91影视

Consider the linear transformation $L\left(A\right)=\frac{1}{2}(A-A^T)$from ${\mathrm{鈩�}}^{n脳m}\mathrm{to}{\mathrm{鈩�}}^{n脳m}$.

Assume $A鈭�{\mathrm{鈩�}}^{n脳n}$such that $L\left(B\right)=A^T$where $A=\left[\begin{array}{cccc}{a}_{11}& a_{12}& ...& a_{1n}\\ a_{21}& a_{22}& ...& a_{2n}\\ 鈰�& 鈰�& 鈰�& 鈰�\\ a_{n1}& a_{n2}& ...& a_{33}\end{array}\right]$and $\left[\begin{array}{cccc}{b}_{11}& b_{12}& ...& b_{1n}\\ b_{21}& b_{22}& ...& b_{2n}\\ 鈰�& 鈰�& 鈰�& 鈰�\\ b_{n1}& b_{n2}& ...& b_{33}\end{array}\right]$.

Substitute the value

$$\begin{gathered} B^T={\left[\begin{array}{cccc}{b}_{11}& b_{12}& ...& b_{1n}\\ b_{21}& b_{22}& ...& b_{2n}\\ 鈰�& 鈰�& 鈰�& 鈰�\\ b_{n1}& b_{n2}& ...& b_{33}\end{array}\right]}^T \\ {B}^T=\left[\begin{array}{cccc}{b}_{11}& b_{12}& ...& b_{1n}\\ b_{21}& b_{22}& ...& b_{2n}\\ 鈰�& 鈰�& 鈰�& 鈰�\\ b_{n1}& b_{n2}& ...& b_{33}\end{array}\right] \end{gathered}$$

Substitute the values $\left[\begin{array}{cccc}{b}_{11}& b_{12}& ...& b_{1n}\\ b_{21}& b_{22}& ...& b_{2n}\\ 鈰�& 鈰�& 鈰�& 鈰�\\ b_{n1}& b_{n2}& ...& b_{33}\end{array}\right]$for $B^T$and $\left[\begin{array}{cccc}{b}_{11}& b_{12}& ...& b_{1n}\\ b_{21}& b_{22}& ...& b_{2n}\\ 鈰�& 鈰�& 鈰�& 鈰�\\ b_{n1}& b_{n2}& ...& b_{33}\end{array}\right]$for B in the equation role="math" localid="1660125174445" $A=\frac{1}{2}(B-B^T)$as follows.

$$\begin{gathered} A=\frac{1}{2}(B-B^T) \\ =\frac{1}{2}\left(\left[\begin{array}{cccc}{b}_{11}& b_{12}& ...& b_{1n}\\ b_{21}& b_{22}& ...& b_{2n}\\ 鈰�& 鈰�& 鈰�& 鈰�\\ b_{n1}& b_{n2}& ...& b_{33}\end{array}\right]+\left[\begin{array}{cccc}{b}_{11}& b_{12}& ...& b_{1n}\\ b_{21}& b_{22}& ...& b_{2n}\\ 鈰�& 鈰�& 鈰�& 鈰�\\ b_{n1}& b_{n2}& ...& b_{33}\end{array}\right]\right) \\ =\left[\begin{array}{cccc}2b_{11}& b_{12}-b_{21}& ...& b_{1n}-b_{1n}\\ b_{21}-b_{12}& 2b_{22}& ...& b_{2n}-b_{n2}\\ 鈰�& 鈰�& 鈰�& 鈰�\\ b_{n1}-b_{1n}& b_{n2}-b_{2n}& ...& 2b_{33}\end{array}\right] \\ \left[\begin{array}{cccc}{b}_{11}& b_{12}& ...& b_{1n}\\ b_{21}& b_{22}& ...& b_{2n}\\ 鈰�& 鈰�& 鈰�& 鈰�\\ b_{n1}& b_{n2}& ...& b_{33}\end{array}\right]=\left[\begin{array}{cccc}{b}_{11}& (b_{12}-b_{21})/2& ...& b_{1n}\\ (b_{21}+b_{12})/2& b_{22}& ...& (b_{2n}-b_{n2})/2\\ 鈰�& 鈰�& 鈰�& 鈰�\\ (b_{n1}-b_{1n})/2& (b_{n2}-b_{2n})/2& ...& b_{33}\end{array}\right] \end{gathered}$$

Compare the equation both side as follows.

$$\begin{gathered} a_{jj}=b_{jj} \\ {a}_{ij}=a_{ji}(b_{ij}-b_{ji})/2 \end{gathered}$$

Therefore, the image of the linear transformation is the set of all skew-symmetric matrix.

Assume $A鈭�ker\left(L\right)$, by the definition L (A) = 0.

Compare the equations $L\left(A\right)=0\mathrm{and}L\left(A\right)=\frac{1}{2}(A-A^T)$and as follows.

$\frac{1}{2}(A-A^T)=0$

Simplify the equation$\frac{1}{2}(A-A^T)=0$as follows.

role="math" localid="1660124838257" $$\begin{gathered} \frac{1}{2}(A-A^T)=0 \\ A-A^T=0 \\ A=A^T \end{gathered}$$

Therefore, the kernel of linear transformation L is set of all $n脳n$symmetric matrix.

Hence, the image of L is skew-symmetric matrices and kernel of L is symmetric matrices.