I have this exersize and we were told these 2 matrices can be diagonalized with no calculations at all.

We need to find P such that $PAP^{-1}=D$ is diagonal (if the matrices are diagonalizable) for both $A_1$ and $A_2$:

$A1= \begin{bmatrix} 2 && 0 && 0 && 0 && 0 \\ 1 && 2 && 0 && 0 && 0 \\ 0 && 1 && 2 && 0 && 0 \\ 0 && 0 && 0 && 1 && 0 \\ 0 && 0 && 0 && 0 && 1 \end{bmatrix} $

$A2= \begin{bmatrix} 1 && 5 && 2 && 1 \\ 1 && 2 && 0 && -1 \\ 0 && 0 && 3 && 2 \\ 0 && 0 && 0 && 4 \end{bmatrix} $

Characteristic polynomial is easy to find for both $A_1$ and $A_2.$

So $\chi_{A_1}(X)=(X-2)^3(X-1)^2$, but how can I see that it isn't diagonalizable without calculating $V_2$?

As for $A_2,$ $\chi_{A_2}(X)=(X-1)(X-2)(X-3)(X-4)$ which is diagonalizable. Now again how do I proceed from here without calculations?

- I know how to find eigenvectors and then build P according to the eigenvectors. I'm just asking if there is a quick way to solve the problem for these two?

Thanks!