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 $\prod\limits_{i=1}^\infty (1+y b_i)$?

  • $\begingroup$ 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. $\endgroup$ Commented Nov 22, 2022 at 18:36
  • $\begingroup$ @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. $\endgroup$ Commented Nov 22, 2022 at 18:44
  • $\begingroup$ How about the $1 \times 1$ matrix $A = (y)$? $\endgroup$ Commented Nov 22, 2022 at 18:47
  • $\begingroup$ @DustanLevenstein- You're right, I should clarify that all the infinite series entries have non-zero constant terms. $\endgroup$ Commented Nov 22, 2022 at 18:50
  • $\begingroup$ How about $A=(1+y/2)$? $\endgroup$ Commented Nov 22, 2022 at 18:57

1 Answer 1


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.


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