Inverse of Plane Rotation Matrix
Theorem
Let $\mathbf R$ be the matrix associated with a rotation of the plane about the origin through an angle of $\alpha$:
- $\mathbf R = \begin{bmatrix}
\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}$
Then its inverse matrix $\mathbf R^{-1}$ is:
- $\mathbf R^{-1} = \begin{bmatrix}
\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix}$
Proof
Let:
- $\mathbf A = \begin{bmatrix}
\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix}$
Consider $\mathbf R \mathbf A$:
| \(\ds \mathbf R \mathbf A\) | \(=\) | \(\ds \begin{bmatrix}
\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix}\) |
||||||||||||
| \(\ds \) | \(=\) | \(\ds \begin{bmatrix}
\map \cos {\alpha} \map \cos {\alpha} + \map \sin {\alpha} \map \sin {\alpha} & \map \cos {\alpha} \map \sin {\alpha} - \map \sin {\alpha} \map \cos {\alpha} \\ \map \sin {\alpha} \map \cos {\alpha} - \map \cos {\alpha} \map \sin {\alpha} & \map \sin {\alpha} \map \sin {\alpha} + \map \cos {\alpha} \map \cos {\alpha} \end{bmatrix}\) |
Definition of Matrix Product (Conventional) | |||||||||||
| \(\ds \) | \(=\) | \(\ds \begin{bmatrix}
\cos^2\alpha + \sin^2\alpha & 0 \\ 0 & \sin^2\alpha + \cos^2\alpha \end{bmatrix}\) |
||||||||||||
| \(\ds \) | \(=\) | \(\ds \begin{bmatrix}
1 & 0 \\ 0 & 1 \end{bmatrix}\) |
Sum of Squares of Sine and Cosine | |||||||||||
| \(\ds \) | \(=\) | \(\ds \mathbf {I_2}\) | Definition of Unit Matrix |
$\Box$
Now consider $\mathbf A \mathbf R$:
| \(\ds \mathbf A \mathbf R\) | \(=\) | \(\ds \begin{bmatrix}
\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix} \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}\) |
||||||||||||
| \(\ds \) | \(=\) | \(\ds \begin{bmatrix}
\map \cos {\alpha} \map \cos {\alpha} + \map \sin {\alpha} \map \sin {\alpha} & -\map \cos {\alpha} \map \sin {\alpha} + \map \sin {\alpha} \map \cos {\alpha} \\ -\map \sin {\alpha} \map \cos {\alpha} + \map \cos {\alpha} \map \sin {\alpha} & \map \sin {\alpha} \map \sin {\alpha} + \map \cos {\alpha} \map \cos {\alpha} \end{bmatrix}\) |
Definition of Matrix Product (Conventional) | |||||||||||
| \(\ds \) | \(=\) | \(\ds \begin{bmatrix}
\cos^2\alpha + \sin^2\alpha & 0 \\ 0 & \sin^2\alpha + \cos^2\alpha \end{bmatrix}\) |
||||||||||||
| \(\ds \) | \(=\) | \(\ds \begin{bmatrix}
1 & 0 \\ 0 & 1 \end{bmatrix}\) |
Sum of Squares of Sine and Cosine | |||||||||||
| \(\ds \) | \(=\) | \(\ds \mathbf {I_2}\) | Definition of Unit Matrix |
$\Box$
Hence, by the definition of the inverse matrix, $\mathbf A$ is the inverse matrix of $\mathbf R$.
$\blacksquare$