# Symmetric matrix and diagonalization

1. Mar 31, 2007

### EvLer

This is a T/F question:
all symmetric matrices are diagonalizable.

I want to say no, but I do not know how exactly to show that.... all I know is that to be diagonalizable, matrix should have enough eigenvectors, but does multiplicity of eigenvalues matter, i.e. can I say that if eignvalue has multiplicity > 1 then the matrix is not diagonalizable?
Thanks

EDIT: zero matrix is diagonalizable, right?

Last edited: Mar 31, 2007
2. Mar 31, 2007

### Dick

The zero matrix is already diagonal. So it's VERY diagonalizable. Think spectral theorem.

3. Mar 31, 2007

### EvLer

ummm... not there yet

4. Mar 31, 2007

### Dick

You must have something like it. Note that it also makes a big difference whether the entries are real or complex. If they are real then it's true, but a proof is just a proof of the finite dimensional version of the spectral theorem. If they are complex, then you just need a counter example. Because it's false.

Last edited: Mar 31, 2007
5. Mar 31, 2007

### EvLer

oh, yeah, looked at further chapters, now I see the spectral theorem. Thanks a lot.

6. Mar 31, 2007

### ZioX

Uhhh. Not true.

Remember eigenvalues have a lot to do with what field you're in.

Principal axis theorem is the theorem you want.

7. Apr 1, 2007

### HallsofIvy

Staff Emeritus
A symmetric matrix is "self-adjoint"- that is, for a symmetric matrix A, the dot product (Ax).y= x.(Ay). It's fairly easy to prove, it may already be proven in your book, that any self-adjoint (and, so, any symmetric) n by n matrix has n independent eigenvectors. Using those eigenvectors as basis gives you a diagonal matrix with the eigenvalues on the diagonal.

8. Apr 1, 2007

### matt grime

....over an algebraically closed field....

9. Apr 1, 2007

### HallsofIvy

Staff Emeritus
Was that in response to my post? The field of real numbers is not algebraically closed but it is certainly true that any symmetric matrix over the real numbers has real eigenvalues. In fact, any self-adjoint linear transformation on a vector space over the real numbers has all real eigenvalues.

10. Apr 1, 2007

### matt grime

I should have said 'not necessarily over the field of definition of the matrix'. Or 'a field with sufficiently many elements'.