Eigenvalues of FIR Convolution Matrix
For any convolution matrix (even truncated), your diagonal entries would be $h_0$; hence, the characteristic polynomial, setting $b_0 = h_0-\lambda$,

\begin{align}|H-\lambda I| = \begin{vmatrix} b_0 &amp; 0  &amp; 0\\h_1 &amp; b_0 &amp; 0\\h_2 &amp; h_1 &amp; b_0 \end{vmatrix} &amp;= b_0\,\begin{vmatrix} b_0 &amp; 0\\h_1 &amp; b_0 \end{vmatrix} - 0\,\begin{vmatrix} h_1&amp;0\\h_2&amp;b_0\end{vmatrix} + 0\,\begin{vmatrix} h_1 &amp; b_0\\h_2&amp;h_1 \end{vmatrix} \\ \\
&amp;=b_0\,\begin{vmatrix} b_0 &amp; 0\\h_1 &amp; b_0 \end{vmatrix}\\
&amp;=b_0\left(b_0\,b_0 - 0\,h_1 \right)\\
&amp;=b_0^3
\end{align}

Little induction is necessary to see that for any square matrix of dimension $N\times N$ with an lower left structure and a constant $d$ in each element of the diagonal, the characteristic polynomial is $(d-\lambda)^N$; the Eigenvalues are roots of that polynomial, and hence there's but one Eigenvalue (with $N$ multiplicity): $d$, which in your case is $h_0$.

So, since not all zeros of $H(z)$ are defined by your first filter tap, no, you can't find all zeros of $H(z)$ through Eigenvalues.