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I want to show that there exists an column or row vector with four entries in $\mathbb{F}_2$ such that there are 64 4 by 4 binary matrices $M$ where $Mv =v$, ie $M$ leaves $v$ fixed. ie, the stabilizer of $v$ has order 64. I have a hunch that the answer is upper triangular matrices; after all, with 4 by 4 matrices, these would leave leave the components above the diagonal to be varied while the ones at the diagonal or below could be fixed. However, I do not know how to express this idea mathematically without resorting to an exhaustive demonstration of matrix multiplication. How can I use group theory to help me out here?

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Stabilizer in what sense? Do you mean centralizer? In the matrix ring? – Mariano Suárez-Alvarez Oct 23 '12 at 5:51
I mean that I want to find an element st the # of elements in the mult group of 4 by 4 matrices that fix the element is 64. – tacos_tacos_tacos Oct 23 '12 at 5:52
Fix in what sense? What is the action of what on what? – Mariano Suárez-Alvarez Oct 23 '12 at 5:53
...Right. I just realized I mis understood what I was trying to prove entirely. I'm actually trying to show there is a vector $v$ such that the number of matrices $M$ where $Mv = v$ is 64. – tacos_tacos_tacos Oct 23 '12 at 5:55
Please add all clarifications to the question itself :-) (and be sure to be explicit about where those matrices $M$ you want to count are taken from) – Mariano Suárez-Alvarez Oct 23 '12 at 5:56
up vote 1 down vote accepted

If $v$ is a non-zero vector, the number of matrices $M$ in $M_2(\mathbb F_2)$ such that $Mv=v$ does not depend on $v$.

Indeed, if $v$ and $w$ are non-zero vectors, there is an invertible matrix $A\in M_4(\mathbb F_2)$ such that $Av=w$, and then the function $$M\in\{X\in M_4(\mathbb F_2):Xv=v\}\longmapsto AMA^{-1}\in\{X\in M_4(\mathbb F_2):Xw=w\}$$ is a bijection.

To count the matrices fixing a non-zero vector, then, we can suppose that $v=(1,0,0,0)^t$. Then the matrices in question are those whose first column is precisely $(1,0,0,0)^t$, and there are $2^{12}$ of them.

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If we were to extend the example to $(1, 1, 0, 0)$, by the same logic, do we have that the first two columns of the matrix would have to be $(1, 0, 0, 0)$ and $(0, 1, 0, 0)$ meaning that you can vary the other $2^8$ entries? – tacos_tacos_tacos Oct 23 '12 at 6:10
No. Write down the set of equations on the entries of a matrix that say that $(1,1,0,0)$ is fixed. Then count the number of solutions. – Mariano Suárez-Alvarez Oct 23 '12 at 6:11
Indeed, the main point of my answer is that the number of matrices does not depend on the vector! – Mariano Suárez-Alvarez Oct 23 '12 at 6:13
I read your answer over a few times, but I still don't believe the result! I know you are correct as sure as 42.7k > 162, but I still don't see how you allow the first two columns to be anything but the given vector expressed with a 1 entry in diagonal for each 1 entry in the given vector – tacos_tacos_tacos Oct 23 '12 at 6:19
Silly reputation points have nothing to do with it. As I said, write down explicitely the 4 equations that say that a matrix $M$ is such that $M(1,1,0,0)=(1,1,0,0)$. Solve them. Count the solutions. – Mariano Suárez-Alvarez Oct 23 '12 at 6:21

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