# Show that for a symmetric or normal matrix

1. May 5, 2010

### MatthewD

Is there anyway to show that for a symmetric or normal matrix A, that det(A) = $$\prod \lambda_i$$ without using Jordan blocks? I want to show this result using maybe unitary equivalence and other similar matrices... any ideas? It's obviously easy with JCF...

2. May 5, 2010

### NaturePaper

Re: Determinants

I don't know what is the function of JCF in it.....it simply follows from the well known Caley-Hamilton Theorem (Every square matrix satisfies its own characteristic equation) and the result holds for any square matrix.

3. May 5, 2010

### MatthewD

Re: Determinants

Do I have to use Cayley-Hamilton? Could I use the fact that A would be orthogonally equivalent to a diagonal matrix by defintion of symmetric, so for some orthogonal matrix Q and diagonal matrix D:
A=Q*DQ
then det(A)=det(Q*DQ)=det(D)
D is diagonal=>det(D)=product of diagonal entries... but how would I show these are the eigenvalues?
if they're the eigenvalues, then i have my result since similar matrices have the same eigenvalues...

4. May 6, 2010

### NaturePaper

Last edited by a moderator: Apr 25, 2017
5. May 6, 2010

### HallsofIvy

Staff Emeritus
Re: Determinants

Every symmetric, or normal, matrix, A, can be diagonalized- that is, there exist an invertible matrix P such that $PAP^{-1}= D$ where D is a diagonal matrix having the eigenvalues of A on its diagonal.

Now $det(PAP^{-1})=$$det(P)det(A)det(P)^{-1}= det(A)= det(D)$ and that last is, of course, the product of the eigenvalues.