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If A={\begin{bmatrix}\cos \theta &\sin \theta \\-\sin \theta &\cos \theta \end{bmatrix}}\, for all integral values of n,show that A^{n}={\begin{bmatrix}\cos n\theta &\sin n\theta \\-\sin n\theta &\cos n\theta \end{bmatrix}}\,

Given A={\begin{bmatrix}\cos \theta &\sin \theta \\-\sin \theta &\cos \theta \end{bmatrix}}\,

Statement P(n)=A^{n}={\begin{bmatrix}\cos n\theta &\sin n\theta \\-\sin n\theta &\cos n\theta \end{bmatrix}}\,

Let n=1 LHS={\begin{bmatrix}\cos \theta &\sin \theta \\-\sin \theta &\cos \theta \end{bmatrix}}\, thus P(n)\, is true for n=1.

Let P(n) be true for n=k

Therefore A^{k}={\begin{bmatrix}\cos k\theta &\sin k\theta \\-\sin k\theta &\cos k\theta \end{bmatrix}}\,

A^{k}\cdot A={\begin{bmatrix}\cos \theta &\sin \theta \\-\sin \theta &\cos \theta \end{bmatrix}}{\begin{bmatrix}\cos k\theta &\sin k\theta \\-\sin k\theta &\cos k\theta \end{bmatrix}}\,

A^{{k+1}}={\begin{bmatrix}\cos(k+1)\theta &\sin(k+1)\theta \\-\sin(k+1)\theta &\cos(k+1)\theta \end{bmatrix}}\,

Thus P(n) is true for n=k+1.

Therefore,by the principle of finite induction,P(n) is true for all the positive integral values of n.

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