Finding a formula for a matrix

Question

$$B:=\begin{bmatrix} -1&9&0&0&0\\ 0&-1&0&0&0\\ 0&3&-1&0&0\\ 0&1&1&1&0\\ 0&-2&-2&0&1\end{bmatrix}$$

Is there any way to compute a formula for $B^n$ for all integers $n\geq5$? I ran across this and was curious.

Answer 1

Not a full answer but this wouldn't format correctly when I put in a comment.

$$B^n=\begin{bmatrix} (-1)^n&-9(-1)^n&0&0&0\\ 0&(-1)^n&0&0&0\\ 0&-3(-1)^n&(-1)^n&0&0\\ 0&\frac{1}{4}\left(5-5(-1)^n+6(-1)^nn\right)&\frac{1}{2}\left(1-(-1)^n\right)&1&0\\ 0&-\frac{1}{2}\left(5-5(-1)^n+6(-1)^nn\right)&-1+(-1)^n&0&1\end{bmatrix}$$

Answer 2

The multiplication below represents $B\cdot B^n$ where $\delta_n=0$ for $n$ even and $1$ for $n$ odd. It is not hard to check that the product has the required form. So we have verified by induction that the form holds for all positive integers $n$.

$$\left(\begin{matrix} -1&9&0&0&0\\ 0&-1&0&0&0\\ 0&3&-1&0&0\\ 0&1&1&1&0\\ 0&-2&-2&0&1\end{matrix}\right)\left(\begin{matrix} (-1)^n&(-1)^{n+1}9n&0&0&0\\ 0&(-1)^n&0&0&0\\ 0&(-1)^{n+1}3n&(-1)^n&0&0\\ 0&(-1)^n\frac{3n}{2}+\frac{5\delta_n}{2}&\delta_n&1&0\\ 0&(-1)^{n+1}3n-5\delta_n&-2\delta_n&0&1\end{matrix}\right)$$ As to how one gets the form, that is simple: work out the first few powers and observe the pattern.

Of course the more sophisticated approach is to find $$P=\left(\begin{matrix} -\frac{1}{15}&0&\frac{1}{5}&0&0\\ \frac{1}{36}&0&0&0&0\\ 0&\frac{1}{4}&0&0&0\\ 0&-\frac{5}{2}&-1&0&1\\ 0&\frac{5}{4}&\frac{1}{2}&1&0\end{matrix}\right)$$ so that $PAP^{-1}=$ $$\left(\begin{matrix} -1&0&0&0&0\\ 0&-1&1&0&0\\ 0&0&-1&0&0\\ 0&0&0&1&0\\ 0&0&0&0&1\end{matrix}\right)$$

Then we can take powers of that to get $$\left(\begin{matrix} 1&0&0&0&0\\ 0&1&-n&0&0\\ 0&0&1&0&0\\ 0&0&0&1&0\\ 0&0&0&0&1\end{matrix}\right)$$ for the even powers and $$\left(\begin{matrix} -1&0&0&0&0\\ 0&-1&n&0&0\\ 0&0&-1&0&0\\ 0&0&0&1&0\\ 0&0&0&0&1\end{matrix}\right)$$ for the odd powers. Then we convert back to get $B^n$ by premultiplying by $P^{-1}$ and postmultiplying by $P$.