Positive operator and inner product

Question

We have a positive and self-adjoint linear operator $T: H\to H$. Moreover, There is $x\in H$ such that $\langle Tx, x\rangle=0.$ How we can prove that $Tx=0?$

Answer 1

By definition $T$ being positive means that \begin{equation} \langle Ty, y \rangle \geq 0 \end{equation} for all $y \in H$. In particular \begin{align} \langle T(x+y), x+y \rangle &\geq 0 \\ \langle T(x-y), x-y \rangle &\geq 0. \end{align} Expanding the above inner products, using that $\langle Tx, x\rangle = 0$ and that $T$ is self-adjoint we get \begin{equation} |\langle Tx, y\rangle| \leq \frac 12 \langle Ty, y\rangle. \end{equation} Now, we can replace $x$ by $\lambda x$ for any $\lambda \in \mathbb{R}$ above and the argument doesn't change. Therefore we must have $\langle Tx, y \rangle = 0$ for all $y \in H$. Therefore $Tx = 0$ as required.

Answer 2

Here are a couple arguments.

1. If you know that positive operators always have a positive square root, you can write $$0=\langle Tx,x\rangle=\langle T^{1/2}x,T^{1/2}x\rangle=\|T^{1/2}x\|^2.$$ Then $T^{1/2}x=0$, and $Tx=T^{1/2}(T^{1/2}x)=0$.

2. Because $\langle Ty,y\rangle\geq0$ for all $y$, the sequilinear form $[y,z]:=\langle Ty,z\rangle$ is positive. So Cauchy Schwarz applies! Then $$ \|Tx\|^2=\langle Tx,Tx\rangle=[x,Tx]\leq [x,x]^{1/2}\,[Tx,Tx]^{1/2}=\langle Tx,x\rangle^{1/2}\,[Tx,Tx]^{1/2}=0. $$ So $Tx=0$.