Finding $B^{225}$ without many computations I have that $B$ is a 4x4 matrix. $B-5I=\begin{pmatrix} -2 & 0 & 2 & 0 \\ 0 & 0 & 0 & 0 \\ 3 & 0 & -3 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$. The question asks to find $B^{225}$ without performing many computations, leaving the answer as a product of 3 matrices (univerted matrices are acceptable).
My thinking is that this problem may have to do with diagonalizing $B$ based on eigenvectors. $\lambda=5$ is clearly an eigenvalue because $det(B-5I)=0$, and I also reasoned that possible eigenvectors could be $\begin{pmatrix} 3 \\ 0 \\ -2 \\ 0 \end{pmatrix}$ and $\begin{pmatrix} -3 \\ 0 \\ 2 \\ 0 \end{pmatrix}$, just based on the way the matrix multiplication would work to yield linear combinations of these eigenvectors. So for diagonalization we need to bases $\alpha=...$ and $\beta=\left\{\begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix},\begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix},\begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix},\begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}\right\}$. I believe these eigenvectors should belong in $\alpha$ but I don't know which other eigenvectors belong or how to find them.
Even beyond that point, I don't quite understand the meaning of this question and how we are supposed to make the logical leap to $B^{225}$. What am I missing?

Let $A=B-5I$. Then $A^2=-5A$. Therefore $A^n=(-5)^{n-1}A$ for $n\ge1$.
Then
$$B^n=(5I+A)^n=5^n I+\sum_{k=1}^n {n\choose k}5^{n-k}A^k
=5^n I+\sum_{k=1}^n (-1)^{k-1}{n\choose k}5^{n-1}A
=5^n I+5^{n-1}A.$$
