Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.