Exercise 10.A.11

Proof. Because $T$ is self-adjoint, $V$ has a basis consisting of eigenvectors of $T$, by the Spectral Theorems (see 7.24 and 7.29). By 7.35 (b), all eigenvalues of $T$ are nonnegative. If their sum equals $0$, then they all equal $0$. Applying $T$ to each of the basis vectors (which are eigenvectors of $T$), we see that $T=0$. □