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?
 A: I found $$B^2=5B$$
Therefore $$B^3=5B^2=25B$$
Continue and you get $$B^{225} = 5^{224}B$$
A: 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
$$B^{225} = 5^{224}B = 5^{224}\begin{bmatrix}
3 & 0 & 2 & 0\\
0 & 5 & 0 & 0\\
3 & 0 & 2 & 0\\
0 & 0 & 0 & 5\end{bmatrix}$$
A: There may be better answers but what comes to my mind immediately is the following:  you notice that last row and last column are zero, so this will be true in all powers of $B-5I$ also. Similarly, the second column and second row are zero so you only have to consider the $2\times 2$ matrix formed by first and third rows/columns. It is very easy to write down any power of this $2 \times 2$ matrix. Now you can compute powers of $B$ by Binomial theorem. 
A: Hint. When $B=uv^T$ is a rank-1 matrix, as $v^Tu$ is a scalar,
$$
B^n
=\underbrace{(uv^T)(uv^T)\cdots(uv^T)}_{n \text{ copies}}
=u\underbrace{(v^Tu)\cdots(v^Tu)}_{n-1 \text{ copies}}v^T
=(v^Tu)^{n-1}uv^T=(v^Tu)^{n-1}B.
$$
A: Let $A=B-5I$. Then $A^2=-5A$. Therefore $A^n=(-5)^{n-1}A$ for $n\ge1$.
Then
$$B^n=(5I+A)^n=5^n I+\sum_{k=1}^n {n\choose k}5^{n-k}A^k
=5^n I+\sum_{k=1}^n (-1)^{k-1}{n\choose k}5^{n-1}A
=5^n I+5^{n-1}A.$$
A: One thing to remember is that $B - 5I$ has the same eigenvectors as $B$, and the corresponding eigenvalues are $5$ less than those of $B$'s. Put simply and clearly,
$$Bv = \lambda v \iff (B - 5I)v = (\lambda - 5)v.$$
If you can diagonalise $B - 5I$ (by the usual methods), then you can find a matrix $P$ (consisting of eigenvectors for $B - 5I$ and hence for $B$ too) such that the matrix
$$P^{-1}(B - 5I) P = P^{-1}BP - 5P^{-1}IP = P^{-1}BP - 5I$$
is diagonal. Note that therefore
$$D:= P^{-1} B P = (P^{-1} B P - 5I) + 5I$$
is the diagonalisation of $B$. You can raise $D$ to any power easily, since $D$ is diagonal. Then,
$$D^{225} = P^{-1} B^{225} P \implies B^{255} = PD^{255}P^{-1}.$$
So, if you diagonalise $B - 5I$, you're half way there!
