OpenStudy (empty):

Looking for a simple proof from linear algebra

2 years ago
OpenStudy (empty):

Here's the theorem I want to prove. If you're given a complex vector space any vector v with an operator T that $v^\dagger T v = 0$ then this implies $T=0$

2 years ago
OpenStudy (empty):

Note that this is not true for real vectors, you could rotate a vector 90 degrees to itself to make its dot product 0, which is quite peculiar!

2 years ago
OpenStudy (anonymous):

Does that dagger denote the transpose, adjoint, adjugate, or something else entirely?

2 years ago
OpenStudy (empty):

Oh sorry so it means: $A^{* \top} = A^\dagger$ So you take the complex conjugate of all the entries and transpose the matrix (in any order, transposing and taking the complex conjugate aren't order dependent).

2 years ago
OpenStudy (anonymous):

Ah ok, adjoint it is. Thanks for clearing that up

2 years ago
OpenStudy (empty):

Yeah I think there are like a ton of different names for this, Hermitian adjoint, Hermitian conjugate, Hermitian transpose, Conjugate transpose, uhhh... I don't know I guess the difference might be between the words representing the actual matrix you get from the operation and one the operation itself so I tried to not say it ambiguously haha.

2 years ago
OpenStudy (empty):

All I am really staring at is taking the conjugate of this and maybe multiplying in complex numbers and stuff: $v^\dagger Tv = 0$$v^\dagger T^\dagger v= 0$ Add 'em together after multiplying by some arbitrary constant idk why not $v^\dagger(T+e^{i \theta}T^\dagger) v = 0$ I don't really feel like I'm closer to T=0 haha.

2 years ago