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!

9 added 9 characters in body

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 $\lambda _i$.

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, $r(x) = \sum_{t = 0}^{l-1} b_t b_t(x) \cdot g_t(x)^T$ for some $b_t b_t(x) \in \mathbf{F}_q$. mathbf{F}_q[x]$. How it can be proved? What technique can be used for that? Thank you! 8 added 2 characters in body Hello, all! I have a polynomial non-singular square matrix over$\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$\lambda _i$. My problem is to prove that if some$l \times 1$- vector of polynomials$r(x)$satisfies$\underset{l \times 1}{r(\lambda _i)} i)}^T \underset{1 \times l}{v _{i,j}^T} {i,j}} = 0\forall i, j$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,$r(x) = \sum_{t = 0}^{l-1} b_t \cdot g_t(x)$g_t(x)^T$ for some $b_t \in \mathbf{F}_q$. How it can be proved? What technique can be used for that?

Thank you!