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May 8th, 2014, 04:57 AM   #1
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linear Operator// scalar produkt

Hi,

Let V be a komplex Vector space, and T a linear Operator on V

i have to Show:

$\displaystyle <T(u),u>=0 \forall u \in V \Rightarrow T=0$

In course we have just discussed Hermitian adjoint

Hope someone could give me a tip for the proof....
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May 8th, 2014, 05:00 AM   #2
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May 8th, 2014, 05:13 AM   #3
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maybe someone with a more constructive answer?
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May 8th, 2014, 10:00 AM   #4
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Consider $\langle T(u+v),u+v\rangle$. Add to it a couple of terms of the form $\pm\langle Tw,w\rangle$ so that the result is $\langle Tu,v\rangle+\langle Tv,u\rangle$. Similarly, consider $-i\langle T(iu+v),iu+v\rangle$. Add to it a couple of terms of the form $\pm i\langle Tw,w\rangle$ so that the result is $\langle Tu,v\rangle-\langle Tv,u)\rangle$.
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May 8th, 2014, 11:22 AM   #5
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But when i look at

$\displaystyle ⟨T(u+v),u+v⟩=<T(u),u>+<T(u),v>+<T(v),u>+<T(v), v>$

which is because of $\displaystyle <T(u),u>=0 \forall u$

$\displaystyle =<T(u),v>+<T(v),u>$

where do i need a $\displaystyle +-<T(w),w>$ ???
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May 8th, 2014, 11:41 AM   #6
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That works too. The idea is to prove that $\langle Tu,v\rangle+\langle Tv,u\rangle=0$ and similarly for the difference. I was suggesting first expressing $\langle Tu,v\rangle+\langle Tv,u\rangle$ as a sum of terms of the form $\langle Tw,w\rangle$ that works always and only then using the fact that $\langle Tw,w\rangle=0$.
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