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

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Discussion Overview

The discussion centers on the proof of whether the product of two full-rank matrices results in a full-rank matrix. The scope includes theoretical considerations and mathematical reasoning related to matrix rank and properties of null spaces.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested, Mathematical reasoning

Main Points Raised

  • One participant questions how to prove that the product matrix C = AB is full-rank given matrices A and B of appropriate dimensions.
  • Another participant suggests exploring the relationship between the null spaces of the original matrices and the product, proposing that demonstrating a nullity of zero would imply full rank.
  • A participant seeks clarification on the definition of "full rank," leading to a definition based on the relationship between rank and the number of rows and columns.
  • One participant asserts that the original claim may be false and encourages finding a counterexample.
  • Another participant mentions that the multiplicative property of determinants could be relevant, but notes that this property applies specifically to square matrices, which may not be the case for the matrices in question.

Areas of Agreement / Disagreement

Participants express differing views on the validity of the original claim regarding the product of full-rank matrices, with some suggesting it may not hold true. There is no consensus on the proof or the conditions under which the claim may be valid.

Contextual Notes

There are unresolved assumptions regarding the dimensions of the matrices involved and the implications of their ranks. The discussion also highlights the need for clarity on definitions and properties applicable to non-square matrices.

leden
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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?
 
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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.
 
What do you mean with "full rank" in the first place?
 
By full-rank, I mean rank(M) = min{num_cols(M), num_rows(M)}.
 
leden said:
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.
 
just off the top of my head, it seems to follow from the multiplicative property of determinants. (assuming you state it correctly.)
 
mathwonk said:
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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