<s>[INST] Exponentiation of a symmetric $2 \times 2$ matrix Why does the following hold?
$$\left[\begin{matrix}a & b\\b & a\end{matrix}\right]^k=\frac{1}{2}\left[\begin{matrix}\left(a - b\right)^{k} + \left(a + b\right)^{k} & - \left(a - b\right)^{k} + \left(a + b\right)^{k}\\- \left(a - b\right)^{k} + \left(a + b\right)^{k} & \left(a - b\right)^{k} + \left(a + b\right)^{k}\end{matrix}\right]$$
 [/INST] You can use the split-complex numbers. What you've got on the LHS is the matrix representation of $a+bj, j^2=+1$. So:
$$
\begin{align}
(a+bj)^k &= \sum_{r=0}^k{k \choose r} a^{k-r}b^rj^r\\
&= \sum_{r=0}^k {n \choose r}a^{k-r}b^r[\text{$r$ even}] + j\sum_{r=0}^k {n \choose r}a^{k-r}b^r[\text{$r$ odd}]\\
&= \sum_{r=0}^k {n \choose r}a^{k-r}b^r\frac{1+(-1)^r}{2} + j\sum_{r=0}^k {n \choose r}a^{k-r}b^r\frac{1-(-1)^r}{2}\\
&= \frac{\sum_{r=0}^k {n \choose r}a^{k-r}b^r+\sum_{r=0}^k {n \choose r}a^{k-r}b^r(-1)^r}{2} + j\frac{\sum_{r=0}^k {n \choose r}a^{k-r}b^r-\sum_{r=0}^k {n \choose r}a^{k-r}b^r(-1)^r}{2}\\
&= \frac{(a+b)^k+(a-b)^k}{2} + j\frac{(a+b)^k-(a-b)^k}{2}
\end{align}$$
On the first line we use the binomial theorem. 
On the second line we decompose into real and imaginary parts using the fact that $j^r=1$ when $r$ is even, and $j^r=j$ when $r$ is odd. We use Iverson Bracket notation.
On the third line we use the fact that $[\text{$r$ even}]=\frac{1+(-1)^r}{2}$ and $[\text{$r$ odd}]=\frac{1-(-1)^r}{2}$.
On the fourth line we distribute the summations. 
On the fifth line we use the binomial theorem.
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