Q: Find all subsets of the set that forms a basis for .
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??
July 15th 2009, 12:06 PM
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 a three dimensional vector space. Do this for each of the four possibilities.
July 15th 2009, 02:35 PM
I see, so something like this...
Therefore, we have three subsets that form a basis for . They are...
, , .
July 15th 2009, 09:20 PM
As long as those determinants are correct your solution is correct.