Exponentiation of a symmetric $2 \times 2$ matrix

Question

Why does the following hold?

$$\left[\begin{matrix}a & b\\b & a\end{matrix}\right]^k=\frac{1}{2}\left[\begin{matrix}\left(a - b\right)^{k} + \left(a + b\right)^{k} & - \left(a - b\right)^{k} + \left(a + b\right)^{k}\\- \left(a - b\right)^{k} + \left(a + b\right)^{k} & \left(a - b\right)^{k} + \left(a + b\right)^{k}\end{matrix}\right]$$

Answer 1

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.

Answer 2

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 $A^k=UD^kU^T$.

Answer 3

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 $$ a(a-b)^k+a(a+b)^k-b(a-b)^k+b(a+b)^k =(a-b)(a-b)^k+(a+b)(a+b)^k =(a-b)^{k+1}+(a+b)^{k+1}. $$

Answer 4

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 $A^k=\alpha A+\beta I$.

Answer 5

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.

Answer 6

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.