Question

If A = $\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right]$, find $\alpha$ satisfying $0<\alpha <\frac{\pi }{2}$ when $A+A^T=\sqrt{2}{I}_2$; where $A^T$ is transpose of A.

Answer 1

We have A = $\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right],$

Now, $A^T=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]$

But given that $A+A^T=\sqrt{2}{I}_2$

$\Rightarrow \left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right]+\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]=\sqrt{2}\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

$\Rightarrow \left[\begin{array}{cc}2\cos \alpha & 0\\ 0& 2\cos \alpha \end{array}\right]=\left[\begin{array}{cc}\sqrt{2}& 0\\ 0& \sqrt{2}\end{array}\right]$

By equality of matrices, we have

$2\cos \alpha =\sqrt{2}\Rightarrow \cos \alpha =\frac{\sqrt{2}}{2}$

$\Rightarrow \cos \alpha =\frac{1}{\sqrt{2}}\Rightarrow \cos \alpha =\cos \frac{\pi }{4}$

$\therefore \alpha =\frac{\pi }{4}$ [$\text{as}0<\alpha <\frac{\pi }{2}$]