Thread: Proof involving invertible matrices

I need to prove that if A is an invertible matrix and B is row equivalent to A, then B is also invertible.

I understand the logic of that statement, but I'm not experienced enough to know how to write it into words that prove it.

Here's my thinking so far:

1.) B is row equivalent to A, therefore there exist a sequence of elementary row operations that can be performed on A to produce B:
(En*...*E2*E1)A=B

2.) A is invertible, therefore there exist a sequence of elementary row operations that can be performed on A to produce the identity matrix:
(En*...*E2*E1)A=I

3.) From 1 & 2, there exist a sequence of elementary row operations that can be performed on B to produce the identity matrix, therefore B is also invertible.


How can I write the above statements so that they flow in a step-by-step proof?

Originally Posted by madmartigano

I need to prove that if A is an invertible matrix and B is row equivalent to A, then B is also invertible.

I understand the logic of that statement, but I'm not experienced enough to know how to write it into words that prove it.

Here's my thinking so far:

1.) B is row equivalent to A, therefore there exist a sequence of elementary row operations that can be performed on A to produce B:
(En*...*E2*E1)A=B

2.) A is invertible, therefore there exist a sequence of elementary row operations that can be performed on A to produce the identity matrix:
(En*...*E2*E1)A=I



Well, yes, but the first do not HAVE to be the same as the second ones, do they? So call the elementary matrices representing the row operations in 2. as (watch the index!), and try again (you're close...!)

Tonio


3.) From 1 & 2, there exist a sequence of elementary row operations that can be performed on B to produce the identity matrix, therefore B is also invertible.


How can I write the above statements so that they flow in a step-by-step proof?

.

Is this a step in the right direction?



I can't figure out how to show that B will equal I after a sequence of elementary row operations.

If A is invertible, it has full rank and hence non-zero determinant. Since row-equivalence is achieved by multiplying by elementary (in particular, invertible) matrices, the determinant of B will be the product of the determinants of a bunch of matrices with non-zero determinants. Therefore, det(B) is non-zero, hence B has full rank and is invertible.

I would be inclined to use the fact that every elementary matrix (except the ones constructed by multiplying a row by 0 which are not relvant here) is itself invertible.

If then so B is invertible.