True or false Self adjoint operator and trace

Question

I have a problem trying to prove the next statement or giving a counterexample

If $T:\mathbb{R}^3 \to \mathbb{R}^3$ is self-adjoint such that trace($T^2$)=$0$, then $T=0$.

I tried with the canonical basis to form the associated matrix to give a counterexample but now I think the statement its true but I don't have idea how to prove it.

Answer 1

True. To see it explicitly diagonalize $T$. Since it is self adjoint it has real eigenvalues. The trace of $T^2$ is the sum of the squares of these eigenvalues. If it equals zero then all eigenvalues must be zero and hence $T=0$.