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?
Element in a 4 by 4 matrix mult-group over $\mathbb{F}_2$ such that it has a stabilizer subgroup of order 64
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group-theory
matrices
binary
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1Stabilizer in what sense? Do you mean centralizer? In the matrix ring? – 2012-10-23
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0I 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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1Fix in what sense? What is the action of what on what? – 2012-10-23
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0...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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0Please 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
1 Answers
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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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0If 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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0No. 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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0Indeed, 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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0I 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 – 2012-10-23
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0Silly 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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0...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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0and I guess I have it since $2^8$ times $2^4$ = $2^{12}$. I see – 2012-10-23