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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: $\displaystyle <T(u),u>=0 \forall u \in V \Rightarrow T=0$ In course we have just discussed Hermitian adjoint Hope someone could give me a tip for the proof.... 
May 8th, 2014, 05:00 AM  #2 
Newbie Joined: May 2014 From: delhi Posts: 1 Thanks: 0  Era business school review
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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 $\displaystyle ⟨T(u+v),u+v⟩=<T(u),u>+<T(u),v>+<T(v),u>+<T(v), v>$ which is because of $\displaystyle <T(u),u>=0 \forall u$ $\displaystyle =<T(u),v>+<T(v),u>$ where do i need a $\displaystyle +<T(w),w>$ ??? 
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 $\langle Tw,w\rangle=0$.


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linear, operator or or, operator or or, produkt, scalar 
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