# Finding a formula of a power of a matrix

Part of a solution I came across of calculating the following matrix:

$$\begin{pmatrix}\frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2}\\ \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \end{pmatrix}^n$$

I'm trying to find a formula for this matrix so I can prove it using induction. I tried to calculate $$M^2,M^3,M^4$$ but I can seem to see the pattern. How should I approach this issue?

• Could you give us a bit more context here? What class is this for? Are you familiar with general linear-algebra techniques such as diagonalization? Are you comfortable using complex numbers? – Omnomnomnom Nov 20 '18 at 15:44

It helps to interpret the matrix geometrically. The matrix of a counterclockwise rotation about the origin by angle $$\theta$$ is given by $$R_\theta = \pmatrix{\cos \theta & -\sin \theta\\ \sin \theta & \cos \theta}$$ Your matrix is simply $$R_{45^\circ}$$. You should find, then, that $$(R_{45^\circ})^n = R_{(45n)^\circ}$$.

• What is the order of $R_{45n}$ (when speaking of groups)? – vesii Nov 20 '18 at 15:47
• $R_{45^\circ}$ has order $8$, so that should tell you everything you need to know – Omnomnomnom Nov 20 '18 at 19:49

HINT

If you diagonalize and write $$A = VDV^{-1}$$ then $$A^n = VD^n V^{-1}$$.

• can it be diagonalized? I get the characteristic polynomial: $p(\lambda)=\lambda^2-\sqrt{2} \lambda+1$. – vesii Nov 20 '18 at 16:01
• @vesii - It can be diagonalized in $\Bbb C$ just fine. You will find that even though $V$ and $D$ are complex matrices, $A^n$ is real for all $n$. – Paul Sinclair Nov 20 '18 at 21:09

Idea $$\\\begin{pmatrix}\frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2}\\ \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \end{pmatrix}=\begin{pmatrix}1 & -1\\ 1 & 1 \end{pmatrix}\cdot\frac{\sqrt{2}}{2} \\\begin{pmatrix}1 & -1\\ 1 & 1 \end{pmatrix}^n\cdot\Big(\frac{1}{\sqrt{2}}\Big)^n=\begin{pmatrix}-i & i\\ 1 & 1 \end{pmatrix}\cdot\begin{pmatrix}(1-i)^n & 0\\ 0 & (1+i)^n \end{pmatrix}\cdot\begin{pmatrix}-i & i\\ 1 & 1 \end{pmatrix}^{-1}\cdot\frac{1}{2^{\frac n 2}}$$