- Calculus, Volume 2: Multi-Variable Calculus and Linear Algebra with Applications to Differential Equations and Probability
- Tom M. Apostol
- Second Edition
- 1991
- 978-1-119-49676-2
2.16 Exercises
7. $\quad$ Let $A = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}.$ Verify that $A^2 = \begin{bmatrix} \cos 2\theta & -\sin 2\theta \\ \sin 2\theta & \cos 2\theta \end{bmatrix}$ and compute $A^n.$
Solution. $\quad$ We first recall from Volume 1, Section 2.5, that for all real $x$ and $y,$ we have the addition formulas: \begin{align*} \cos (x + y) &= \cos x \cos y - \sin x \sin y \\ \sin (x + y) &= \sin x \cos y + \cos x \sin y \end{align*} In addition, we recall that $\cos 0 = 1,$ and $\sin 0 = -\sin 0 = 0,$ which means we can express the identity matrix in terms of the sine and cosine: \begin{align*} I &= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} \cos 0 & -\sin 0 \\ \sin 0 & \cos 0 \end{bmatrix} \end{align*}
$\quad$ To verify that $A^2 = \begin{bmatrix} \cos 2\theta & -\sin 2\theta \\ \sin 2\theta & \cos 2\theta \end{bmatrix},$ we use the integral powers of square matrices to give $A^2 = AA,$ or: \begin{align*} A^2 &= \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix} \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix} \\ &= \begin{bmatrix} \cos \theta \cos \theta - \sin \theta \sin \theta & - \sin \theta \cos \theta - \cos \theta \sin \theta \\ \sin\theta \cos\theta + \cos \theta \sin \theta & \cos \theta \cos \theta -\sin \theta \sin \theta \end{bmatrix} \\ &= \begin{bmatrix} \cos 2\theta & -\sin 2\theta \\ \sin 2\theta & \cos 2\theta \end{bmatrix} \end{align*} As such, we have proven for $n = 0, 1, 2$ that $A^n = \begin{bmatrix} \cos n\theta & -\sin n\theta \\ \sin n\theta & \cos n\theta \end{bmatrix}.$ To show that this is the case for any integer $n \geq 0,$ we will do an induction for $n \geq 2.$ Assume this equation is true for some $k \geq 2,$ we then have: \begin{align*} A^{k+1} &= AA^k \\ &= \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} \\ &= \begin{bmatrix} \cos \theta \cos k\theta - \sin \theta \sin k\theta & - \cos \theta \sin k\theta - \sin \theta \cos k\theta \\ \sin\theta \cos k\theta + \cos \theta \sin k\theta & \cos \theta \cos k\theta -\sin \theta \sin k\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*} By induction, because this is true for $k + 1,$ it is true for any $n \geq 0. \quad \blacksquare$