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 May 8th, 2014, 04:57 AM #1 Newbie   Joined: Nov 2013 Posts: 26 Thanks: 0 linear Operator// scalar produkt Hi, Let V be a komplex Vector space, and T a linear Operator on V i have to Show: $\displaystyle =0 \forall u \in V \Rightarrow T=0$ In course we have just discussed Hermitian adjoint Hope someone could give me a tip for the proof....
 May 8th, 2014, 05:00 AM #2 Newbie   Joined: May 2014 From: delhi Posts: 1 Thanks: 0 Era business school review Era business school is the best option.
 May 8th, 2014, 05:13 AM #3 Newbie   Joined: Nov 2013 Posts: 26 Thanks: 0 maybe someone with a more constructive answer?
 May 8th, 2014, 10:00 AM #4 Senior Member   Joined: Dec 2013 From: Russia Posts: 327 Thanks: 108 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 $\langle Tu,v\rangle-\langle Tv,u)\rangle$.
 May 8th, 2014, 11:22 AM #5 Newbie   Joined: Jan 2014 Posts: 6 Thanks: 0 But when i look at $\displaystyle ⟨T(u+v),u+v⟩=+++$ which is because of $\displaystyle =0 \forall u$ $\displaystyle =+$ where do i need a $\displaystyle +-$ ???
 May 8th, 2014, 11:41 AM #6 Senior Member   Joined: Dec 2013 From: Russia Posts: 327 Thanks: 108 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 $\langle Tw,w\rangle=0$. Thanks from Sandra93

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