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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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    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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    and I guess I have it since $2^8$ times $2^4$ = $2^{12}$. I see2012-10-23