How to find a power of a matrix?

Question

Let $$ B = \begin{bmatrix}1&1&0&0\\1&1&0&0\\0&0&0&-1\\0&0&-1&0\end{bmatrix} $$ How to find the following matrix $B^{2023}$?

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I have tried to diagonalize it, and have found the eigenvalues and eigenvectors, but as I get a product of three $4 \times 4$ matrices, I thought there might be an easier solution. Also how can I prove that $C = A B A^{T}$ is symmetric, where $A$ is a $4 \times 4$ orthogonal matrix (i.e., $A A^T = 1$)?

Answer 1

Let

$$B=\left[\begin{matrix}X&Y\\Z & T\end{matrix}\right]$$

where

$$X=\left[\begin{matrix}1&1\\1 & 1\end{matrix}\right], \qquad Y=Z=\left[\begin{matrix}0&0\\0 & 0\end{matrix}\right], \qquad T= \left[\begin{matrix}0&-1\\-1 & 0\end{matrix}\right]$$

Since $T^2=I$, we have $T^{2023}=T$ and since $X^2=2X$ we have $X^{2023}=2^{2022}X$. Now, by block matrix multiplication,

$$B^{2023}=\left[\begin{matrix}X^{2023}&O\\O & T^{2023}\end{matrix}\right]=\left[\begin{matrix}2^{2022}&2^{2022}&0&0\\2^{2022}&2^{2022}&0&0\\0&0&0&-1\\0&0&-1&0\end{matrix}\right].$$

Answer 2

In answer to your second question, which is how to show that $C$ is symmetric, note that \begin{alignat*}{2} C^T &= \left(ABA^T\right)^T \\ &= AB^TA^T && \text{since } (ABC)^T = C^TB^TA^T\\ &= A BA^T && \text{since $B$ is symmetric} \\ &= C. \end{alignat*}