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Math Help - Expressing a non-trivial linear dependency relationship

  1. #1
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    Expressing a non-trivial linear dependency relationship

    I've been ask to see if a set of vectors is linearly independent, and if not, find a non-trivial linear dependency relationship.

    I believe I've correctly determined that the set is linearly dependent and found the solution, but I don't know how I'm meant to express the linearly dependency relationship??

    This is what I have so far:

    <br />
v_1 = \begin{pmatrix}2 \\ 1 \\ 2\end{pmatrix}<br />
v_2 = \begin{pmatrix}4 \\ 4 \\ 5\end{pmatrix}<br />
v_3 = \begin{pmatrix}6 \\ 1 \\ 5\end{pmatrix}<br />

    As a matrix, then reducing:

    <br />
\begin{bmatrix}<br />
2 & 4 & 6 \\<br />
1 & 4 & 1 \\<br />
2 & 5 & 5<br />
\end{bmatrix}<br />

    <br />
\begin{bmatrix}<br />
1 & 0 & 5 \\<br />
1 & 4 & 1 \\<br />
2 & 5 & 5<br />
\end{bmatrix}<br />

    <br />
 \begin{bmatrix}<br />
1 & 0 & 5 \\<br />
 1 & 4 & 1 \\<br />
0 & -3 & 3<br />
 \end{bmatrix}<br />

    <br />
 \begin{bmatrix}<br />
1 & 0 & 5 \\<br />
 1 & 1 & 4 \\<br />
0 & -3 & 3<br />
 \end{bmatrix}<br />

    <br />
  \begin{bmatrix}<br />
 1 & 0 & 5 \\<br />
0 & 1 & -1 \\<br />
 0 & -3 & 3<br />
  \end{bmatrix}<br />

    <br />
   \begin{bmatrix}<br />
  1 & 0 & 5 \\<br />
0 & 1 & -1 \\<br />
  0 & 0 & 0<br />
   \end{bmatrix}<br />

    If we let x_4 = t, then x_1 = -5t and x_2 = t. I believe this is the non-trivial linear dependency relationship, but I don't know if that the right way to leave it? Any help would be greatly appreciated.
    Last edited by drew.walker; September 8th 2009 at 04:56 AM. Reason: Accidentally clicked submit
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  2. #2
    Super Member
    Joined
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    If the vectors are linearly dependent there will be a set of coefficients, \alpha, \beta, \gamma, not all zero, such that

    \alpha v_1 + \beta v_2 + \gamma v_3 = 0.

    This you can translate into a set of three linear equations.

    For non-trivial solutions, it is necessary that the determinant of coefficients is zero, and this you have demonstrated, in which case the vectors are indeed linearly dependent.

    Easier though is to simply solve the equations using elementary row operations. I get

    \alpha = -5t, \beta = t, \gamma = t,

    where t is some arbitrary number.

    (If the equations turn out to be inconsistent, it means that the vectors are linearly independent.)
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  3. #3
    Junior Member
    Joined
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    So when it says "find a non-trivial linear dependency relationship", would this satisfactorily answer that part of the question:

    <br />
\alpha = -5t, \beta = t, \gamma = t,<br />
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