How to find $A^5$ in matrix $A$ with eigenvalues and eigenvectors given

Question

Let $A$ be a $3 \times 3$ diagonalizable matrix whose eigenvalues are $\lambda_1=2, \lambda_2=4$, and $\lambda_3=3$. If $$v_1=[1, 0, 0], v_2=[1, 1, 0], v_3=[0, 1, 1]$$ are eigenvectors of $A$ corresponding to $\lambda_1, \lambda_2$, and $\lambda_3$, respectively, then factor $A$ into a product $XDX^{−1}$ with $D$ diagonal, and use this factorization to find $A^5$.

I can't wrap my head in this. If I diagonalize this I get $A$ as $$ \begin{bmatrix} 2 & 2 & 0 \\ 0 & 4 & 3 \\ 0 & 0 & 3 \end{bmatrix} $$ if I do $A^5$ I don't get the right answer. How do I do this?

Answer 1

If $A$ is the matrix of which you want to calculate its $k$-th power then if $S$ is the matrix of the eigenvectors as columns and $D$ the diagonal matrix of eigenvalues, note that it is:

$$A^{k}=(SDS^{-1})^{k}=\underset{k}{\underbrace{(SDS^{-1})(SDS^{-1})...(SDS^{-1})}} = SD^k S^{-1}.$$

Essentialy what Paul said in the comments but just providing a wider look to it.

Answer 2

Following Paul's comment and Rebellos more general answer; which you should accept:

$$ \begin{align} A^{5}&=(XDX^{-1})(XDX^{-1})(XDX^{-1})(XDX^{-1})(XDX^{-1})\\ &=XD(X^{-1}X)D(XX^{-1})D(X^{-1}X)D(X^{-1}X)DX^{-1}\\ &=XDDDDDX^{-1}\\ &=XD^{5}X^{-1} \end{align} $$