Exponentiation of a symmetric $2 \times 2$ matrix Why does the following hold?
$$\left[\begin{matrix}a & b\\b & a\end{matrix}\right]^k=\frac{1}{2}\left[\begin{matrix}\left(a - b\right)^{k} + \left(a + b\right)^{k} & - \left(a - b\right)^{k} + \left(a + b\right)^{k}\\- \left(a - b\right)^{k} + \left(a + b\right)^{k} & \left(a - b\right)^{k} + \left(a + b\right)^{k}\end{matrix}\right]$$
 A: Note that 
$$
\pmatrix{a&b\\b&a} = \frac 12 \pmatrix{1&1\\1&-1}\pmatrix{a+b&0\\0&a-b} \pmatrix{1 & 1\\1&-1}
$$
where we observe that 
$$
\pmatrix{1 & 1\\1&-1}^2 = 2I
$$
With that, we compute
$$
\begin{align*}
\pmatrix{a&b\\b&a}^k &= \left[\frac 12 \pmatrix{1&1\\1&-1}\pmatrix{a+b&0\\0&a-b} \pmatrix{1 & 1\\1&-1}\right]^k\\
&= 
\frac 12 \pmatrix{1&1\\1&-1}\left[\pmatrix{a+b&0\\0&a-b}  \underbrace{\left(\frac 12 \pmatrix{1 & 1\\1&-1}\pmatrix{1&1\\1&-1}\right)}_I\right]^{k-1}
\\ & \qquad \cdot \pmatrix{a+b&0\\0&a-b} \pmatrix{1&1\\1&-1} 
\\ & = \frac 12 \pmatrix{1&1\\1&-1}\pmatrix{a+b&0\\0&a-b}^{k}\pmatrix{1&1\\1&-1}
\end{align*}
$$
I'll let you figure out the rest.
A: Guide:
Given a symmetric $2 \times 2$ matrix $A$, one way to commpute $A^k$ is as follows:
$$\begin{bmatrix} a & b \\ b & a \end{bmatrix}\begin{bmatrix} 1 \\ 1 \end{bmatrix} = (a+b) \begin{bmatrix} 1 \\ 1 \end{bmatrix}$$
$$\begin{bmatrix} a & b \\ b & a \end{bmatrix}\begin{bmatrix} 1 \\ -1 \end{bmatrix} = (a-b) \begin{bmatrix} 1 \\ -1 \end{bmatrix}$$
Diagonalize the matrix in the form of $A=UDU^T$ where $U$ is orthogonal and $D$ is diagonal, then $A^k=UD^kU^T$.
A: 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.
A: You can prove it by induction. The induction step would be 
\begin{align}
\begin{bmatrix}a&b\\ b&a\end{bmatrix}^{k+1}
&=\begin{bmatrix}a&b\\ b&a\end{bmatrix}\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]\\ \ \\
&=\left[\begin{matrix}\left(a - b\right)^{k+1} + \left(a + b\right)^{k+1} & - \left(a - b\right)^{k+1} + \left(a + b\right)^{k+1}\\- \left(a - b\right)^{k+1} + \left(a + b\right)^{k+1} & \left(a - b\right)^{k} + \left(a + b\right)^{k+1}\end{matrix}\right].
\end{align}
The computations are straightforward: for example for the 1,1 entry you have 
$$
a(a-b)^k+a(a+b)^k-b(a-b)^k+b(a+b)^k
=(a-b)(a-b)^k+(a+b)(a+b)^k
=(a-b)^{k+1}+(a+b)^{k+1}.
$$
A: The characteristic polynomial of this matrix say $A$ is $\chi_A=(x-a-b)(x-a+b)$ so the Euclidean division of $x^k$ by $\chi_A$ gives
$$x^k= \chi_A Q(x)+\alpha x+\beta$$
We find $\alpha$ and $\beta$ by substituting  $x$ by $a+b$ and $b-a$.
Finaly using Cayley-Hamilton theorem we get $A^k=\alpha A+\beta I$.
A: Starting from Siong Thye Goh’s observation, there are two possibilities. If $a+b=a-b$, then $b=0$, so $A=aI$ and it’s easy to see that $A^k=a^kI$. If $a+b\ne a-b$, then $A$ can be decomposed into the sum $(a+b)P_1+(a-b)P_2$, where $P_1$ and $P_2$ are matrices with the properties that $P_1^2=P_1$, $P_2^2=P_2$ and $P_1P_2=P_2P_1=0$. (For those of you keeping score at home, they are projections onto the eigenspaces of $a+b$ and $a-b$.) It’s fairly straightforward to prove that if there is such a decomposition, then $$A^k=(a+b)^kP_1+(a-b)^kP_2\tag{*}.$$ All of the terms in the expansion that involve both $P_1$ and $P_2$ vanish.  
Let $\lambda_1=a+b$ and $\lambda_2=a-b$. You can verify that the matrices $$P_1 = {1\over\lambda_1-\lambda_2}(A-\lambda_2 I) = \frac12 \begin{bmatrix}1&1\\1&1\end{bmatrix} \\ P_2 = {1\over\lambda_2-\lambda_1}(A-\lambda_1I) = \frac12 \begin{bmatrix}1&-1\\-1&1\end{bmatrix}$$ satisfy the above conditions. It’s easy to see by inspection that with these matrices $P_1$ and $P_2$, the right-hand side of (*) is exactly the required result.
