Let $A_{\alpha}$ be the $\alpha$-rotation matrix. Prove $A_{\alpha}^T = (A_{\alpha})^{-1}$

Question

Let $A_{\alpha}$ be the alpha-rotation matrix. Prove $A_{\alpha}^T = (A_{\alpha})^{-1}$

In other words, prove $A_{\alpha}$ transpose = $A_{\alpha}$ inverse.

First of all, what is a rotation-matrix? And what does that imply about the $\alpha$-rotation matrix? What does that make $A_{\alpha}$?

Answer 1

Hint: the rotation matrix is given by:

$$M = \begin{bmatrix} \cos \alpha & -\sin \alpha\\ \sin \alpha&\cos \alpha \end{bmatrix}$$

and its transpose by:

$$M^{T} = \begin{bmatrix} \cos \alpha & \sin \alpha\\ -\sin \alpha&\cos \alpha \end{bmatrix}$$

Now can you show $MM^{T} = I$? (This will imply $M^{T} = M^{-1}$! Also be sure to think about WHY this is the rotation matrix)

Answer 2

A rotation matrix is an element of the class of orthogonal matrices. Since all orthogonal matirices' transpose is equal to their inverses, this applies to all elements of the set, including the rotation matrix.