\ -7 & 8 \\ 2 & 0\end{bmatrix}\boldsymbol{x}T(x)=T(x1e1+x2e2)=x1T(e1)+x2T(e2