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**[SOLVED] Two square invertible matrices, prove product is invertible**

## Homework Statement

If

*A*and

*B*are

*n*x

*n*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?

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