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

  • Thread starter leden
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  • #1
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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?
 

Answers and Replies

  • #2
chiro
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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.
 
  • #3
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What do you mean with "full rank" in the first place?
 
  • #4
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By full-rank, I mean rank(M) = min{num_cols(M), num_rows(M)}.
 
  • #5
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By full-rank, I mean rank(M) = min{num_cols(M), num_rows(M)}.

In that case, the statement in the OP is false. Try to find a counterexample.
 
  • #6
mathwonk
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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.)
 
  • #7
AlephZero
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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.)

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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