A is a hermitian matrix such that $A^2 = \mathbf 0$.Prove that $A = \mathbf 0$

Question

A is a hermitian matrix such that $A^2 = \mathbf 0$.Prove that $ A = \mathbf 0$

I first assumed that $A =(a_{ij} )_{mXn} $ Then $\overline A =(\overline a_{ij} )_{mXn} $ Then $$\overline A^t =(\overline a_{ji} )_{nXm} $$ Now, $$\biggl((\overline A)^t\biggr)^2=(b_{pq})_{nXn} $$ where $$(b_{pq})=\sum_{i=1}^n \overline a_{pi}\overline a_{iq} $$ So, $$Tr\biggl(\biggl((\overline A)^t\biggr)^2\biggr) =\sum_{i=1}^n \overline a_{pi}^2$$ Also, given $$\overline A^2 =\mathbf 0$$ and A is hermitian. Then $$\biggl((\overline A)^t\biggr)=A$$ So, $$\biggl((\overline A)^t\biggr)^2=\overline A^2 =\mathbf 0$$ So, $$Tr\biggl(\biggl((\overline A)^t\biggr)^2\biggr) =\sum_{i=1}^n \overline a_{pi}^2 = 0 $$

From now where to go.

Answer 1

If $ \mu$ is an eigenvalue of $A$, then $\mu^2$ is an eigenvalue of $A^2$, hence $ \mu=0$.

Now use the spectral theorem to deduce that $A=0$.

Answer 2

Compute the trace of $A^2=A^HA$.

The trace is $\sum_{i=1}^n\sum_{j=1}^n |a_{ij}|^2$.

Since $A^2=0$, the trace is $0$ and $|a_{ij}|^2=0$ for all $i,j$, hence $A=0$.