[Math] Is the zero matrix the only symmetric, nilpotent matrix with real values

linear algebramatrices

My intuition tells me that the zero matrix is the only matrix that is symmetric and nilpotent with real values, but I'm having trouble proving it (or finding a counterexample.)

I have searched for related problems, but I've found only one where nilpotent was defined as any matrix $A$ where $A^2=0$; using this definition, the problem is pretty easy. I'm using the more general definition that $A$ is nilpotent if and only if there exists a positive integer $k$ such that $A^k=0$.

Based on my observations while trying to find a counterexample, I've been trying to formulate some argument about the positive semi-definiteness of the entries on the main diagonal, but I'm not getting very far with it. Is this the right approach? Is my gut feeling even true?

Best Answer

Hint: Real symmetric matrices are (orthogonally) diagonalisable. And all eigenvalues of nilpotent matrices are zero.