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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 $(Tx,x)=0$?

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3 Answers 3

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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).

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    $\begingroup$ 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. $\endgroup$
    – Ian Hincks
    Commented Oct 19, 2016 at 20:18
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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 $0$?

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  • $\begingroup$ :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؟ $\endgroup$
    – rese
    Commented Feb 23, 2013 at 22:52
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$T$ 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 $T \neq 0$.

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

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