# Generalizing the fact that symmetric matrices have only real eigenvalues.

We know that if $$A$$ is a symmetric matrix, then $$det (A+xI)$$ is a polynomial that only has real roots. Let us assume that we know that this polynomial has non-zero roots. Then, taking $$y=\frac{1}{x}$$, we deduce that $$det (I+yA)$$ has only real roots.

Now let $$A$$ be a symmetric matrix such that each entry is a (possibly infinite) series in $$y$$ (EDIT: with a non-zero constant term). Can $$det (I+yA)$$ be written as an infinite product of the form $$\prod\limits_{i=1}^\infty (1+y b_i)$$?

• I don't think every infinite series can be factored that way? Which means you choke at the $1 \times 1$ case, or similarly the diagonal case. Commented Nov 22, 2022 at 18:36
• @DustanLevenstein- The given formula includes the possibility that some factors are repeated, etc. I am just hoping that all "roots" of the determinant are real. Commented Nov 22, 2022 at 18:44
• How about the $1 \times 1$ matrix $A = (y)$? Commented Nov 22, 2022 at 18:47
• @DustanLevenstein- You're right, I should clarify that all the infinite series entries have non-zero constant terms. Commented Nov 22, 2022 at 18:50
• How about $A=(1+y/2)$? Commented Nov 22, 2022 at 18:57

The $$1 \times 1$$ matrix $$A=\begin{pmatrix}1+y/2\end{pmatrix}$$ is a counterexample, because $$\det(I+yA) = 1+y(1+y/2) = 1+y+y^2/2 = \frac{1}{2}( (y+1)^2+1 )$$ has no real roots.