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.



An $n$ eigenvalue multiplicity
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 $n$.


