Linear Algebra: Hermitian Matrices

[Niles]

Homework Statement

Hi all.

Let's say that I have a Hermitian 2x2 matrix A with two distinct eigenvalues, and thus two eigenvectors.

Question 1: What space is it they span? Is it R²?

Now let us say I have another Hermitian 2x2 matrix B with two distinct eigenvalues, and thus two eigenvectors.

Question 2: Do the eigenvectors of B span the same space as the eigenvectors of A?

Thanks in advance.


Niles.

 

[Dick]

If A has two distinct eigenvalues, then the corresponding eigenvectors must be linearly independent, right? Isn't this true regardless of whether A is hermitian or not?

 

[Niles]

Yes, you are correct. A minor detail I was not aware of. Can you confirm me in that the answers to question 1 and 2 are "yes"?

By the way, what name is "Dick" short for? I mean, Richard Feynman was called "Dick Feynman", but they never explain in any of the books (at least the ones I've read) what it is short for.

Thanks in advance.

Regards,
Niles.

EDIT: I just need to get this confirmed. A Hermitian nxn matrix will always have n (not necessarily distinct) eigenvalues, since it is diagonalizable, right?

 

[Dick]

Yes, the answers to A and B are 'yes'. And yes, a hermitian nxn matrix has n (not necessarily distinct) eigenvalues. In the sense that it's characteristic equation has n real linear factors. 'Dick' is a nickname for 'Richard' (just like 'Rick'). I have no idea where the 'D' came from, it's just what people call me.

 

[Niles]

Yes, the answers to A and B are 'yes'. And yes, a hermitian nxn matrix has n (not necessarily distinct) eigenvalues. In the sense that it's characteristic equation has n real linear factors.

Great. Thanks.

'Dick' is a nickname for 'Richard' (just like 'Rick'). I have no idea where the 'D' came from, it's just what people call me.

Ahh, I see. This I found from Wikipedia (http://en.wikipedia.org/wiki/Richard):

"The first or given name Richard comes from the Germanic elements "ric" (ruler, leader, king) and "hard" (strong, brave)." Interesting.

Thanks again.