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Two square invertible matrices, prove product is invertible

  1. Apr 29, 2008 #1
    [SOLVED] Two square invertible matrices, prove product is invertible

    1. The problem statement, all variables and given/known data
    If A and B are nxn matrices of rank n, prove that AB has rank n.


    2. Relevant equations
    There is a list in my textbook outlining equivalent statements, such as:
    - A is invertible
    - rank(A) = n
    - nullity(A) = 0
    - The column vectors of A are linearly independent.
    - and many others...


    3. The attempt at a solution
    I've been staring at equivalent statements, theorems, and examples, and cannot seem to think of an equivalency once I consider multiplying AB. I guess the main feat of this part would be to prove that, given rank(A) = n and rank(B) = n, AB has, once reduced, neither a row nor column that is zero, which would ultimately lead to rank(AB) = n. However, I'm not sure how to get there.
    I've tried aplying the statement of "The reduced row echelon form of A is the indentity matrix," but realized that A would be row equivalent but not equal to the identity matrix. Anyone have any ideas? Thanks.

    EDIT: I realized I can solve this using determinants, but up to the point in the textbook where this proof is requested, they had not been covered, so the question still stands.

    2nd edit: I had another realization: Since rank(A) = n and rank(B) = n, then by the Fundamental Theorem of Invertible Matrices, A is a product of elementary matrices and B is a product of elementary matrices. Therefore, AB is a product of elementary matrices and therefor rank(AB) = n. Is this right?
     
    Last edited: Apr 29, 2008
  2. jcsd
  3. Apr 29, 2008 #2
    I think I got it. See my second edit above. Comments welcome.
     
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