Let $A_0, A_1$ be two square symmetric matrix over real numbers.
Consider $A_0 + zA_1$, where $z$ is a real number. Assume also that $A_0 + zA_1 = S(z)S(z)^t.$
I would like to know if I can write it as $$A_0 + zA_1 = \Sigma D(z) \Sigma^t, $$ where $D$ is a diagonal matrix.
I will post my ideas on the subject.
Imagine I can do so, then I say I can tell what the matrix $D$ is, in fact: $$A_0 + zA_1 = \Sigma D(z) \Sigma^t \Rightarrow A_1 = \Sigma D'(z) \Sigma^t, $$ so that $D'$ must be the diagonal matrix with eigenvalues of $A_1$.
Clearly we can do the following:
$$A_0 + zA_1 = \Sigma(z) D(z) \Sigma(z)^t, $$
now if one take the determinant of these two matrix one gets $$p(z) = \det(\Sigma(z))^2q(z), $$
but both $p$ and $q$ are polynomials of grade $n$, so that $\Sigma$ would be a very strange matrix.
Write $$A_0 + zA_1 = A_0 + \Sigma zT \Sigma^t, $$ where $T$ contains eigenvalues of $A_1$.
There are matrices such that $A_0 = QCQ^t = \Sigma \Sigma^{-1}QCQ^t \Sigma^{-1t}\Sigma^t$, where C is diagonal.
Taking all together it looks like one has $\Sigma^{-1}Q$ to be diagonal as a sufficient (and necessary?) condition.