0

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!

user26857
  • 53,190
Mathman
  • 7
  • For an approach that avoids the spectral theorem: note that $A$ always has the same rank as $AA^T$. However, if $A$ is non-zero and nilpotent, then the rank of $A^2$ must be smaller than that of $A$. – Ben Grossmann Jun 16 '21 at 17:17
  • That was actually a question I was asking myself and couldn’t find a clear answer to. If $A$ is nilpotent: is it always: $Rank(A^{k})>Rank(A^{k+1})$ or could it also be be $Rank(A^{k})=Rank(A^{k+1})$? – Mathman Jun 16 '21 at 23:01

1 Answers1

1

Since $A^T=-A$, $A$ and $A^T$ commute. In other words, $A$ is a normal matrix. Therefore, by the spectral theorem, it is diagonalizable. And a diagonalizable matrix is nilpotent if and only if it is the null matrix.

José Carlos Santos
  • 440,053