If $A$ is an $n \times n$ matrix such that $A^5 = 0$, which of the following is true?

Question

I have this question from my textbook:

Given $A$ an $n \times n$ matrix such that $A^5 = 0$, which of the following is true?

$(a)$ $A$ is invertible.

$(b)$ $A = 0$.

$(c)$ $A$ is diagonalizable.

$(d)$ If $A$ is diagonalizable then $A = 0$.

$(e)$ None of the above.

If we take determinant both sides, we have $|A^5| = |0| = 0 \implies |A|^5 = 0 \implies det(A) = 0$. So it's not invertible.

How to continue?

Answer 1

For (b) and (c) consider the matrix

$$\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}.$$

For (d) note the following: If $A^5 = 0$ holds, the only eigenvalue of $A$ is $0$ (see this answer). Suppose now that $A$ is diagonalizable, say $A = SDS^{-1}$ for an invertible matrix $S$ and a diagonal matrix $D$. Since all eigenvalues of $A$ are $0$, what can you conclude about $D$ and therefore about $A$?

Answer 2

Suppose $A$ is diagonalizable, say $A=PDP^{-1}$ for $D$ diagonal and $P$ invertible. Then $$ 0=A^5=P^{-1}D^5P $$ Do you see how to continue?