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Bachelier
Nov5-10, 03:15 AM
A,B are nxn. If AB is invertible. Show that A and B are invertible.

I know how to prove it by determinant, using linear transformations and contradictions.
I am looking for a direct way using a proof by matrices. Can anyone think of one?

Thank you.

Outlined
Nov5-10, 05:24 AM
AB is invertible so there exist a matrix C such that

I = (AB)C = A(BC)

so A is invertible and A-1 = BC.

Try B yourself.

Bachelier
Nov5-10, 08:32 AM
I thought about that but I thought it looked too simple.
Thanks. I think this completes all possible ways to prove this problem.
Take a look at my Challenge problem. :)

http://www.physicsforums.com/showthread.php?t=444941

arkajad
Nov5-10, 07:17 PM
There is another way if you know that a matrix X is not invertible is equivalent to saying that there is a nonzero vector v such that Xv=0.

Then suppose B is not invertible with Bv=0. Then AB is not invertible since ABv=0. Now, suppose B is invertible but A is not invertible with Av=0. Then ABB^-1 v = 0.

It follows that if A or B are not invertible then AB is not invertible.

Bachelier
Nov5-10, 09:05 PM
There is another way if you know that a matrix X is not invertible is equivalent to saying that there is a nonzero vector v such that Xv=0.

Then suppose B is not invertible with Bv=0. Then AB is not invertible since ABv=0. Now, suppose B is invertible but A is not invertible with Av=0. Then ABB^-1 v = 0.

It follows that if A or B are not invertible then AB is not invertible.

Good one.