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Thread: 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 $\displaystyle A=\left[ \begin{array}{ccc}
    x_{1} & a_{1} & b_{1} \\
    x_{2} & a_{2} & b_{2} \\
    x_{3} & a_{3} & b_{3} \end{array} \right]
    \in M_{3}(k)
    $
    where the $\displaystyle a$'s and $\displaystyle b$'s are fixed and the $\displaystyle x$'s may vary. Show that a set of all $\displaystyle (x_{1} , x_{2}, x_{3})$ such that $\displaystyle det(A)=0$ is a subspace of $\displaystyle k^{3}$ 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; Apr 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) $\displaystyle \vec a = (a_1, a_2, a_3)$ and $\displaystyle \vec b =(b_1, b_2, b_3)$ are linearly independent : then the subspace contains the span of $\displaystyle \vec a, \vec b$, which is two dimensional (in fact, try to show that it's equal to the span of $\displaystyle \vec a, \vec b$);

    (2) $\displaystyle \vec a, \vec b$ are linearly dependent : then the space is all of $\displaystyle k^3$! (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; Apr 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; Apr 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) $\displaystyle \vec a = (a_1, a_2, a_3)$ and $\displaystyle \vec b =(b_1, b_2, b_3)$ are linearly independent : then the subspace contains the span of $\displaystyle \vec a, \vec b$, which is two dimensional (in fact, try to show that it's equal to the span of $\displaystyle \vec a, \vec b$);

    (2) $\displaystyle \vec a, \vec b$ are linearly dependent : then the space is all of $\displaystyle k^3$! (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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