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Math Help - Find a basis for the subspace R3 defined by {(3s,5b,2a-b):a,b R}

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    Find a basis for the subspace R3 defined by {(3s,5b,2a-b):a,b R}

    Find a basis for the subspace R3 defined by

    {(3a,5b,2a-b): a,b are members of R}
    Last edited by pasleycakes; October 27th 2009 at 09:50 AM.
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  2. #2
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by pasleycakes View Post
    Find a basis for the subspace R3 defined by

    {(3s,5b,2a-b): a,b are members of R}
    What is " s"?

    When you are given a set in a general form you can quite easily come up with a spanning set. You just split up the general form into vectors containing one and only one of the variables. For instance, " (0, 5b, -b)", and then take out the variables and you will have yous spanning set. So here " (0, 5b, -b)" would become " (0, 5, -1)" which is a member of your spanning set.

    You then have to ask if this spanning set is a basis or not. That is, is it linearly independent.
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  3. #3
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    Quote Originally Posted by pasleycakes View Post
    Find a basis for the subspace R3 defined by

    {(3a,5b,2a-b): a,b are members of R}
    So x= 3a, y= 5b, z= 2a- b.


    Letting a= 1, b= 0, x= 3, y= 0, z= 2 so <3, 0, 2> is such a vector.

    Letting a= 0, b= 1, x= 0, y= 5, z= -5 so <0, 1, -5> is such a vector.

    I will leave it to you to show that those two vectors span the subspace and are independent.
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    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    Letting a= 0, b= 1, x= 0, y= 5, z= -5 so <0, 1, -5> is such a vector.
    This should read "Letting a= 0, b= 1, x= 0, y= 5, z= -1 so <0, 5, -1> is such a vector."
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    \left[\begin{array}{c} 3a \\ 5b \\2a-b \end{array}\right]=\left[\begin{array}{c} 3a \\ 0 \\2a \end{array}\right]+\left[\begin{array}{c} 0 \\ 5b \\-b \end{array}\right]=a\left[\begin{array}{c} 3 \\ 0 \\2 \end{array}\right]+b\left[\begin{array}{c} 0 \\ 5 \\-1 \end{array}\right]

    So basis \mathfrak{B}=span(\left[\begin{array}{c} 3 \\ 0 \\2 \end{array}\right], \left[\begin{array}{c} 0 \\ 5 \\-1 \end{array}\right])
    Last edited by math2009; October 28th 2009 at 04:48 PM.
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    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by math2009 View Post

    So basis \mathfrak{B}=span(\left[\begin{array}{c} 3 \\ 0 \\2 \end{array}\right], \left[\begin{array}{c} 0 \\ 5 \\-1 \end{array}\right])
    No, for two reasons. Firstly, you assume that the set is linearly independent. This needs to be proven. Secondly, the basis is not the span of those vectors - that's the whole vector space! V=span(\mathfrak{B}).
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    Quote Originally Posted by math2009 View Post
    \left[\begin{array}{c} 3a \\ 5b \\2a-b \end{array}\right]=\left[\begin{array}{c} 3a \\ 0 \\2a \end{array}\right]+\left[\begin{array}{c} 0 \\ 5b \\-b \end{array}\right]=a\left[\begin{array}{c} 3 \\ 0 \\2 \end{array}\right]+b\left[\begin{array}{c} 0 \\ 5 \\-1 \end{array}\right]

    So basis \mathfrak{B}=span(\left[\begin{array}{c} 3 \\ 0 \\2 \end{array}\right], \left[\begin{array}{c} 0 \\ 5 \\-1 \end{array}\right])
    Sorry, it should be \mathfrak{B}=(\left[\begin{array}{c} 3 \\ 0 \\2 \end{array}\right], \left[\begin{array}{c} 0 \\ 5 \\-1 \end{array}\right]).
    It's obvious, vectors are linear independent.
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  8. #8
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by math2009 View Post
    Sorry, it should be \mathfrak{B}=(\left[\begin{array}{c} 3 \\ 0 \\2 \end{array}\right], \left[\begin{array}{c} 0 \\ 5 \\-1 \end{array}\right]).
    It's obvious, vectors are linear independent.
    Well yeah, but you have to say it!
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