Congruence transformation

[thoughtgaze]

If A diagonalizes V under a congruence transformation (instead of a similarity transformation), are the diagonal elements, the eigenvalues of V?

If A is orthogonal, it is obvious that this is true, but what if A is not orthogonal?

 

[jshtok]

First glance:

There are some conditions on A to allow diagonal of D=A'VA be composed of eigenvalues of V. If this is the case, V and D have the same determinant and trace (first is a product of eig.values and the second is the sum). Same determinant implies
det(A'VA) = det(V),
which means det(A) =1 or det(A)=-1. Same trace implies
tr(V)=tr(A'VA)=tr(AA'V).
I don't immediately see what does this condition mean for A.