If $(Tx \mathbin{|} x) = 0$ for all $x$ then $T = 0$

Let $T$ be a linear operator on an inner product vector space $V$. I'd like to prove that if $$(Tx|x)=0 \quad \forall x \in V$$ then $T$ is the null operator.
I can't figure out this proof using contradiction, do I have to choose an appropriate vector?

Solution 1:

You can prove this for a $\Bbb C$omplex inner-product space $V$.

But if $\,V\,$ is an inner-product space over $\mathbb R$, then the implication doesn't hold.
Crostul gives in a comment the concrete example $$T:\Bbb{R}^2\to\Bbb R^2, (x_1,x_2)\mapsto(-x_2,x_1)$$ of a non-zero operator satisfying the initial condition. $T$ rotates by $+90°$, hence $\,x\perp Tx\,$ for all $\,x\in\Bbb R^2$.

Let $\,V\,$ be an inner-product space over $\mathbb C$.
Every sesquilinear form on $V$ satisfies the polarisation identity. For the sesquilinear form $(T(\,\cdot\,)|\,\cdot\,)$ it reads $$(Tx|y)\:=\:\frac14\sum_{k=0}^3\:i^k\left(\,T(x+i^ky)\,\big|\, x+i^ky\,\right).$$ Thus by assumption, $(Tx|y)=0\,$ for all $\,x,y\in V$, hence $T$ must be zero.