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Math Help - Determinant/Dimension problem

  1. #1
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    Determinant/Dimension problem

    This is from my text. My instructor listed this as one for us to think about to help us prep for the exam on Tuesday. I really am stuck though, which is scaring me.

    Here it goes:

    Consider the matrix A=\left[ \begin{array}{ccc}<br />
x_{1} & a_{1} & b_{1} \\<br />
x_{2} & a_{2} & b_{2} \\<br />
x_{3} & a_{3} & b_{3} \end{array} \right]<br />
\in M_{3}(k)<br />
    where the a's and b's are fixed and the x's may vary. Show that a set of all (x_{1} , x_{2}, x_{3}) such that det(A)=0 is a subspace of k^{3} of dimension at least 2.

    I know that the determinant is a linear transformation in each column, but I don't really know how that helps me. I asked a classmate who said she thought Rank-Nullity theorem might help.. but I am still clueless. Any and all explanation on how to prove this would be great. Thanks!
    Last edited by sabrepride; April 12th 2010 at 11:32 AM.
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    Can you show it's a subspace?

    Assuming that you can, if you want to show that the dimension is at least 2, consider the following two cases :

    (1) \vec a = (a_1, a_2, a_3) and \vec b =(b_1, b_2, b_3) are linearly independent : then the subspace contains the span of \vec a, \vec b, which is two dimensional (in fact, try to show that it's equal to the span of \vec a, \vec b);

    (2) \vec a, \vec b are linearly dependent : then the space is all of k^3! (Can you see why?)

    Can you generalize to higher dimensions?
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  3. #3
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    I feel really bad, but I don't see how to do any of that. How do I show a column of a matrix is a subspace, and of what?

    1. I understand that the span of {a,b} is two dimensional as then it would be a basis (smallest spanning set, largest linearly independent set). Since I am not sure on what the subspace is, I dont really know how it would be equal.

    2. Maybe once I get the first parts cleared up this will become more apparent. I dont see how this is true, and how {x1, x2, x3} or the determinant help me.
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  4. #4
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    DISREGARD: My computer posted multiple times
    Last edited by sabrepride; April 12th 2010 at 04:41 PM.
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  5. #5
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    DISREGARD: My computer posted multiple times
    Last edited by sabrepride; April 12th 2010 at 04:40 PM. Reason: duplicate posts
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  6. #6
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    Quote Originally Posted by Bruno J. View Post
    Can you show it's a subspace?

    Assuming that you can, if you want to show that the dimension is at least 2, consider the following two cases :

    (1) \vec a = (a_1, a_2, a_3) and \vec b =(b_1, b_2, b_3) are linearly independent : then the subspace contains the span of \vec a, \vec b, which is two dimensional (in fact, try to show that it's equal to the span of \vec a, \vec b);

    (2) \vec a, \vec b are linearly dependent : then the space is all of k^3! (Can you see why?)

    Can you generalize to higher dimensions?
    I feel really bad, but I don't see how to do any of that. How do I show a column of a matrix is a subspace, and of what?

    1. I understand that the span of {a,b} is two dimensional as then it would be a basis (smallest spanning set, largest linearly independent set). Since I am not sure on what the subspace is, I dont really know how it would be equal.

    2. Maybe once I get the first parts cleared up this will become more apparent. I dont see how this is true, and how {x1, x2, x3} or the determinant help me.
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  7. #7
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    Figured it out... don't need any help.
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