# Setup help

• Jul 15th 2009, 11:58 AM
Danneedshelp
Setup help
Q: Find all subsets of the set that forms a basis for $\mathbb{R}^{3}$.

$S=\{(1,3,-1),(-4,1,1),(-2,7,-3),(2,1,1)\}$

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??
• Jul 15th 2009, 12:06 PM
Gamma
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 $\mathbb{R}^3$ a three dimensional vector space. Do this for each of the four possibilities.
• Jul 15th 2009, 02:35 PM
Danneedshelp
I see, so something like this...

$S=\{(1,3,-2),(-4,1,1),(-2,7,-3),(2,1,1)\}$

$\det{\begin{pmatrix}1&-4&-2\\3&1&7\\-2&1&-3\end{pmatrix}}=0$

$\det{\begin{pmatrix}1&-2&2\\3&7&1\\-2&-3&1\end{pmatrix}}=30$

$\det{\begin{pmatrix}1&-4&2\\3&1&1\\-2&1&1\end{pmatrix}}=30$

$\det{\begin{pmatrix}-4&-2&2\\1&7&1\\1&-3&1\end{pmatrix}}=-60$

Therefore, we have three subsets that form a basis for $\mathbb{R}^{3}$. They are...

$S_{1}=\{(1,3,-2),(-2,7,-3),(2,1,1)\}$, $S_{2}=\{(1,3,-2),(-4,1,1),(2,1,1)\}$, $S_{3}=\{(-4,1,1),(-2,7,-3),(2,1,1)\}$.
• Jul 15th 2009, 09:20 PM
Gamma
Right on
As long as those determinants are correct your solution is correct.