Is there a relationship between the eigenvalues of convex combinations of two matrices and roots of convex combination of their polynomials?

Question

Let $A_1, A_2 \in M_n(\mathbb R)$ and denote $A := (1-\lambda)A_1 + \lambda A_2$ where $\lambda \in (0,1)$. Let $p_1(t) = \chi_{A_1}(t)$ and $p_2(t) = \chi_{A_2}(t)$ be the characteristic polynomials. Let us denote the convex combination of these two polynomials by $p(t) = (1-\lambda)p_1(t) + \lambda p_2(t)$. I am wondering whether there is some relation between the eigenvalues of $A$ and the roots of $p(t)$ (ideally some containment in either direction). Or they are completely unrelated?

Answer 1

There is no clear relation and at least no inclusion. Take for example $A_1=\mu_1 Id$ and $A_2=\mu_2 Id$.

Then $p_1(t)=(X-\mu_1)^n$, $p_2(t)=(X-\mu_2)^n$ and $\xi_A(t)=(X-\mu)^n$ where $\mu=(1-\lambda) \mu_1 +\lambda \mu_2$.

But $p(t)=(1-\lambda) (X-\mu_1)^n+\lambda (X-\mu_2)^n$ and one can easily check that outside degenerate cases ($\lambda=0$ or $1$, $\mu_1=\mu_2$) $\mu$ is not a root of $p$.

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Note that as your question can be seen as "is there a relation between the convex combination of roots and the roots of the convex convex combination" there seems to be results if you have additional assumptions (for example if the roots are in the unit circle).