Power of a non diagonalizable matrix

Question

For $A= \begin{bmatrix}3&1&0&0\\0&2&0&0\\0&1&-1&4\\0&2&3&-2\end{bmatrix}$

Without computing $A^4$ compute the vector $A^4 \begin{bmatrix}-1\\0\\4\\3\end{bmatrix}$

I have found the eigenvectors and eigenvalues for matrix A

ƛ₁=3 , ƛ₂=2 , ƛ₃=-5

$E_3=\begin{bmatrix}1\\0\\0\\0\end{bmatrix}$ , $E_2=\begin{bmatrix}0\\0\\4/3\\1\end{bmatrix}$ , $E_{-5}=\begin{bmatrix}0\\0\\-1\\1\end{bmatrix}$

Since this matrix is not diagonalizable is there some other way of calculating $A^n$ other than using $A^n=PD^nP^{-1}$?

Answer 1

The vector you are being asked to compute is equal to $A^4(3E_2-E_3)$. We know $AE_2=2E_2$, so $A^4E_2=2^4E_2$, etc.

Hence the answer is given by $$3\cdot 2^4 E_2 - 3^4E_3$$

Here we didn't have to compute $A^4$ because the vector we were acting on was in the span of our eigenvectors. However, if we wanted to compute $A^4$, we could do so using Jordan normal forms