Consider that I have N different matrices $A_1,A_2,....A_N$ over finite fields, and all are non-singular. There exists a non-zero vector $x$ such that

$A_ix = A_jx$

for any (i,j) pair in the matrices.

Can you prove (or give a counter example) for the following claim

$x$ is an eigenvector for all $A_i$ for the eigen value 1.

Edit 1: I am looking at the case where the linear span of $A_i$ are not the entire of $k^{d\times d}$

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    $\begingroup$ I mean, if the linear span of those matrices is the whole $k^{d\times d}$ (which is entirely possible), then no such vector will exist. $\endgroup$ – Saucy O'Path Sep 14 '18 at 11:27
  • $\begingroup$ I agree. I will add that condition in my question. The linear span of the matrices need not be the whole of $k^{d\times d}$ $\endgroup$ – RaM1188 Sep 14 '18 at 11:29
  • $\begingroup$ My learned opinion is that ruling out objection as other people point them out is not the kind of procedure that leads somewhere. $\endgroup$ – Saucy O'Path Sep 14 '18 at 11:33
  • $\begingroup$ I am sorry. I dint intend that. I genuinely missed that point when i framed the question. $\endgroup$ – RaM1188 Sep 14 '18 at 11:35
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    $\begingroup$ Why should the eigenvalue be $1$? Consider matrices of the form $$\pmatrix{2&b\cr 0&1\cr}$$ with $b$ ranging over the field. $x=(1,0)^T$ is an eigenvector for all those matrices, but belongs to eigenvalue two. Obvious any other non-zero element can also serve as the eigenvalue. $\endgroup$ – Jyrki Lahtonen Sep 14 '18 at 11:47

Given any two vectors $x,y$ such that $x\ne 0$, the set of matrices $U(x,y)=\{A\in k^{d\times d}\,:\, Ax=y\}$ is an affine subspace of dimension $d^2-d$. What you are asking is whether there are some $x,y$ such that $\{A_1,\cdots, A_N\}\subseteq U(x,y)$. It is thus apparent that this cannot be the case if $\dim\operatorname{Span}(A_2-A_1,\cdots, A_N-A_1)>d^2-d$.

  • $\begingroup$ How is $U(x,y)$ a subspace? Take any two matrices $A,B \in U$. The addition of matrices $A+B \notin U$. Correct me if i am wrong. $\endgroup$ – RaM1188 Sep 14 '18 at 12:13
  • $\begingroup$ @RaM1188 Yes, I was wrong. It's an affine subspace, though, and the same considerations hold. $\endgroup$ – Saucy O'Path Sep 14 '18 at 12:14

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