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Parabolox
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[SOLVED] Two square invertible matrices, prove product is invertible
If A and B are nxn matrices of rank n, prove that AB has rank n.
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...
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?
Homework Statement
If A and B are nxn matrices of rank n, prove that AB has rank n.
Homework 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...
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: