# 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?

Sincerely,
Niles.

Homework Helper
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.

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?

Homework Helper
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): [Broken]

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

Thanks again.

Last edited by a moderator: