Let $A$ be an $n\times n$ real matrix which is nilpotent. Prove that if for the entries of the matrix: $a_{ij}=-a_{ji}$ then $A$ has to be $0$.  I know that this is equivalent to saying that $A^T=-A$ but I don't see how to prove $A=0$ yet. Does anyone have an idea? Thanks a lot!
Given a matrix $A \in R^{n \times n}$ which is normal ($AA^H=A^HA$ where $A^H$ is hermitian of $A$) and nilpotent ($A^k=0$ for some $k$). Now we need to show that $A=0$. (This is essentially exercise 5(b) in sec. 80 on p.162 of Paul R. Halmos' Finite-Dimensional Vector Spaces.) I tried to show in the following way, we know that, $AA^H=A^HA$ pre-multiply by $A^{k-1}\implies A^kA^H=A^{k-1}A^HA$ Now, we have $0 = A^{k-1}A^HA$, since $A$ nilpotent. I am not sure how to proceed from here to show $A=0$. Can someone help me in this problem?
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