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edited Jan 19 2012 at 9:28 spk
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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!
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edited Jan 18 2012 at 10:29 spk
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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!
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edited Jan 18 2012 at 10:20 spk
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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!
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edited Dec 23 2011 at 12:26 spk
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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 $\underset{l \times 1}{r(\lambda _i)} \underset{1 \times l}{v _{i,j}^T} = 0$ $\forall i, j$ then it must belongs to space of rows of $G(x)$.
G(x) = (\underset{1 \times l}{g_0(x)}, \ldots, \underset{1 \times l}{g_{l-1}(x)})$: so, $r(x) = \sum_{t = 0}^{l-1} b_t \cdot g_t(x)$ for some $b_t \in \mathbf{F}_q$.
How it can be proved? What technique can be used for that?
Thank you!
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edited Dec 23 2011 at 12:18 spk
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Hello, all!
I have a polynomial non-singular square matrix over $\mathbf{F} _q[x]$,
$$G(x) $\underset{l \times l}{G(x)} = \left( \begin{matrix} g _{0,0}(x) & g _{0,1}(x) & \ldots & g _{0,l-1}(x) \\ \vdots & \vdots & \vdots & \vdots \\ g _{l-1,0}(x) & g _{l-1,1}(x) & \ldots & g _{l-1,l-1}(x) \end{matrix} \right).$$
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 $v \underset{l \times 1}{v _{i,j}$ {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 $r(\lambda_i) v_{i,j} \underset{l \times 1}{r(\lambda _i)} \underset{1 \times l}{v _{i,j}^T} = 0$ $\forall i, j$ then it must belongs to space of rows of $G(x)$.
How it can be proved? What technique can be used for that?
Thank you!
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edited Dec 23 2011 at 9:37 spk
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Hello, all!
I have a polynomial non-singular square matrix over finite field $\mathbf{F} _q$,
q[x]$,
$$G(x) = \left( \begin{matrix} g _{0,0}(x) & g _{0,1}(x) & \ldots & g _{0,l-1}(x) \\ \vdots & \vdots & \vdots & \vdots \\ g _{l-1,0}(x) & g _{l-1,1}(x) & \ldots & g _{l-1,l-1}(x) \end{matrix} \right).$$
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 $v _{i,j}$ of system of equations $G(\lambda _i) v _{i,j} = 0$.
My problem is to prove that if some vector of polynomials $r(x)$ satisfies $r(\lambda_i) v_{i,j} = 0$ $\forall i, j$ then it must belongs to space of rows of $G(x)$.
How it can be proved? What technique can be used for that?
Thank you!
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edited Dec 23 2011 at 9:18 |
Hello, all!
I have a polynomial non-singular square matrix over finite field $\mathbf{F} q$ _q$,
$G(x) $G(x) = \left( \begin{matrix} g {0,0}(x) _{0,0}(x) & g_{0,1}(x) g _{0,1}(x) & \ldots & g_{0,l-1}(x) g _{0,l-1}(x) \\ \vdots & \vdots & \vdots & \vdots \g_{l-1,0}(x) \ g _{l-1,0}(x) & g_{l-1,1}(x) g _{l-1,1}(x) & \ldots & g_{l-1,l-1}(x) g _{l-1,l-1}(x) \end{matrix} \right)$.
right).$$
I call an eigenvalue of $G(x)$ roots of equation $\det G(x) = 0$. It can be founded from some extension $\mathbf{F}_{q^r}$ \mathbf{F} _{q^r}$ of $\mathbf{F}_q$. \mathbf{F} _q$. I call an eigenvector corresponding to eigenvalue $\lambda_i$ \lambda _i$ a solution $v_{i,j}$ v _{i,j}$ of system of equations $G(\lambda_i) v_{i,j} G(\lambda _i) v _{i,j} = 0$.
My problem is to prove that if some vector of polynomials $r(x)$ satisfies $r(\lambda_i) v_{i,j} = 0$ $\forall i, j$ then it must belong belongs to space of rows of $G(x)$.
How it can be proved? What technique can be used for that?
Thank you!
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edited Dec 23 2011 at 9:16 spk
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Hello, all!
I have a polynomial non-singular square matrix over finite field $\mathbf{F}q$ $G(x) = \left( \begin{matrix} g{0,0}(x) & g_{0,1}(x) & \ldots & g_{0,l-1}(x) \ \vdots & \vdots & \vdots & \vdots \ g_{l-1,0}(x) & g_{l-1,1}(x) & \ldots & g_{l-1,l-1}(x) \end{matrix} \right)$.
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 $\mathbf{F}_q$. I call an eigenvector corresponding to eigenvalue $\lambda_i$ a solution $v_{i,j}$ of system of equations $G(\lambda_i) v_{i,j} = 0$.
My problem is to prove that if some vector of polynomials $r(x)$ satisfies $r(\lambda_i) v_{i,j} = 0$ $\forall i, j$ then it must belongs belong to space of rows of $G(x)$.
How it can be proved? What technique can be used for that?
Thank you!
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edited Dec 23 2011 at 9:06 spk
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Hello, all!
I have a polynomial non-singular square matrix over finite field $\mathbf{F}q$ $G(x) = \begin{pmatrix} left( \begin{matrix} g{0,0}(x) & g_{0,1}(x) & \ldots & g_{0,l-1}(x) \ \vdots & \vdots & \vdots & \vdots \ g_{l-1,0}(x) & g_{l-1,1}(x) & \ldots & g_{l-1,l-1}(x) \end{pmatrix}$.
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 $\mathbf{F}_q$. I call an eigenvector corresponding to eigenvalue $\lambda_i$ a solution $v_{i,j}$ of system of equations $G(\lambda_i) v_{i,j} = 0$.
My problem is to prove that if some vector of polynomials $r(x)$ satisfies $r(\lambda_i) v_{i,j} = 0$ $\forall i, j$ then it must belongs to space of rows of $G(x)$.
How it can be proved? What technique can be used for that?
Thank you!
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asked Dec 23 2011 at 8:58 spk
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polynomial matrices and its spectrum
Hello, all!
I have a polynomial non-singular square matrix over finite field $\mathbf{F}q$ $G(x) = \begin{pmatrix} g{0,0}(x) & g_{0,1}(x) & \ldots & g_{0,l-1}(x) \ \vdots & \vdots & \vdots & \vdots \ g_{l-1,0}(x) & g_{l-1,1}(x) & \ldots & g_{l-1,l-1}(x) \end{pmatrix}$.
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 $\mathbf{F}_q$. I call an eigenvector corresponding to eigenvalue $\lambda_i$ a solution $v_{i,j}$ of system of equations $G(\lambda_i) v_{i,j} = 0$.
My problem is to prove that if some vector of polynomials $r(x)$ satisfies $r(\lambda_i) v_{i,j} = 0$ $\forall i, j$ then it must belongs to space of rows of $G(x)$.
How it can be proved? What technique can be used for that?
Thank you!
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