Question

If\[A = \begin{bmatrix}\cos \alpha & \sin \alpha \\ - \sin \alpha & \cos \alpha\end{bmatrix}\] , then verify that A^T A = I₂.

Answer 1

\[Given: A = \begin{bmatrix}\cos \alpha & \sin \alpha \\ - \sin \alpha & \cos \alpha\end{bmatrix} \]

\[ A^T = \begin{bmatrix}\cos \alpha & - \sin \alpha \\ \sin \alpha & \cos \alpha\end{bmatrix}\]

\[Now, \]

\[ A^T A = I_2 \]

\[Consider: LHS = A^T A\]

\[ = \begin{bmatrix}\cos \alpha & - \sin \alpha \\ \sin \alpha & \cos \alpha\end{bmatrix}\begin{bmatrix}\cos \alpha & \sin \alpha \\ - \sin \alpha & \cos \alpha\end{bmatrix}\]

\[ = \begin{bmatrix}\cos^2 \alpha + \sin^2 \alpha & \cos \alpha \sin \alpha - \sin \alpha \cos \alpha \\ \sin \alpha \cos \alpha - \cos \alpha \sin \alpha & \sin^2 \alpha + \cos^2 \alpha\end{bmatrix}\]

\[ = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix} = RHS\]

Hence proved.