Suppose that of dimension of $V$ is finite and let $T\in L(V)$ be a non-negative linear operator. Prove that if $v\in V$ is such that $\left< T(v),v\right>=0$, then $T(v)=0$.
My attempt:
I know that if $T$ is non-negative , we can find a linear operator $S$ such that $T=S^*S$, then, we have:
$0=\left< T(v),v\right>=\left< S^*S(v),v\right>=\left< S(v),S(v)\right>=||S(v)||^2$
Can I conclude then that $S(v)=0$ and so $T(v)=S^*S(v)=0$?. Why? I think so but I don't know what conditions have to satify $s$ in order to have $(||S(v)||^2\to S(v)=0)$. Thanks!