Two square invertible matrices, prove product is invertible

In summary, the question is asking for a proof that the product of two nxn matrices of rank n is also of rank n. The proposed solution involves using equivalent statements and the Fundamental Theorem of Invertible Matrices to show that the rank of AB is equal to n.
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[SOLVED] Two square invertible matrices, prove product is invertible

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?
 
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I think I got it. See my second edit above. Comments welcome.
 

1. What is the definition of an invertible matrix?

An invertible matrix is a square matrix that has a unique inverse matrix. This means that when multiplied together, the two matrices will result in the identity matrix.

2. How can you prove that a matrix is invertible?

A matrix can be proved to be invertible by showing that it has a unique inverse matrix. This can be done by using the determinant of the matrix, which must be non-zero in order for the matrix to have an inverse. Additionally, the matrix can be reduced to row-echelon form to determine if it is invertible.

3. Can a product of two invertible matrices also be invertible?

Yes, the product of two invertible matrices will also be invertible. This is because the inverse of the product is equal to the product of the inverses in reverse order.

4. How does the inverse of a product of matrices relate to the individual inverses?

The inverse of a product of matrices is equal to the product of the individual inverses in reverse order. This means that (AB)^-1 = B^-1A^-1. This relationship is what allows us to prove that the product of two invertible matrices is also invertible.

5. Are there any exceptions to the rule that the product of two invertible matrices is also invertible?

No, there are no exceptions to this rule. As long as the matrices are square and have non-zero determinants, their product will be invertible. If one of the matrices is not invertible, then the product will also not be invertible.

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