
Setup help
Q: Find all subsets of the set that forms a basis for $\displaystyle \mathbb{R}^{3}$.
$\displaystyle S=\{(1,3,1),(4,1,1),(2,7,3),(2,1,1)\}$
A: well, I did a similar one of these and got the same answer as book, but with a bunch of negative 1's unlike their strictly positive solution. I figured it was ok, because subspaces are closed under scalar multiplication (then again, we are just talking about a subset). I don't know if that’s alright.
My approach is to set up a homogenous linear combination and solve for the constants. Then I set up my new linear combination and then solve for each vector in the set S. Then that gives me all my subspaces. I have to hope that S is linearly dependent for this to work. Is that the correct method??

Well there are 4C3=4 choices of 3 elements from that set of 4 elements (recall the ordering of the basis does not affect spanning or linear independence). So pick three of them and make each vector a column of a 3x3 matrix. Take the determinant, if it is 0 then they are not a basis since they are not linearly independent. If the determinant is not zero then they are linearly independent and there are three of them, so they span $\displaystyle \mathbb{R}^3$ a three dimensional vector space. Do this for each of the four possibilities.

I see, so something like this...
$\displaystyle S=\{(1,3,2),(4,1,1),(2,7,3),(2,1,1)\}$
$\displaystyle \det{\begin{pmatrix}1&4&2\\3&1&7\\2&1&3\end{pmatrix}}=0$
$\displaystyle \det{\begin{pmatrix}1&2&2\\3&7&1\\2&3&1\end{pmatrix}}=30$
$\displaystyle \det{\begin{pmatrix}1&4&2\\3&1&1\\2&1&1\end{pmatrix}}=30$
$\displaystyle \det{\begin{pmatrix}4&2&2\\1&7&1\\1&3&1\end{pmatrix}}=60$
Therefore, we have three subsets that form a basis for $\displaystyle \mathbb{R}^{3}$. They are...
$\displaystyle S_{1}=\{(1,3,2),(2,7,3),(2,1,1)\}$, $\displaystyle S_{2}=\{(1,3,2),(4,1,1),(2,1,1)\}$, $\displaystyle S_{3}=\{(4,1,1),(2,7,3),(2,1,1)\}$.

Right on
As long as those determinants are correct your solution is correct.