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 $n$.