Invertible Matrices

• Feb 23rd 2014, 07:59 AM
davidciprut
Invertible Matrices
So there is a preposition that says;
If AB is invertible, then A is invertible and B is invertible too.

I uploaded the proof that I made, however I think it's not enough.

The problem with my proof, in my opinion, is that I showed that B and A is invertible from one side (either from left or right), so I think that in order to show that they are invertible I have to show that they have inverses from both side and they are the same, am I mistaken?

I would appreciate if you can verify, and tell me if my proof is correct, however if it's not don't tell me what's the right proof, I want to work on it. Thank you!
• Feb 23rd 2014, 10:45 AM
romsek
Re: Invertible Matrices
seems to me you can just say

Because AB is given to be invertible

\$\$(AB)^{-1}=B^{-1}A^{-1}\$\$

and because A is given to be invertible

\$\$(AB)^{-1}A=B^{-1}A^{-1}A\$\$

\$\$B^{-1}=(AB)^{-1}A\$\$
• Feb 23rd 2014, 11:27 AM
Deveno
Re: Invertible Matrices
You can simplify it a great deal:

If AB is invertible, there is a matrix C such that C(AB) = (AB)C = I.

By associativity, we have: (CA)B = I, and A(BC) = I.

You are correct in that you have only shown B is left-invertible, and A is right-invertible.

We have 4 cases to consider:

1) A is invertible, B is not,

2) B is invertible, A is not,

3) Neither are invertible,

4) Both are invertible.

Let's look at case (1):

We know there is a matrix C such that (AB)C = I. Multiplying by A-1, we get:

A-1[(AB)C] = A-1I = A-1.

(A-1A)(BC) = A-1

BC = A-1

(BC)A = A-1A = I

B(CA) = I, showing B is right-invertible, with the same right-inverse as left-inverse, that is: B is invertible, contradicting our assumption. So case 1 cannot occur.

I think you can easily do case (2).

For case (3) show that B singular implies AB singular, which is impossible. What's left?
• Feb 23rd 2014, 01:18 PM
davidciprut
Re: Invertible Matrices
This is my the new proof. The credit goes to Deveno.
Though I am curious , is it possible to improve my first proof and get to the conclusion that A and B is invertible?
• Feb 23rd 2014, 01:41 PM
Deveno
Re: Invertible Matrices
There is a problem with your proof for case (3), can you see what it is? (Hint: suppose A was the 0-matrix...does it follow from D ≠ I that AD ≠ A?).

You could modify your original proof by using column reduction instead of row reduction, this would give you the inverses on "the other side".
• Feb 23rd 2014, 11:03 PM
davidciprut
Re: Invertible Matrices
Quote:

Originally Posted by Deveno
There is a problem with your proof for case (3), can you see what it is? (Hint: suppose A was the 0-matrix...does it follow from D ≠ I that AD ≠ A?).

But that is the case when A is the zero matrix, and we know for sure that A can't be the zero matrix, otherwise AB wouldn't be invertible, am I mistaken?