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Math Help - Finding a two-dimensional span given three vectors.

  1. #1
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    Finding a two-dimensional span given three vectors.

    Find the values of k so that the vectors [3 -k -1 0}^T, [-1 2 -k -1]^T, [0 -1 3 -k]^T span a two-dimensional space.


    My intuition says that I should form a 4x3 matrix out of these vectors with the first vector as the first column, the second vector as the second column, etc.

    Then I should row-reduce until I have only 2 leading ones. This will give me the two vectors that span the two dimensional space, but I am having trouble reducing.

    Is this the right approach, or am I doing this incorrectly.
    Thanks very much for any help on this topic.
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  2. #2
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    k+1

    Quote Originally Posted by shonucic View Post
    Find the values of k so that the vectors [3 -k -1 0}^T, [-1 2 -k -1]^T, [0 -1 3 -k]^T span a two-dimensional space.


    My intuition says that I should form a 4x3 matrix out of these vectors with the first vector as the first column, the second vector as the second column, etc.

    Then I should row-reduce until I have only 2 leading ones. This will give me the two vectors that span the two dimensional space, but I am having trouble reducing.

    Is this the right approach, or am I doing this incorrectly.
    Thanks very much for any help on this topic.
    So if you want to span to be 2D then you need the matrix have exactly 2 ones when the matrix is in reduced row form.

    \begin{bmatrix} 3 & -1 & 0 \\ -(k+1) & 2 & -1 \\ 0 & -(k+1) & -(k-3)\end{bmatrix}

    Now (k+1)R_1+3R_2 gives the matrix

    \begin{bmatrix} 3 & -1 & 0 \\ 0& -k+5 & -1 \\ 0 & -(k+1) & -(k-3)\end{bmatrix}

    What value must k take to make the last to row dependant?
    Last edited by TheEmptySet; February 10th 2011 at 10:32 AM.
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  3. #3
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    Another way of looking at it: the span of a set of vectors has dimension two if there are two independent vectors such that all of the other vectors can be written as a linear combination of those two. Here, that just means that one is a linear combination of the other two.

    Let's try
    a\begin{bmatrix}3 \\ -k \\ -1 \\ 0\end{bmatrix}+ b\begin{bmatrix}-1 \\ 2 \\ -k \\ -1\end{bmatrix}= \begin{bmatrix} 0 \\ -1 \\ 3 \\ -k\end{bmatrix}
    \begin{bmatrix}3a- b \\ -ak+ 2 \\ -a- bk \\ -b\end{bmatrix}= \begin{bmatrix}0 \\ - 1 \\ 3 \\ -k\end{bmatrix}

    so we have four equations, 3a- b= 0, -ak+ 2= -1, a- bk= 3, and -b= -k to solve for a, b, and k. From the last equation, k= b. Then 3a- k= 0 so a= k/3 and a- bk= a- k^2= 3 so that a= k/3= 3- k^2.
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    http://www.mathhelpforum.com/math-he...dbea0df917.png

    How did you get that matrix from the initial three vectors?
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