Is the Following Proof Correct?
Proposition. Let $T$ be a linear operator on an inner product space $V$.Given that we have $\langle T(x),y\rangle = 0,\forall x,y\in \beta$, where $\beta$ is some basis for $V$ prove that $T = T_0$. ($T_0$ is the zero map on $V$).
Proof. Assume $x,y\in V$ since $\beta$ is a basis for $V$ we have $y=\sum_{v\in\beta}\alpha_vv$ therefore $$\langle T(x),y\rangle = \left\langle T(x),\sum_{v\in\beta}\alpha_vv\right\rangle = \sum_{v\in\beta}\overline{\alpha_v}\langle T(x),v\rangle$$ and from hypothesis it follows that $ \sum_{v\in\beta}\overline{\alpha_v}\langle T(x),v\rangle = 0$ then since our choice of $y$ was arbitrary it follows that $T(x)\in V^{\perp} = \{0\}$ and so $T(x) = 0$ and since $x$ was arbitrarily chosen it follows that $T=T_0$.
$\blacksquare$