Determinant of a Hermitiain matrix

[kingwinner]

Q: Suppose A is a Hermitiain matrix, prove that det A is real.

Note: I know nothing about inner products yet.

Some thoughts:
Perhaps proving that det A = conjugate of (det A) ? But how?


Can someone please help me? Thanks!

 

[morphism]

Try proving that det(conjugate of A) = conjugate of det(A) (use the formula for det(.)).

 

[kingwinner]

How can I do that? Can you please tell me more about it?

 

[Kreizhn]

I'm not terribly sure as to what formula morphism is referring to. There is a process/algorithm for finding the determinant of a general matrix but no closed formula.

You know that by the definition of a Hermitian Matrix, it's equal to it's transpose conjugate. Furthermore, we know that the the determinant of the transpose is equal to the determinant of the original matrix. Thus using this bit of information, this is a one line proof.

 

[kingwinner]

I'm not terribly sure as to what formula morphism is referring to. There is a process/algorithm for finding the determinant of a general matrix but no closed formula.

You know that by the definition of a Hermitian Matrix, it's equal to it's transpose conjugate. Furthermore, we know that the the determinant of the transpose is equal to the determinant of the original matrix. Thus using this bit of information, this is a one line proof.

But this just gives det A = det (conjugate of A)

 

[morphism]

The formula is here: http://planetmath.org/encyclopedia/Determinant2.html

 

[Kreizhn]

It may be considered a formula semantically in that following the outlined process will give you the determinant, but I don't believe that we can consider it a formula in the classical sense. Rather it's a procedure, which logically is a very different thing.

Furthermore, the determinant of the conjugate is the conjugate of the determinant.

 

[morphism]

Furthermore, the determinant of the conjugate is the conjugate of the determinant.

Yes, and using that formula (or procedure - call it whatever you want) gives a very easy proof of that, like I stated in post #2.