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Linear dependence

  1. Nov 2, 2008 #1
    (1,0,0) (-3,7,0) and (1,1,0)

    I'm trying to work out if these vectors are linearly independent or not.
    Intuitively i believe they are dependent as they span the xy-plane.. but then how do i work out the linear combinations.

    e.g:
    (1,0,0) = a(-3,7,0) + b(1,1,0) where a and b are real numbers.

    Ive tried writing the vectors as column vectors, then row reducing and i end up with

    [1 -10 0;0 7 1;0 0 0] (which has rank(2) which means they are linearly dependent)

    but as for where to go from here I'm lost.

    Any help greatly appreciated.
     
  2. jcsd
  3. Nov 2, 2008 #2

    HallsofIvy

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    Okay, so 1= -3a+ b and 0= 7a+b. Solve those two equations for a and b. I would recommend subtracting one equation from the other!

     
  4. Nov 2, 2008 #3
    Is there a way of doing this with matrices?

    So writing the vectors as columns and row reducing, and from here deciding what are the linear combinations?

    so [1 -3 1;0 7 1;0 0 0]

    becomes [1 -10 0;0 7 1;0 0 0]


    from here is there a way of getting the answer directly from this matrix?
     
  5. Nov 3, 2008 #4

    HallsofIvy

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    You prefer a harder way? Writing the two equations, 1= -3a+ b and 0= 7a+ b as an "augmented" matrix gives
    [tex]\left[\begin{array}{ccc}-3 & 1 & 1\\ 7 & 1 & 0\end{array}\right][/tex]
    where, as you can see, the vectors form the rows. That row reduces to
    [tex]\left[\begin{array}{ccc} 1 & 0 & -\frac{1}{10} \\ 0 & 1 & \frac{7}{10}\end{array}\right[/tex]
    showing that the equations have a solution (a= -1/10 and b= 7/10) and so the vectors are dependent. As for your initial method, the very fact that your matrix has rank only two tells you that the vectors are dependent.

    Writing the
     
    Last edited: Nov 3, 2008
  6. Nov 3, 2008 #5
    Once again thank you! you are my new Walter Lewin!
     
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