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Homework Help: Linear Algebra Proof (rank)

  1. Feb 8, 2010 #1
    1. The problem statement, all variables and given/known data
    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.

    2. Relevant equations

    3. The attempt at a solution
    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? It says in the question B is d x k! Also if A is not zero then why can't you say right off the bat that AB = 0 implies B =0?
  2. jcsd
  3. Feb 8, 2010 #2
    because these are matrices not numbers. for example
    Code (Text):

    A= [0 1
       0 0]

    B=[1 0
       0 0]
    AB=0 yet neither A or B are 0.

    as to why they say 'write B as 1xk', maybe they mean write Bv (i.e. B times an arbitrary vector) as a 1xk?
    Last edited: Feb 8, 2010
  4. Feb 8, 2010 #3
    But when A is in row echelon form and you multiply it by some B, the because the solutions are zero the entries of B must be zero??
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