# Showing that B is invertible if A = BAB, where A is invertib

1. Oct 26, 2016

### Mr Davis 97

1. The problem statement, all variables and given/known data
Let A and B be square matrices. Suppose that A is invertible and A = BAB. Show that B is invertible.

2. Relevant equations

3. The attempt at a solution
First, since BAB = A, and A is invertible, this means that B(ABA-1) = I. However, to show that B is invertible, I also need to show that (ABA-1)B = I, but I don't see how to do this...

2. Oct 27, 2016

### Staff: Mentor

With the same method you can construct a (possibly different) left inverse C of B.
Then you have CB = 1 = BD. (In your case you already have D = ABA-1.)
Now show that in general C=D has to hold.

3. Oct 27, 2016

### Mr Davis 97

Well ABA-1 = A-1BA only if A = A-1, right? But we aren't given that in the problem.

4. Oct 27, 2016

### Staff: Mentor

And it is not necessary. (Probably it isn't true either, but I haven't thought about a counterexample.)

So forget now the A and prove: CB=BD implies C=D, which is useful to know in other cases, too.

Edit: I forgot to repeat that CB=BD=1 which is important here.

Last edited: Oct 27, 2016
5. Oct 27, 2016

### Mr Davis 97

I'm not seeing how to do it. Since we aren't given that C, B, or D have an inverse, it doesn't seem possible to manipulate the equation CB = BD to get C = D,

6. Oct 27, 2016

### Staff: Mentor

You have found a left inverse C of B and a right inverse D of B. (C=A-1BA and D=ABA-1)
Thus you know CB=1=BD. What is CBD?

7. Oct 27, 2016

### Staff: Mentor

8. Oct 27, 2016

### PeroK

There's also that useful little concept called the determinant!

9. Oct 27, 2016

### Mr Davis 97

So are you saying that to find an inverse for an nxn matrix I only need to show that AC = I, then automatically CA = I too?

10. Oct 27, 2016

### PeroK

You should be able to prove that easily enough. And, again, I would like to mention the d-word: "determinant".

11. Oct 27, 2016

### Mr Davis 97

Well, using determinants, on $AC = CA$, we get that $det(A) det(C) = det(C) det(A)$, which is a true equation, but I don't see how we can then go back to say that AC = CA as result

12. Oct 27, 2016

### PeroK

The determinant is related to the existence of an inverse.

13. Oct 27, 2016

### Mr Davis 97

Well we know that $det(A) det(C) = det(C) det(A) = 1$, so both AC and CA have an inverse. Does this imply that A has an inverse and that inverse is C?

14. Oct 27, 2016

### PeroK

If $det(A) det(C) = 1$, what does that tell you about $det(A)$ and $det(C)$?

15. Oct 27, 2016

### Mr Davis 97

That neither is zero, and thus they're both invertible. Thus, C is the inverse of A, and A is the inverse of C, since CA = I. Is that right?

Also, is this only true for n x n matrices because of general m x n matrices we can't take determinants?

16. Oct 27, 2016

### PeroK

You might like to tidy it up by saying that if $AC = I$ then $det(C) \ne 0$ hence $C^{-1}$ exists and:

$ACC^{-1} = IC^{-1}$, hence $A = C^{-1}$

For general m x n matrices, $AC$ and $CA$ are generally not defined.

17. Oct 27, 2016

### Mr Davis 97

Just out of curiosity, is there any simple way to prove it without determinants?

18. Oct 27, 2016

### PeroK

It's not generally true in a ring that a left inverse implies a right inverse. So, you need more than the ring axioms.

19. Oct 27, 2016

### Staff: Mentor

If you already have a left inverse $CB = I$ and a right inverse $BD = I$ as in your case, then all you need is associativity to prove they are equal: $C = C \cdot I = C\cdot (BD) = (CB)\cdot D = I\cdot D = D$

And to not confuse you with what @PeroK has said: You cannot automatically assume that the existence of a one-sided inverse implies one from the other side. However, you've proven the existence of both inverses to $B$ in posts #1, #3. Strictly speaking, that's all you needed to do. The equality of both hasn't been required. But as you can see, it's easy to do in this case.

20. Oct 27, 2016

### Mr Davis 97

Ah, okay, thanks for clearing that all up. So just to be clear, if I have found that AC= I, then this does not guarantee that A has an inverse. I need to show also that there exists a D such that DA = I, and if this is the case, it must be so that D = C.