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Hello, all!

Let $\underset{l \times l}{A(x)}$ be square polynomial matrix over $GF(q)[x]$, where $q$ is a prime power. Let $x_i \in GF(q^m)$ ($x_i \not= 0$) be eigenvalue of $A(x)$: $det(A(x_i)) = 0$. Let $\underset{l \times 1}{\mathbf{v}_{i, j}} \in GF(q^m)^l$ be an right eigenvector that corresponds to $x_i$. So we have $A(x_i) \cdot \mathbf{v}_{i,j} = \underset{l \times 1}{\mathbf{0}}$.

How it could be proved that there are existed no polynomial vector $\mathbf{c}(x) = \left( c_0(x), c_1(x), \ldots, c_{l-1}(x) \right)$ that for all $x_i$ and $\mathbf{v}_{i,j}$ $\mathbf{c}(x_i) \cdot \mathbf{v}_{i,j} = 0$ and that does not belong to $GF(q)[x]$-linear space generated by $A(x)$?

I suppose, existence of no such polynomial should be a nice guess because of correspondence to eigendecomposition notion from classic linear algebra. So I suppose that system of polynomial matrix eigenvalues and its right eigenvectors and original polynomial matrix are in one-to-one correspondence. But proof for this is not clear.

Thank you!

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    $\begingroup$ You are really going to need to give a ton of additional material. What is $q,$ what is the overall setting, what work have you done so far, why do you think "are existed no..." might be true? If you have no idea what is going on, it is simply unfair of you to ask strangers to put their effort into this. $\endgroup$
    – Will Jagy
    Commented Sep 4, 2011 at 19:29
  • $\begingroup$ Is it true? Take in the scalar case $A(x)=x^2$, $x_1=1$, $v_1=1$, $c(x)=x$. Then $c(x)v$ vanishes, but $c(x)$ is not spanned by $x^2$. $\endgroup$
    – Federico Poloni
    Commented Sep 7, 2011 at 12:25
  • $\begingroup$ Do you mean $x_1 = 0$? I have to consider only non-zero elements that vanishes the determinant of $A(x)$. Thank you, I put fix to main text. $\endgroup$
    – spk
    Commented Sep 7, 2011 at 13:34
  • $\begingroup$ Yup, I meant $x_1=0$. In answer to your fix, the point here is not an element being zero (you can get other counterexamples by a change of variable $y=x+a$ for each $a$), but rather multiple roots. $\endgroup$
    – Federico Poloni
    Commented Sep 7, 2011 at 20:00
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    $\begingroup$ Incidentally, I hope you know (or realized) that eigenvectors of matrix polynomials are not linearly independent (a degree-$d$ has $dl$ eigenpairs, and thus its eigenvectors are simply too many to be linearly independent). Maybe this answers your question. $\endgroup$
    – Federico Poloni
    Commented Sep 7, 2011 at 20:03

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