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Let $S$ be the set of all $2 × 3$ real matrices each of whose entries is $1, 0,$ or $−1.$ (There are $36$ matrices in $S.$) Recall that the column space of a matrix $M$ in $S$ is the subspace of $R^ 2$ (the vector space of $2×1$ real matrices) spanned by the three columns of $M.$ For two elements $M$ and $M'$ in $S$, let us write $M ∼ M'$ if $ M$ and $M'$ have the same column space. Note that $∼$ is an equivalence relation. How many equivalence classes are there in $S$?

Doubt:-Is the number of equivalence classes the same as the number of Subaspaces of dimension $1$ or $2$? I used this formula: Here Number of $k$-dimensional subspaces in $V$

$|V|=2$ and $|F|=q=3$ $k=1,2$. Am I doing it correctly?

Math geek
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    I think there are $3^6$ matrices in $S$, not $36$. Also, that counting-subspaces-post probably isn't going to help you, as that's specifically for vector spaces over finite fields, while you have vector spaces over what seems to be $\Bbb R$. – Arthur Dec 05 '19 at 18:38
  • But Answer given is that $6$. – Math geek Dec 05 '19 at 18:43
  • @Mathgeek Counting the number of matrices in $S$ is not the same as the number of equivalence classes. While there may be $6$ equivalence classes (I haven't bothered trying to solve this yet), the number of matrices in your set can be found quite easily using introductory counting techniques using in particular the rule of product. For each unknown, count how many options there are. Multiply the number of unknowns for each step together. So, pick the first row first column entry, there are three options. Then pick the first row second column entry, again 3 options, so on so forth – JMoravitz Dec 05 '19 at 18:48
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    The subspaces are 0,$\mathbb{R}^2$,$\langle(1,0)\rangle$,$\langle(0,1)\rangle$,$\langle(1,1)\rangle$,$\langle(1,-1)\rangle$. – almagest Dec 05 '19 at 18:48

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The only column vectors that it is possible to get in the matrix are $a=(0,0),b=(1,0),c=(0,1)$, $d=(1,1),e=(-1,1)$ or one of these times -1. If all the entries are $a$, then we get the subspace 0. If we have both $b$ and $c$, then we get $\mathbb{R}^2$. If we have nothing but $a,b,-b$ then we get $\langle(1,0)\rangle$. If we have nothing but $a,c,-c$ we get $\langle(0,1)\rangle$. If we have nothing but $a,\pm d$ we get $\langle(1,1)\rangle$. If we have nothing but $a,\pm e$ we get $\langle(-1,1)\rangle$. Other combinations (like $b$ and $d$) give us $\mathbb{R}^2$. It is easy to see there are no other possibilities. So we have 6 equivalence classes.

almagest
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