91影视

Let A and B are two square matrices, then the matrix A is said to be similar to the matrix B if there exists an invertible matrix P such that

$A=P^{-1}BP$

The matrix of the linear transformation $T:{\mathrm{鈩�}}^2鈫�{\mathrm{鈩�}}^2$which rotates points in the plane by $胃$radian is$\left[\begin{array}{cc}cos鈦�胃& -sin鈦�胃\\ sin鈦�胃& cos鈦�胃\end{array}\right]$

For$胃=\frac{蟺}{2},$ we have

$A=\left[\begin{array}{cc}cos鈦�\frac{蟺}{2}& -sin鈦�\frac{蟺}{2}\\ sin鈦�\frac{蟺}{2}& cos鈦�\frac{蟺}{2}\end{array}\right]=\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]$

For $胃=\frac{蟺}{4},$we have

$B=\left[\begin{array}{cc}cos鈦�\frac{蟺}{4}& -sin鈦�\frac{蟺}{4}\\ sin鈦�\frac{蟺}{4}& cos鈦�\frac{蟺}{4}\end{array}\right]=\left[\begin{array}{cc}\frac{1}{\sqrt{2}}& -\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{array}\right]$

Let there exist an invertible matrix$P=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ such that

$A=P^{-1}BP$

$鈬�\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\left[\begin{array}{cc}\frac{1}{\sqrt{2}}& -\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{array}\right]$

$鈬�\left[\begin{array}{cc}-c& -d\\ a& b\end{array}\right]=\left[\begin{array}{cc}\frac{a}{\sqrt{2}}+\frac{b}{\sqrt{2}}& \frac{-a}{\sqrt{2}}+\frac{b}{\sqrt{2}}\\ \frac{c}{\sqrt{2}}+\frac{d}{\sqrt{2}}& \frac{-c}{\sqrt{2}}+\frac{d}{\sqrt{2}}\end{array}\right]$

$$\begin{gathered} 鈬�AP=PB \\ 鈬�-c=\frac{a}{\sqrt{2}}+\frac{b}{\sqrt{2}} \\ -d=\frac{-a}{\sqrt{2}}+\frac{b}{\sqrt{2}} \\ a=\frac{c}{\sqrt{2}}+\frac{d}{\sqrt{2}} \\ b=\frac{-c}{\sqrt{2}}+\frac{d}{\sqrt{2}} \end{gathered}$$

$鈬�a=0,b=0,c=0,d=0$

Thus, we have$P=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$ , which is not an invertible matrix.

Since there doesn鈥檛 exist any invertible matrix P such that$A=P^{-1}BP$ , therefore, matrix A is not similar to matrix B.

Hence, if matrix A represents a rotation through $\frac{蟺}{2}$and matrix B a rotation through $\frac{蟺}{4}$, then A is not similar to B.