<s>[INST] 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?
 [/INST] Let $C = B - 5I$. Notice it can be written as a outer product of two column vectors (or matrix product between a column and a row vector):
$$C = B - 5 I  = u \otimes v = u v^T\quad\text{ where }\quad u = \begin{bmatrix}-2\\ 0 \\ 3 \\ 0\end{bmatrix}\quad\text{ and }\quad v = \begin{bmatrix}1 \\ 0 \\ -1  \\0\end{bmatrix}$$
This leads to
$$C^2 = (u v^T)(u v^T) = u(v^Tu) v^T = u((-2)(1) + (3)(-1))v^T = -5uv^T = -5C$$
As a result,
$$B^2 = (5I+C)^2 = 25I +10C + C^2 = 25I + 5C = 5B$$
Start from this, we can use induction to prove $B^n = 5^{n-1}B$ for all $n > 1$.
In particular, we have
$$B^{225} = 5^{224}B = 5^{224}\begin{bmatrix}
3 & 0 & 2 & 0\\
0 & 5 & 0 & 0\\
3 & 0 & 2 & 0\\
0 & 0 & 0 & 5\end{bmatrix}$$
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