Hello, all!

I have a polynomial non-singular square matrix over $\mathbf{F} _q[x]$, $$\underset{l \times l}{G(x)} = \left( \begin{matrix} g _{0,0}(x) & g _{0,1}(x) & \ldots & g _{0,l-1}(x) \\\ \vdots & \vdots & \vdots & \vdots \\\ g _{l-1,0}(x) & g _{l-1,1}(x) & \ldots & g _{l-1,l-1}(x) \end{matrix} \right).$$ I call an eigenvalue of $G(x)$ roots of equation $\det G(x) = 0$. It can be founded from some extension $\mathbf{F} _{q^r}$ of finite field $\mathbf{F} _q$. I call an eigenvector corresponding to eigenvalue $\lambda _i$ a solution $\underset{l \times 1}{v _{i,j}}$ of system of equations $G(\lambda _i) v _{i,j} = 0$. So $v _{i,j}$ is the $j$-th eigenvector corresponding to eigenvalue $\lambda _i$.

I suppose, eigenvectors of $G(x)$ have equal algebraic and geometric multiplicities.

My problem is to prove that if some $l \times 1$ - vector of polynomials $r(x)$ satisfies $\underset{l \times 1}{r(\lambda _i)}^T \underset{1 \times l}{v _{i,j}} = 0$ $\forall i, j$ then it must belongs to space of rows of $G(x) = (\underset{1 \times l}{g_0(x)}, \ldots, \underset{1 \times l}{g_{l-1}(x)})$: so, $r(x) = \sum_{t = 0}^{l-1} b_t(x) \cdot g_t(x)^T$ for some $b_t(x) \in \mathbf{F}_q[x]$. How it can be proved? What technique can be used for that?

Thank you!