Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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!

share|improve this question
1  
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. –  Will Jagy Sep 4 '11 at 19:29
    
Thank you! I put some fixes to my quesitions. –  spk Sep 5 '11 at 10:30
    
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 $x^2$. –  Federico Poloni Sep 7 '11 at 12:25
    
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. –  spk Sep 7 '11 at 13:34
1  
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. –  Federico Poloni Sep 7 '11 at 20:03
show 1 more comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.