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Proving that the product of two full-rank matrices is full-rank

  1. Sep 19, 2012 #1
    Say I have a mxn matrix A and a nxk matrix B. How do you prove that the matrix C = AB is full-rank, as well?
  2. jcsd
  3. Sep 19, 2012 #2


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    Hey leden and welcome to the forums.

    For this problem, I'm wondering if you can show that the null-space for the final space is directly related to the null-spaces of the original two systems (like for example if they are additive or have some other relation).

    If this is possible, then if you can show that the nullity is zero then you have shown it has full rank.
  4. Sep 19, 2012 #3
    What do you mean with "full rank" in the first place?
  5. Sep 19, 2012 #4
    By full-rank, I mean rank(M) = min{num_cols(M), num_rows(M)}.
  6. Sep 19, 2012 #5
    In that case, the statement in the OP is false. Try to find a counterexample.
  7. Sep 22, 2012 #6


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    just off the top of my head, it seems to follow from the multiplicative property of determinants. (assuming you state it correctly.)
  8. Sep 22, 2012 #7


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    That's true for square matrices, but the OP's matrices are not necessariily square (again, assuming it was stated correctly).
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