Prove that a non-zero matrix satisfying $A^N = \textbf0$ for some $N$ cannot be diagonalized.

Question

Let a non-zero matrix $A$ satisfy $A^5 = \textbf0$. Prove that A cannot be diagonalized. More generally, any non-zero nilpotent matrix, i.e. a non-zero matrix satisfying $A^N = \textbf0$ for some $N$ cannot be diagonalized.

My understanding of how to show that $A$ cannot be diagonalized is that $A = PDP^{-1} \rightarrow D = P^{-1}AP$ where $D = \lambda I_5$. Since $A^5 = 0$, $D = P * 0 * P^{-1} = 0$, meaning that the only eigenvalue for $A$ is $\lambda = 0$, and $A$ cannot be diagonalized.

Is this correct and can this be extended to $A^N$ as a complete proof?

Answer 1

Given that $$A^N=0$$, $A$ is nilpotent matrix. If $A$ is diagonalizatble then for some invertible matrix $A$, we have $$P^{-1} A P= D~~~~(1)$$ Here $D$ is a diaginal matrix, we know that $$D^N=(P^{-1} A~P)^N= P^{-1} A^N P =0.$$ This proves that the Diagonal matrix $D$ is actually a null matrix meaning that all eigenvalues of $A$ are zero. This proves that $A$ (nilpotent matrix) is not diagonalizable.

Answer 2

If

$A \ne 0 \tag 1$

is diagonalizable, then there exists a non-singular matrix $P$ and a diagonal matrix $D$ such that

$D = PAP^{-1}; \tag 2$

thus,

$D^N = (PAP^{-1})^N = PA^NP^{-1} = P0P^{-1} = 0; \tag 3$

if $d$ is one of the diagonal entries of $D$, then (3) implies

$d^N = 0 \Longrightarrow d = 0; \tag 4$

since this holds for every diagonal entry of $D$, we have

$D = 0 \Longrightarrow A = P^{-1}DP = 0, \tag 5$

which contradicts (1); thus $A$ cannot be a diagonalizable matrix.

Note that $D$, though diagonal, need not be of the form $\lambda I$; though the off-diagonal entries of $D$ all vanish, the diagonal entries need not be the same; indeed, they may all be distinct. Furthermore, even though it is correct to say the only eigenvalue of $A$ is $0$, this in and of itself does not forbid the diagonalizability $A$; in the present instance it is the condition (1) which ultimately obstructs diagonalizability, since $A$ diagonalizable implies $A = 0$ as we have seen. Nevertheless a correct proof, such as the one given here, extends to all $N$.