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Math Help - Setup help

  1. #1
    Senior Member Danneedshelp's Avatar
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    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??
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  2. #2
    Super Member Gamma's Avatar
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    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.
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  3. #3
    Senior Member Danneedshelp's Avatar
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    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)\}.
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  4. #4
    Super Member Gamma's Avatar
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    Right on

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