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Math Help - linear algebra

  1. #1
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    linear algebra

    if u v and w are vectors that are linearly independent will u-v v-w and u-w be linearly independent?
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  2. #2
    Senior Member tukeywilliams's Avatar
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    So you are given  a \vec{u} + b \vec{v} + c \vec{w} = 0 \Rightarrow a = b = c = 0 .

    Then you want to show that  x(\vec{u} - \vec{v}) + y(\vec{v}- \vec{w}) + z(\vec{u} - \vec{w}) = 0  \Rightarrow x = y = z = 0 is true or false.


    Expand out:  x \vec{u} - x \vec{v} + y \vec{v} - y \vec{w} + z \vec{u} - z \vec{w} = 0 .


    Then  (x+z) \vec{u} + (y-x) \vec{v} + (-y-z) \vec{w} = 0
    But that doesn't imply that  x = y = z = 0 . Only that  a = b = c = 0 , where  a  = x+z, \ b = y-x, \ c = -y-z . So they are linearly dependent.
    Last edited by tukeywilliams; September 12th 2007 at 10:00 PM.
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  3. #3
    Senior Member tukeywilliams's Avatar
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     A = \begin{bmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 2 & 1 \\ 0 & 3 & 2 & 1 \\ 0 & 1 & -1 & 1 \end{bmatrix}



     \det A = -1 \begin{bmatrix} 0 & 0 & 2 \\ 0 & 3 & 2 \\ 0 & 1 & -1 \end{bmatrix} = -1(-2)(0) =0


    Linearly dependent.
    Last edited by tukeywilliams; September 12th 2007 at 10:05 PM.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by tukeywilliams View Post
     A = \begin{bmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 2 & 1 \\ 0 & 3 & 2 & 1 \\ 0 & 1 & -1 & 1 \end{bmatrix}



     \det A = -1 \begin{bmatrix} 0 & 0 & 2 \\ 0 & 3 & 2 \\ 0 & 1 & -1 \end{bmatrix} = -1(-2)(0) = 0


    Linearly independent.
    i believe they are dependent if the Wronskian is zero.

    recall that a set of two or more vectors is linearly dependent if and only if at least one of the vectors in the set can be expressed as the linear combination of finitely many other vectors in the set.

    now, note that:

    u - v = u - v + w - w = (u - w) - (v - w)

    v - w = v - w + u - u = (u - w) - (u - v)

    and

    u - w = u - w + v - v = (u - v) + (v - w)

    so we see that each of the vectors can be expressed as a linear combination of the other two, thus the vectors u - v, v - w, and u - w form a linearly dependent set of vectors
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  5. #5
    Senior Member tukeywilliams's Avatar
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    yeah I did it a different way and got that they were linearly dependent also (my edit). You are correct, I forgot to edit the last sentence.
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by tukeywilliams View Post
    yeah I did it a different way and got that they were linearly dependent also (my edit). You are correct, I forgot to edit the last sentence.
    how did you set up those matrices?
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  7. #7
    Senior Member tukeywilliams's Avatar
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    that was a different problem. My solution to the first problem is post #2.
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  8. #8
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    Quote Originally Posted by tukeywilliams View Post
    that was a different problem. My solution to the first problem is post #2.
    o ok. yes, post #2 is a nice solution. did the poster delete the problem you responded to in post #3?
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  9. #9
    Senior Member tukeywilliams's Avatar
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    yeah I think he did.
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  10. #10
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    FIND the coefficients of the linear combination.
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  11. #11
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by lord12 View Post
    FIND the coefficients of the linear combination.
    what are you talking about? what linear combination?
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  12. #12
    Senior Member tukeywilliams's Avatar
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    in post #3, the determinant is 0, which implies that those vectors are linearly dependent.

    Hes asking how you would find the coefficients.
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  13. #13
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    Quote Originally Posted by tukeywilliams View Post
    in post #3, the determinant is 0, which implies that those vectors are linearly dependent.

    Hes asking how you would find the coefficients.
    well, that question is for you then, i don't even know what the original question was
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  14. #14
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    the linear combination of the values in the matrix
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  15. #15
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    Quote Originally Posted by lord12 View Post
    the linear combination of the values in the matrix
    put the matrix in reduced row echelon form and you can read off the solutions from there. as the matrix looks now, it seems there will be a parameter for the first column. you really shouldn't delete your question
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