# Symmetric matrix and diagonalization

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?

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Homework Helper
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?

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

EvLer
...Think spectral theorem.

ummm... not there yet

Homework Helper
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.

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

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.

Homework Helper
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.

Homework Helper
...over an algebraically closed field...