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Linear Algebra:Rank

  1. Feb 21, 2013 #1
    1. The problem statement, all variables and given/known data
    Let A be the following 3 × 3 matrix:
    A =([4 2 6],[2 1 3],[2 1 3])
    i) Find the rank of A
    ii) Show that there exists an 3 × 2 matrix W, of rank 2, such that AW = 0.
    iii) Construct one such matrix W.



    2. Relevant equations




    3. The attempt at a solution

    I think the answer to part 1 is rank(A)=1 since row 2 and row 3 are the same and row 1 is just twice row 2 or twice row 3. So there is only one independent row.

    I'm not sure how to prove the existence of W however. Do I use the rank-nullity theorem in some way? i.e. rank(a)+nullity(a)=n where A is a mxn matrix. In this case the nullity(A)=1. Any help would be much appreciated.
     
    Last edited: Feb 21, 2013
  2. jcsd
  3. Feb 21, 2013 #2

    HallsofIvy

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    Yes, the rank of A is 1. Frankly, I would "prove" the existence of W by finding W! (Especially since the third part of the problem asks you to find it.)

    You want a "3 x 2" matrix, W, such that AW= 0. Okay:
    [tex]\begin{bmatrix}4 & 2 & 6 \\ 2 & 1 & 3 \\ 2 & 1 & 3\end{bmatrix}\begin{bmatrix}a & b \\ c & d \\ e & f\end{bmatrix}= \begin{bmatrix}4a+ 2c+ 6e & 4b+ 2d+ 6f \\ 2a+ c+ 3e & 2b+ d+ 3f \\ 2a+ c+ 3e & 2b+ d+ 3f \end{bmatrix}= \begin{bmatrix}0 & 0 \\ 0 & 0 \\ 0 & 0\end{bmatrix}[/tex]

    So we have the equations 4a+ 2c+ 6e= 0, 4b+ 2d+ 6f= 0, 2a+ c+ 3e= 0, 2b+ d+ 3f= 0, 2a+ c+ 3e= 0, and 2b+d+ 3f= 0. Of course, the third and fifth equations are the same as are the fourth and sixth equation. In fact, those reduce to only two independent equations. Solve those so that the W matrix has rank 2.
     
  4. Feb 21, 2013 #3

    jbunniii

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    Yes, that's right.
    You have ##\textrm{rank}(A) = 1##, and ##n = 3##, so ##\textrm{nullity}(A) = 2##, not ##1##. That means that you can find two linearly independent vectors, say ##w_1## and ##w_2##, in the kernel (null space) of ##A##. That is to say, they will satisfy ##Aw_1 = 0## and ##Aw_2 = 0##. If you construct your ##W## matrix so that its two columns consist of ##w_1## and ##w_2##, then you can easily check that this will give you ##AW = 0##.
     
  5. Feb 21, 2013 #4

    jbunniii

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    Halls beat me to it! And his method will be more useful for the next part as it shows you how to find ##w_1## and ##w_2##. Mine is just an existence proof, but hopefully it makes it clear why it should be possible to find such a ##W##.
     
  6. Feb 23, 2013 #5
    THat's great, thank you both for your help
     
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