Finding $A^n$ in terms of $P$ and $D$ (diagonalized)

Question

My question is regarding the last two parts. I have Found $D$ and $P$, how can I obtain $A^{200}$ and det $(A^{200})$ form $PDP^{-1}?$ Thanks!

Answer 1

Finding $A^{n}$ of a diagonal matrix is actually pretty trivial, but I'll show you why (don't worry, I was confused when I first did this).

$$A=PDP^{-1}$$ $$A^{n}=(PDP^{-1})^{n}$$ $$=PDP^{-1}PDP^{-1}PDP^{-1}...$$ Obviously $PP^{-1}=I$, because they are inverses of one another. We can keep doing this as many times and we end up with... $$A^{n}=PD^{n}P^{-1}$$ Now this is easy because $D$ is a diagonal matrix and $D^{n}$ is just the matrix composed of $D_{ij}^{n}$, where $D_{ii}$ is the $ii^{th}$ entry of of D. Finding the determinant is also easy.
$$det(A^{n})=det(PD^{n}P^{-1})$$ $$=det(P)det(D^{n})det(P^{-1})$$ $$=det(P)\frac{1}{det(P)}det(D^{n})$$ $$det(A^{n})=det(D^{n})$$ We are able to do step 2 and three because determinants are numbers and thus can commute, and three because of the inverse property.