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Linear Algebra: Linear Independence and writing Matrices as linear combinations

  1. Mar 23, 2009 #1
    1. The problem statement, all variables and given/known data

    If linearly dependent, write one matrix as a linear combination of the rest.

    [tex]\left[\begin{array}{cc} 1&1 \\ 2&1 \end{array}\right][/tex] [tex]\left[\begin{array}{cc} 1&0 \\ 0&2 \end{array}\right][/tex] [tex]\left[\begin{array}{cc} 0&3 \\ 2&1 \end{array}\right][/tex] [tex]\left[\begin{array}{cc} 4&6 \\ 8&6 \end{array}\right][/tex]

    2. Relevant equations

    3. The attempt at a solution

    Choosing coefficients a, b, c, d for the matrices, respectively, I set up 4 equations and 4 unknowns:

    row 1 column 1: [tex]a + b + 4d = 0[/tex]
    row 1 column 2: [tex]a+ 3c + 6d = 0[/tex]
    row 2 column 1: [tex]2a + 2c + 8d = 0[/tex]
    row 2 column 2: [tex]a + 2b + c + 6d = 0[/tex]

    From this, I create a 4x5 matrix and using Gauss-Jordan elimination arrive at:

    Edit: this is wrong. See below

    [tex]\left[\begin{array}{ccccc}
    1&0&0&2&0 \\
    0&1&0&2&0 \\
    0&0&1&\frac{4}{3}&0 \\
    0&0&0&0&0 \end{array}\right][/tex]

    From this, it is clear that the 4 matrices are linearly dependent.

    What I do not understand: how do I, given this last matrix, write one of the matrices as a linear combination of the others?

    Thanks
     
    Last edited: Mar 23, 2009
  2. jcsd
  3. Mar 23, 2009 #2

    Mark44

    Staff: Mentor

    Your last matrix means this:
    a = -2d
    b = -2d
    c = -4/3 d

    From this we can deduce that d is arbitrary, so let d = 1.
    Then a = -2, b = -2, and c = -4/3.

    Put these values into the equation you set up to determine linear dependence and you'll see that one of them is a particular linear combination of the other three matrices.

    BTW, when you solved the 4 equations in 4 unknowns, you could have used a 4 x 4 matrix instead of a 4 x 5 augmented matrix. Your 5th column started as all zeroes and never changed.
     
  4. Mar 23, 2009 #3
    I attempted to check this, but it appears to not work, as the equation generated is not true.

    [tex]-2a - 2b + \frac{4}{3}c = d[/tex]

    Have I reduced the 4x5 matrix incorrectly or is the problem more fundamental?
     
    Last edited: Mar 23, 2009
  5. Mar 23, 2009 #4
    edit: read the equations wrong
     
  6. Mar 23, 2009 #5

    Mark44

    Staff: Mentor

    It's possible you made a mistake in reducing your 4x5 matrix. If so, that will affect the values of a, b, c, and d.

    Assuming for the moment that your work was correct, these values are solutions of the equation
    a*M1 + b*M2 + c*M3 + d*M4 = 0, where M1 etc are the 2x2 matrices in your first post, and NOTof the equation -2a -2b + 4/3 c = d.

    The a, b, c, and d values should tell you which matrix is a linear combination of the others.
     
  7. Mar 23, 2009 #6
    Going from the 4 equations/unknowns to the matrix I made a silly mistake causing the row reduced matrix to be incorrect.

    The (I think) correct row reduced matrix is

    [tex]
    \left[\begin{array}{ccccc}
    1&0&0&3 \\
    0&1&0&1 \\
    0&0&1&1 \\
    0&0&0&0 \end{array}\right]
    [/tex]

    From here, I thought I would read out of the matrix the following equations:

    [tex]a = -3d[/tex]
    [tex]b = -d[/tex]
    [tex]c = -d[/tex]
    where d is any arbitrary constant.

    Assigning d = 1, and checking to see if [tex]a * M1 + b * M2 + c * M3[/tex] was in fact [tex]d * M4[/tex], I end up with exactly the wrong signs:

    [tex]-3\left[\begin{array}{cc}1&1\\2&1\end{array}\right] + -1\left[\begin{array}{cc}1&0\\0&2\end{array}\right] + -1\left[\begin{array}{cc}0&3\\2&1\end{array}\right] = \left[\begin{array}{cc}-4&-6\\-8&-6\end{array}\right][/tex]

    This must mean that a should equal 3d, b should equal d, c should equal d, without the negatives, but I do not understand why.
     
  8. Mar 23, 2009 #7
    After thinking about it more, I guess that the a, b, c columns should be partitioned from the d column since I intend to write d * M4 in terms of the other matrices and their respective coefficients.

    This would explain the sign error from before.

    Can someone confirm?
     
  9. Mar 24, 2009 #8

    Mark44

    Staff: Mentor

    Yes, that's the problem. The a, b, c, and d are coefficients in this equation:
    a*M1 + b*M2 + c*M3 + d*M4 = 0. I checked and they work out.
     
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