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!

too manyto be linearly independent). Maybe this answers your question. – Federico Poloni Sep 7 '11 at 20:03