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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?2012-10-23
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    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 64.2012-10-23
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    Fix in what sense? What is the action of what on what?2012-10-23
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    ...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 64.2012-10-23
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    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)2012-10-23

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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?2012-10-23
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    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.2012-10-23
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    Indeed, the main point of my answer is that *the number of matrices does **not** depend on the vector*!2012-10-23
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    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 vector2012-10-23
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    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.2012-10-23
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    ...and I get to $a + b = 1, e + f = 1, i + j = 0, m + n = 0$. This allows 8 vars to be free and each of the other equations are independent and 2 solutions each...2012-10-23
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    and I guess I have it since $2^8$ times $2^4$ = $2^{12}$. I see2012-10-23