An $n$ eigenvalue multiplicity

Question

Suppose we have $A_i$, $i=1\ldots n$, $n\times n$ complex matrices linearly independent. It may be conjectured that there exist $(a_1,\ldots,a_n) \in \mathbb{C}^n$ not all zero such that $\sum_{i=1}^na_iA_i$ has a one eigenvalue of (algebraic) multiplicity $n$. I didn't try many things as the linearly independent condition is necessary to assume (hard to use). The question is it provable or do any related facts exist. Thank you.

Answer 1

This is an elaboration on the comment of Alexandre Eremenko. Algebraic multiplicity $n$ means that we have the equality of polynomials $$ \det(t I_n -a_1A_1+\cdots+a_nA_n)=(t-\lambda)^n $$ for some $\lambda$. Comparing coefficients of $t^{n-1},t^{n-2},\ldots,1$, we find a system of $n$ equations
$$ \begin{cases} \mathrm{tr}(a_1A_1+\cdots+a_nA_n)=n\lambda,\\ \ldots\\ \mathrm{tr}_k(a_1A_1+\cdots+a_nA_n)=\binom{n}{k}\lambda^k,\\ \ldots\\ \mathrm{det}(a_1A_1+\cdots+a_nA_n)=\lambda^n. \end{cases} $$ Here $\mathrm{tr}_k$ denotes the sum of principal $k\times k$-minors; this is the trace of the action on the $k$-th exterior power. From the first equation, we find $\lambda=\mathrm{tr}(a_1A_1+\cdots+a_nA_n)/n$, and we can substitute this to all other equations. We now have a system of $n-1$ homogeneous equations for parameters $a_1,\ldots,a_n$. Such a system always has a solution different from the zero vector.

Note that your assumption on linear independence is not needed: if matrices are linearly dependent, their nonzero linear combination giving the zero matrix whose zero eigenvalue is of multiplicity $n$.

Answer 2

Let each $A_i$ be a matrix with all entries 0 except for the $(i,i+1)$ entry which is 1, where $i+1=1$ if $i=n$. The characteristic polynomial of $\sum_j a_j A_j$ is $x^n - \prod_j a_j$. I believe that has one zero of multiplicity $n$ if $\prod_j a_j=0$ and $n$ distinct zeros otherwise.