Is it true that there are $a \in \mathbb{R}$, $k \in \mathbb{N}$, $k>0$ such $(A+aI_4)^k=0$

Question

$A = \begin{pmatrix} 1 & 3 & -1 & 1 \\ 1 & -5 & 1 & -2 \\ -1 & 3 & 1 & 1 \\ -4 & 24 & -4 & 9 \end{pmatrix}$. Is it true that there are $a \in \mathbb{R}$, $k \in \mathbb{N}$, $k>0$ such $(A+aI_4)^k=0$? I begin with $A = \begin{pmatrix} 0 & 3 & 1 & -1 \\ -1 & 1 & -1 & 0 \\ 0 & 3 & 0 & 1 \\ 3 & 0 & 4 & 0 \end{pmatrix}\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 2 \end{pmatrix}\begin{pmatrix} 0 & 3 & 1 & -1 \\ -1 & 1 & -1 & 0 \\ 0 & 3 & 0 & 1 \\ 3 & 0 & 4 & 0 \end{pmatrix}^{-1}$. I guess due to the fact that we have $2$ blocks with eigenvalue $1$, and $2$ blocks with eigenvalue $2$ the answer is false. Am I right?

Answer 1

If $A$ is diagonalizable with $2$ distinct eigenvalues ($1$ and $2$ in this case), then $$\operatorname{rank}((A+aI_4)^k)=\operatorname{rank}(\operatorname{diag}((1+a)^kI_2, (2+a)^kI_2)) \ge2, $$ as we can't make $1+a$ and $2+a$ to be both equal to zero simultaneously. Hence no such $a$ exists as the rank can't be equal to $0$.