Question

If $A=\left[\begin{array}{cc}0& -\tan \frac{\alpha }{2}\\ \tan \frac{\alpha }{2}& 0\end{array}\right]$ and $I$ the identity matrix of order $2$, show that $I+A=(I-A)\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]$.

Answer 1

$A-\left[\begin{array}{cc}0& -\tan \alpha /2\\ \tan \alpha /2& 0\end{array}\right]$

$I+A=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]+\left[\begin{array}{cc}0& -\tan \alpha /2\\ \tan \alpha /2& 0\end{array}\right]$

$=\left[\begin{array}{cc}1& -\tan \alpha /2\\ \tan \alpha /2& 1\end{array}\right]$

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

$=[\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]-\left[\begin{array}{cc}0& -\tan \alpha /2\\ \tan \alpha /2& 0\end{array}\right]]\times \left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]$

$=\left[\begin{array}{cc}1& -\tan \alpha /2\\ \tan \alpha /2& 1\end{array}\right]\times \left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]$

$=\left[\begin{array}{cc}\cos \alpha +\tan \alpha /2\sin \alpha & -\sin \alpha 2+\tan \alpha /2\cos \alpha \\ \tan \alpha /2\cos \alpha +\sin \alpha & \sin \alpha \tan \alpha /2+\cos \alpha \end{array}\right]$

$=\left[\begin{array}{cc}1-\tan \frac{\alpha }{2}+\frac{\tan \frac{\alpha }{2}}{1+\tan \frac{2\alpha }{2}}\times 2\tan \frac{\alpha }{2}& -2\tan \frac{\alpha }{2}+\frac{\tan \frac{\alpha }{2}}{1+\tan \frac{2\alpha }{2}}-\tan \frac{3\alpha }{2}\\ 2\tan \frac{\alpha }{2}-\frac{\tan \frac{\alpha }{2}}{1+\tan \frac{2\alpha }{2}}\tan \frac{3\alpha }{2}& 1-\tan \frac{\alpha }{2}+\frac{\tan \frac{\alpha }{2}}{1+\tan \frac{2\alpha }{2}}\times 2\tan \frac{\alpha }{2}\end{array}\right]$

$=\left[\begin{array}{cc}1& -\tan \alpha /2\\ \tan \alpha /2& 1\end{array}\right]$

$=I+A$