Can someone please explain hermitian conjugates and unitary matrixes?

What do you want to know?

>>8894801

They're like orthogonal matrices but you conjugate transpose them rather than just transpose.

>>8894803
So if a matrixes complex conjugate is also it's inverse then it's a unitary matrix?

>>8894821
If the conjugate transpose of a matrix it's the matrix inverse, it's called unitary. I don't understand the point of thread when your doubt is literally the definition of something. Except you have more interesting discussion I suggest you to kys.

>>8894821
That's right. And they have a lot of really nice properties too.

One cool thing is that they preserve the standard inner product. This gives them a group structure.

>>8894821
Yes.

https://en.wikipedia.org/wiki/Unitary_matrix

>>8894828
I'm revisiting linear algebra after a hiatus and was seeking some definitions and insights not in my textbook, it helps me learn ya cock

>>8894829
>>8894815

Thank you

>>8894850
Well there's the most famous equivalence which is
>>8894829. Also, the columns (and rows) form an orthonomal base of C^n consiering the hermitian inner product.
And over the complex numbers, you can always diagonlize it.

The set of its eigenvectors is linearly independent. They are actually orthogonal.

>>8894903
>They are actually orthogonal.
More than that, they are orthonormal.

>>8894907
I was thinking they might be, but I didn't want to say that without being 100% sure.

>>8894907
I'm also stupid for forgetting that linear independence implies orthogonality.

>>8894801
U=I/U
Fucking brainlets.

File: vectors.png (10KB, 640x480px) Image search: [Google]

10KB, 640x480px

>>8895005
> linear independence implies orthogonality
Wrong.

Orthogonal means that the inner product between two vectors is zero.

Linearly independent means that their is only one way to write zero as a linear combination of the vectors.

As an example, look at the picture. The top left is a linearly independent set, but it is neither orthogonal (to the standard inner product), nor are the vectors unit length.

>>8895005
I should add that orthogonality doesn't imply linear independence either, assuming you take the zero vector to be orthogonal to all other vectors. That's because you can multiply the zero vector by anything and get the zero vector back, meaning any set with the zero vector is not linearly independent.

>>8895012
No shit bud

The group U(n) is compact for n>0, can be easily proved by Heine-Borel and using det function

>>8895016
>their
opinion discarded