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Proof of the Power Formula for Rotation Transformation Matrices 📂Matrix Algebra

Proof of the Power Formula for Rotation Transformation Matrices

Theorem

For every natural number nn, the following holds. [cos⁡θ−sin⁡θsin⁡θcos⁡θ]n=[cos⁡nθ−sin⁡nθsin⁡nθcos⁡nθ] \begin{bmatrix} { \cos \theta }&{ -\sin \theta } \\ { \sin \theta }&{ \cos \theta } \end{bmatrix} ^{n} = \begin{bmatrix} { \cos n\theta }&{ -\sin n\theta } \\ { \sin n\theta }&{ \cos n\theta } \end{bmatrix}

Explanation

A linear transformation matrix that rotates by θ\theta around the origin, when squared, results in a linear transformation that rotates by nθn\theta.

Proof

Strategy: It’s intuitively obvious, and can be easily proven using mathematical induction.

(ㄱ):[cos⁡θ−sin⁡θsin⁡θcos⁡θ]n=[cos⁡nθ−sin⁡nθsin⁡nθcos⁡nθ] (ㄱ) : \begin{bmatrix} { \cos \theta }&{ -\sin \theta } \\ { \sin \theta }&{ \cos \theta } \end{bmatrix}^{ n }= \begin{bmatrix} { \cos n\theta }&{ -\sin n\theta } \\ { \sin n\theta }&{ \cos n\theta } \end{bmatrix}

When n=1n=1, [cos⁡θ−sin⁡θsin⁡θcos⁡θ]=[cos⁡θ−sin⁡θsin⁡θcos⁡θ] \begin{bmatrix} { \cos \theta }&{ -\sin \theta } \\ { \sin \theta }&{ \cos \theta } \end{bmatrix} = \begin{bmatrix} { \cos \theta }&{ -\sin \theta } \\ { \sin \theta }&{ \cos \theta } \end{bmatrix} Therefore, condition (ㄱ) is satisfied. Now, assuming condition (ㄱ) holds when n=kn=k, [cos⁡θ−sin⁡θsin⁡θcos⁡θ]k=[cos⁡kθ−sin⁡kθsin⁡kθcos⁡kθ] \begin{bmatrix} { \cos \theta }&{ -\sin \theta } \\ { \sin \theta }&{ \cos \theta } \end{bmatrix}^{ k }= \begin{bmatrix} { \cos k\theta }&{ -\sin k\theta } \\ { \sin k\theta }&{ \cos k\theta } \end{bmatrix} Multiplying both sides by [cos⁡θ−sin⁡θsin⁡θcos⁡θ]\begin{bmatrix} { \cos \theta }&{ -\sin \theta } \\ { \sin \theta }&{ \cos \theta }\end{bmatrix} results in [cos⁡θ−sin⁡θsin⁡θcos⁡θ]k+1=[cos⁡kθ−sin⁡kθsin⁡kθcos⁡kθ][cos⁡θ−sin⁡θsin⁡θcos⁡θ]=[cos⁡kθcos⁡θ−sin⁡kθsin⁡θ−(sin⁡kθcos⁡θ+cos⁡kθsin⁡θ)sin⁡kθcos⁡θ+cos⁡kθsin⁡θcos⁡kθcos⁡θ−sin⁡kθsin⁡θ]=[cos⁡(k+1)θ−sin⁡(k+1)θsin⁡(k+1)θcos⁡(k+1)θ] \begin{align*} \begin{bmatrix} { \cos \theta }&{ -\sin \theta } \\ { \sin \theta }&{ \cos \theta } \end{bmatrix}^{ k+1 } =& \begin{bmatrix} { \cos k\theta }&{ -\sin k\theta } \\ { \sin k\theta }&{ \cos k\theta } \end{bmatrix} \begin{bmatrix} { \cos \theta }&{ -\sin \theta } \\ { \sin \theta }&{ \cos \theta } \end{bmatrix} \\ =& \begin{bmatrix} { \cos k\theta \cos \theta -\sin k\theta \sin \theta }&{ -(\sin k\theta \cos \theta +\cos k\theta \sin \theta ) } \\ { \sin k\theta \cos \theta +\cos k\theta \sin \theta }&{ \cos k\theta \cos \theta -\sin k\theta \sin \theta } \end{bmatrix} \\ =& \begin{bmatrix} { \cos (k+1)\theta }&{ -\sin (k+1)\theta } \\ { \sin (k+1)\theta }&{ \cos (k+1)\theta } \end{bmatrix} \end{align*} Therefore, condition (ㄱ) is satisfied.

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