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Math Help - Matrix Rank Proof - I don't understand the textbook solution

  1. #1
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    Matrix Rank Proof - I don't understand the textbook solution

    The question:
    A is an c x d matrix. B is a d x k matrix.

    If rank(A) = d and AB = 0, show that B = 0.


    Solution given:
    My textbook has a solution but I don't understand it:

    The rank of A is d, therefore A is not the zero matrix. (I asked my prof why d can't be equal to zero, he said it just couldn't...?)

    If you left multiply A by some elementary matrix to bring it to row echelon form, you get a matrix that looks like:
    [ 1 * * * ... *
    0 1 * * ... *
    0 0 1 * ... *
    0 0 0 0 ... 0] (NOTE: * are arbitrary numbers)

    And we will write B as a column (1 x k), consisting of [B1, ... , Bd]T

    Multiply A and B together, and you get a column that looks like [R1, R2, ... 0, 0, 0]T

    For AB = 0, then Ri = 0. Then since A is not zero, B is 0.

    This proof seems to make no sense. Why are we writing B as 1 x k, when the question says B is d x k?

    And I don't really get how this actually proves anything... =\
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    An easier proof is to notice that the linear transformation represented by A is injective; hence the only solution to Ax=0 is x=0. Since AB=0, B must take everything to the zero vector.
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  3. #3
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    Oh yes, I did notice that. I figured understanding the solution given might come in helpful though! I'd still like to try and understand it
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  4. #4
    MHF Contributor Bruno J.'s Avatar
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    Good! Then you are on the right track.

    I think that when they write B as 1 \times k, the elements B_1, \dots, B_d are actually the columns of B. See if that helps...

    What is your textbook? Probably not the best. In my opinion there is no reason to see it in any other way than the one I stated, because it really points out what is actually going on theoretically underneath. The proof you quote is ugly and unilluminating, in my opinion.
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