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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?

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  • $\begingroup$ 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? $\endgroup$ – Omnomnomnom Nov 20 '18 at 15:44
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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 $(R_{45^\circ})^n = R_{(45n)^\circ}$.

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

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

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  • $\begingroup$ can it be diagonalized? I get the characteristic polynomial: $p(\lambda)=\lambda^2-\sqrt{2} \lambda+1$. $\endgroup$ – vesii Nov 20 '18 at 16:01
  • $\begingroup$ @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 $n$. $\endgroup$ – Paul Sinclair Nov 20 '18 at 21:09
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Idea $$ \\\begin{pmatrix}\frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2}\\ \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \end{pmatrix}=\begin{pmatrix}1 & -1\\ 1 & 1 \end{pmatrix}\cdot\frac{\sqrt{2}}{2} \\\begin{pmatrix}1 & -1\\ 1 & 1 \end{pmatrix}^n\cdot\Big(\frac{1}{\sqrt{2}}\Big)^n=\begin{pmatrix}-i & i\\ 1 & 1 \end{pmatrix}\cdot\begin{pmatrix}(1-i)^n & 0\\ 0 & (1+i)^n \end{pmatrix}\cdot\begin{pmatrix}-i & i\\ 1 & 1 \end{pmatrix}^{-1}\cdot\frac{1}{2^{\frac n 2}} $$

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