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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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  • $\begingroup$ This seems a reasonable question, but I can't understand your notation. Is $j$ the coordinate index of your vector $(v_i)$? What is the shape of the product $r(\lambda_i)v_{i,j}$? Could you write down explicitly the sizes of the involved vectors? What is the "space of rows"? $\endgroup$
    – Federico Poloni
    Commented Dec 23, 2011 at 12:14
  • $\begingroup$ Ok, now it is better. Two more points: I assume the condition is $r(\lambda_i)^T v_{i,j}=0$. with the transpose in this position.Second point,how do your definitions of eigenvalue/eigenvector extend in the case that there are generalized eigenvalues and the matrix is not diagonalizable?Think for instance $G(x)=xI-A$, where $A=\begin{bmatrix}0 & 1\\\\ 0 & 0\end{bmatrix}$. Do you have one or two eigenvectors? This might be crucial. $\endgroup$
    – Federico Poloni
    Commented Dec 23, 2011 at 13:12
  • $\begingroup$ I'm sorry for big delay. Ok, for simplicity I can assume that matrix $G(x)$ has equal algebraic and geometric multiplicities for its eigenvalues. $\endgroup$
    – spk
    Commented Jan 18, 2012 at 10:24
  • $\begingroup$ @spk eigenvalues roots of det(y-M)=0. Your definition of eig.val is somewhat strange for me unless G(x) = x-M... $\endgroup$
    – Alexander Chervov
    Commented Jan 18, 2012 at 10:24
  • $\begingroup$ @Alexander: it a strange thing, I'm ok with your claim. But I consider a matrix of polynomials that is unusual case too. $\endgroup$
    – spk
    Commented Jan 18, 2012 at 10:35

1 Answer 1

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Unless I misunderstand the question, this seems to be false for $1\times 1$ matrices. In this case you are asking whether every polynomial $p(x)$ which vanishes whenever $q(x)$ vanishes is a constant multiple of $q(x).$ This is obviously false. Is there some condition missing?

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  • $\begingroup$ As I understood the question, in the $1\times 1$ case $v_{i,j}=1$ for every $i,j$, so the question becomes "is every polynomial $r(x)$ such that $r(x) \cdot 1=0$ a constant multiple of $g(x)$?", which is trivially true. $\endgroup$
    – Federico Poloni
    Commented Dec 23, 2011 at 20:56
  • $\begingroup$ @Federico: My understanding is that $r(x)$ has to be $0$ ONLY at the roots of $g(x)$ (the OP only has equalities at the eigenvalues). $\endgroup$
    – Igor Rivin
    Commented Dec 23, 2011 at 21:34
  • $\begingroup$ @Igor: you are perfectly right, sorry. $\endgroup$
    – Federico Poloni
    Commented Dec 23, 2011 at 23:56
  • $\begingroup$ @Igor: I'm sorry for big delay. Thank you for this point. Yes, one should consider polynomial coefficients for linear combination of rows of $G(x)$ but not constants as it were earlier. $\endgroup$
    – spk
    Commented Jan 18, 2012 at 10:32
  • $\begingroup$ Igor's point still holds. With the new version, you are asking whether every $p(x)$ which vanishes whenever $q(x)$ vanishes is a polynomial multiple of $q(x)$. This is still false, take e.g. $p(x)=x$, $q(x)=x^2$. Can you please double-check everything so that it works at least in the $1\times 1$ case? $\endgroup$
    – Federico Poloni
    Commented Jan 18, 2012 at 11:36

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