# Proof of equivalent matrices

Printable View

• Sep 8th 2008, 03:42 PM
Brokescholar
Proof of equivalent matrices
Let A and B be m x n matrices. Show that A is equivalent to B if and only if AT is equivalent to BT.

So far I've got this. I'm not totally sure if it's correct or if I'm missing something major. Thanks!

AT = PBTQ for some nonsingular matrices P and Q.

Taking the transpose of both sides I get

A = PTBQT

(PT)-1A(QT)-1 = B

By definition A is equiv. to B because we can get B from A by a sequence of elem. row/column operations.

• Sep 8th 2008, 06:08 PM
ThePerfectHacker
Quote:

Originally Posted by Brokescholar
Let A and B be m x n matrices. Show that A is equivalent to B if and only if AT is equivalent to BT.

So far I've got this. I'm not totally sure if it's correct or if I'm missing something major. Thanks!

AT = PBTQ for some nonsingular matrices P and Q.

Taking the transpose of both sides I get

A = PTBQT

(PT)-1A(QT)-1 = B

By definition A is equiv. to B because we can get B from A by a sequence of elem. row/column operations.

I do it for $n\times n$ matrices to give you an idea. If $A$ is equivalent to $B$ then it means $A=MBM^{-1}$ for an invertible matrix $M$. This means $A^{T} = (MBM^{-1})^T = (M^T)^{-1} B^T M^T$. Thus, $A^T$ is equivalent to $B^T$.
• Sep 9th 2008, 04:29 PM
Brokescholar
Thank you very much!