If $A^*= A$ and $A^m= 0$, then $A= 0$.

Question

If $A^*= A$ and $A^m= 0$, then $A= 0$.

My attempt:

If $m = 2$, then $\text{tr}(A^*A)= 0 \Rightarrow A=0$ and then the result follows for $m = \{2,4,8,16,32,\ldots\}$. But I don't know how it works in general.

Answer 1

Consider a diagonalization of the Hermite matrix $A$ (i.e. $A = A^*$) by a unitary matrix $U$ (with complex coordinates). Then, $U^* A U = D$, where $D$ is a diagonal (complex) matrix.

Let $D = \text{diag}(\lambda_1, \cdots, \lambda_d)$, where $d$ is the size of $A$. ($\lambda_1, \cdots, \lambda_d$ are the eigenvalues of $A$.)

Since $U U^* = U^* U = I$, $$D^m = (U^* A U)^m = U^* A^m U = U^* O U = O.$$ We also see that $D^m = \text{diag}(\lambda_1^m, \cdots, \lambda_d^m)$. Hence, $\lambda_1 = \cdots = \lambda_d = 0$, that is, $D = O$. Hence, $$A = U D U^* = U O U^* = O.$$

Answer 2

Here is an answer without diagonalization.

We will show that $A^m = 0 \implies A^{m-1} = 0$. In particular, let $B = A^{m-1}$, and consider $||B^*Bx|| = (B^*Bx)^*(B^*Bx) = x^*BB^*B^*Bx$; since $A$ is Hermitian, this middle product is just a product of $A$'s. Thus, this must vanish and $||Bx|| = x^*B^*Bx = 0$ as well, so we may conclude $B = 0$ since $x$ was arbitrary.