Question: Suppose $A$ is a square diagonalizable matrix with this property
$\exists k\in \mathbb Z_{>0}, A^k = 0$
I am to prove that $A$ must be the zero matrix.
I tried using property that $A^kx=\lambda^kx$ and $A^k=PB^kP^{-1}$ separately but couldn't come to any logical conclusion. Looking for advice on how to go about this.
Best Answer
If $A$ is diagonalizable, then we can write $A=PDP^{-1}$.
Also, $A^k=PD^kP^{-1}$
If $A^k=0$, then $D^k=0$. Note that $D=diag(d_1, d_2, \ldots, d_k)$ is a diagonal matrix.
If $d_i^k=0$, try to deduce the value of $d_i$ and should be able to complete the proof.