Understanding this Linear Algebra Theorem

Question

Let $U$ be a self-adjoint operator on a finite-dimensional inner product space $V$. If <$x, U(x)$> = $0$ for all $x \in V$, then $U = T_0$.

What is $T_0$ in this case? Thank you.

Answer 1

$T_0$ here is the $0$-map. That is, if $\langle x,Ux \rangle = 0$ for all $x$ (in a real vector space) and $U$ is self-adjoint, then we must have $U = 0$.

Notably: over a complex vector space (with a Hermitian inner product), we need not assume that $U$ is self-adjoint. Also, the theorem still holds for infinite dimensional spaces.