# Relation between $T=0$ and $(Tx,x)=0$

Let $X$ be a vector space and (.,.) be an inner product on $X$ also if we have a linear operator $T:X\rightarrow X$, then in both cases real and complex for inner product what is the relation between $T=0$ and $(Tx,x)=0$?

If $V$ is complex, then if $\langle Tv,v\rangle=0$ for all $v\in V$, then $T=0$. The proof comes from the identity:

$\langle Tu,w\rangle=\frac{\langle T(u+w),u+w\rangle-\langle T(u-w),u-w\rangle}{4}+\frac{\langle T(u+iw),u+iw\rangle-\langle T(u-iw),u-iw\rangle}{4}i$

For all $u$ and $w$ it is $\langle Tu,w\rangle=0$. This gives $T=0$.

In the real case you need to assume for example that $T$ is self-adjoint (the above identity then just lacks the imaginary part).

• This identity, I think, assumes conjugate linearity in the second argument. Beware that you need to switch some minus signs around if your conjugate linearity is in the first argument. Commented Oct 19, 2016 at 20:18

Hints: If $T=0$, what is $Tx$? what is $(0,x)?$, what do you conclude? In the other direction, what is the geometric interpretation of $(x,y)=0$? Can you think of a linear transformation $T:\mathbb R^2 \to \mathbb R^2$ such that $(Tx,x)=0$ without $T$ being $0$?

• :If $T=0$ then $Tx=0$ and $T(x,0)=0$.(for both case real and complex inner product). Now in the other direction what happen؟
– rese
Commented Feb 23, 2013 at 22:52

$T$ is a bounded linear operator on a complex vector space with $<Tx,x> = 0$ $\forall$ $x$ gives $T = 0$. This is not true in real vector space. Consider rotation in $\mathbb{R}^2$ in an angle $\frac{\pi}{2}$. Then $<Tx,x> = 0$ $\forall x$, but $T \neq 0$.

• what about such a rotation in complex plane? I can't conclude that it is non zero. Commented May 7, 2020 at 4:22