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

In summary: So the multiplicative property of determinants may not apply. In summary, to prove that the matrix C = AB is full-rank, we need to show that the null-space of C is directly related to the null-spaces of the matrices A and B, and that the nullity of C is equal to zero. This follows from the multiplicative property of determinants, but may not apply if the matrices A and B are not square.
  • #1
leden
7
0
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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  • #2
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
What do you mean with "full rank" in the first place?
 
  • #4
By full-rank, I mean rank(M) = min{num_cols(M), num_rows(M)}.
 
  • #5
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.
 
  • #6
just off the top of my head, it seems to follow from the multiplicative property of determinants. (assuming you state it correctly.)
 
  • #7
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).
 

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

1. How do you define a full-rank matrix?

A full-rank matrix is a square matrix in which all the columns (or rows) are linearly independent, meaning that no column (or row) can be written as a linear combination of the other columns (or rows). This results in the matrix having the maximum possible rank, which is equal to the number of rows (or columns).

2. What is the significance of proving that the product of two full-rank matrices is full-rank?

Proving that the product of two full-rank matrices is full-rank is important in various fields such as linear algebra, statistics, and machine learning. It allows for the creation of new matrices with the desired properties, as well as the development of algorithms and techniques for solving complex problems.

3. Can two matrices with different dimensions both be full-rank?

Yes, two matrices with different dimensions can both be full-rank. The rank of a matrix is determined by its number of linearly independent columns (or rows), not its size. Therefore, a square matrix and a rectangular matrix can both be full-rank if they meet the criteria of having linearly independent columns (or rows).

4. Is the product of two full-rank matrices always full-rank?

Yes, the product of two full-rank matrices is always full-rank. This is because the product of two matrices can only have a rank equal to or less than the minimum of the ranks of the individual matrices. Since the individual matrices are both full-rank, their minimum rank is equal to their maximum rank, which is equal to their number of rows (or columns).

5. How do you prove that the product of two full-rank matrices is full-rank?

To prove that the product of two full-rank matrices is full-rank, you can use the concept of linear independence and rank. First, show that the columns (or rows) of the individual matrices are linearly independent. Then, use the fact that the product of two matrices can only have a rank equal to or less than the minimum of the ranks of the individual matrices. By showing that the minimum rank is equal to the maximum possible rank, which is equal to the number of rows (or columns), you can conclude that the product of the two matrices is full-rank.

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