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$T$ is a continuous linear operator on complex vector space with $\langle Tx,x \rangle = 0$ for all $x$. Then $T = 0$. How to prove it? One related question is here, without applying continuity.

Some hints will be sufficient. Thank you.

Community
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Supriyo
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    Please refer to this post.

    http://math.stackexchange.com/questions/524970/if-the-expectation-langle-v-mv-rangle-of-an-operator-is-0-for-all-v-is-t/525054#525054

     –  Oct 25 '13 at 06:25
  • The post you referred wasn't just related, it had an answer. – Robert Israel Oct 25 '13 at 06:28
  • Oops sorry, I didn't see that he is asking for a hint. –  Oct 25 '13 at 06:33
  • Consider, for all $x,y$, the two data points $\langle T(x+y),x+y\rangle = 0$ and $\langle T(x+iy), x+iy\rangle = 0$. – Daniel Fischer Oct 27 '13 at 09:40

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Every non-zero linear operator on a complex vector space has a non-zero eigenvalue. This should be sufficient to prove your result.

I am assuming finite dimensional space as there is no mention of norm.

Vishal Gupta
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    Yes. Because otherwise that's definitely not true. One can for example consider the linear map $M_{\text{id}}: C[0,1] \to C[0,1]$ defined by $(M_{\text{id}}f)(x) = xf(x)$ – kahen Oct 25 '13 at 06:21
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    @Vishal But there is a mention of a continuous operator. It looks like the space is infinite dimensional with an inner-product. – Yurii Savchuk Oct 25 '13 at 08:09
  • @Vishal, To write the proof completely you shall use the matrix representation of $T$, then non existence of non-zero eigenvalue will produce the proof. Well. But where is the use of continuity of the linear operator $T$? Please expand the answer. – Supriyo Oct 25 '13 at 10:45
  • @Samprity In finite dimensions, every linear map is continuous. If you are working in infinite dimensions, use polarization identity. You still do not need the continuity condition. – Vishal Gupta Oct 25 '13 at 13:55
  • This answer is wrong. Every linear operator on a complex vector space has an eigenvalue, but they could all be zero without the operator being zero. – JLA Nov 05 '13 at 05:48
  • @JLA Could you give an example? – Vishal Gupta Nov 05 '13 at 06:56
  • $(0,1;0,0)$ first two are top row second two are bottom row. – JLA Nov 05 '13 at 07:04
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You can use the sesquilinear forms.Especially a square form.

So let's say that $f:H \times H-> \Bbb C$ is a square form that $f(x,x)=<Tx,x>$ then you can so that $\lVert f \rVert \leq \lVert T \rVert \leq 2\lVert f \rVert$.

Because $<Tx,x>=0=>\lVert f \rVert=>\lVert T \rVert=>T=0$.

Haha
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