I have the following matrix:
$$A=\begin{bmatrix} 3 & 0 & 0 \\ 5 & -2 & 0 \\ 0 & 4 & 1\end{bmatrix}$$
Is it diagonalizable? I think it is, but when I try to test the eigenvalues (which are $3$, $-2$, $-1$), I get stuck. Specifically, on the test of eigenvalue $-2$, I cannot seem to find a basis for it.
Any help would be appreciated.
Since all eigenvalues ($-2,-1,3$) are different, yes it is.
We can see this fact also knowing with the information of Jordan canonical form. It says that all of eigenvalues squares have to be $1\times 1$ matrices.