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Suppose that for all $u \in V, \langle T(u), u \rangle =0$. How to prove that T is the 0 operator, that is $T(v)=0$ for all $v \in V$? Hint: Apply information on my last question about polarization identity to obtain that $\langle T(u), v \rangle = 0$ for all $u, v \in V$ and then let $v=T(u)$. this is a homework in the area of linear algebra.

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Hey, you really want us to do ALL your homework? How are you going to learn something if you don't even try? –  Djaian Nov 4 '10 at 14:27
    
I have already but am stuck. So some start kick is needed. –  laovultai Nov 4 '10 at 14:30
    
Good site, is it not? –  AD. Nov 4 '10 at 14:38
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This one isn't absolutely trivial. It's essential here that one is working with a complex vector space. There are counterexamples over real vector spaces. –  Robin Chapman Nov 4 '10 at 14:39
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Alvoutila, The hint tells you exactly what to do. Have you looked at the polarization other then when you posted a question about it? –  Matt E Nov 4 '10 at 14:51

1 Answer 1

You've left out some important information. It's not enough to guarantee T(v) is the zero operator just to know T(v) orthogonal to v for all v. (Think about a right angle rotation in the plane.)

Of course if T(v) is orthogonal to u for all u, then (in particular) T(v) must be zero since < T(v),T(v)> = 0 implies that.

So look for the missing extra condition(s)...

Edited per Robin Chapman's comments, to incorporate the OP's earlier question:

How to prove polarization identity?

The polarization identity expresses <T(v),u> for any v,u in a complex inner product space as a linear combination of terms of the form <T(w),w>. Thus if all of the latter are zero, so must <T(v),u> be always zero, and the result follows.

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It is true when the product is a positive definite Hermitian form over a complex vector space. That is the missing condition (see the OP's previous post). –  Robin Chapman Nov 4 '10 at 15:02

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