Show that for a symmetric or normal matrix

[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...

 

[NaturePaper]

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.

 

[MatthewD]

Do I have to use Cayley-Hamilton? Could I use the fact that A would be orthogonally equivalent to a diagonal matrix by definition 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...

 

[NaturePaper]

If you are trying some short cuts (, what wrong with considering Spectral form of http://en.wikipedia.org/wiki/Spectral_theorem#Normal_matrices"?

 

[HallsofIvy]

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.