Question

If matrix A represents a rotation through $\frac{\pi }{2}$and matrix B a rotation through$\frac{\pi }{4}$ , then A is similar to B.

Short answer

The above statement is false.

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

Step-by-step solution

Definition of similar matrix

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

$A=P^{-1}BP$

Matrix form of rotation by θradian

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

Finding the matrix A and B for θ=π/2 and θ=π/4

For$\theta =\frac{\pi }{2},$ we have

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

For $\theta =\frac{\pi }{4},$we have

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

To find an invertible matrix P

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

$A=P^{-1}BP$

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

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

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

$\Rightarrow a=0,b=0,c=0,d=0$

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.

Final Answer

Since there doesn’t 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 $\frac{\pi }{2}$and matrix B a rotation through $\frac{\pi }{4}$, then A is not similar to B.