# Homework Help: 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.

21. Oct 27, 2016

### Staff: Mentor

Yes.
But "this does not guarantee that A has an inverse" only in the sense of a two-sided inverse. With "AC=I" it has a right inverse.

Let's consider functions $f : \mathbb{N} \rightarrow \mathbb{N}$.
Then for $f_1(x) := x+1$ and $f_2(x) := \begin{cases} x-1 & \text{ for } x \geq 2 \\ 1 & \text{ for } x=1 \end{cases}$
we get $f_2 \circ f_1 = id_\mathbb{N}$ and $f_1 \circ f_2 \neq id_\mathbb{N}$. This is an example where $f_2$ has a right inverse but no left inverse.

22. Oct 28, 2016

### PeroK

Perhaps to summarise:

1) In this particular problem, you are dealing with square matrices and given that $A = BAB$ and $A$ is invertible.

The simplest way to solve this is to consider determinants, as you can see immediately that $B$ must be invertible.

As has been shown, however, you only need the ring axioms to show that $B$ is invertible in this case.

2) There was then a related question:

For square matrices if $AB = I$ then $A$ and $B$ are invertible and, of course, $A = B^{-1}$.

Again, considering determinants shows this immediately.

But, if $A$ & $B$ are elements of an arbitrary ring, then it does not follow that $BA = I$. As shown by the above counterexample.

3) The moral is that matrices have specific properties, in particular in terms of inverses, beyond those that all rings have. And these properties are often encapsulated by the equivalence of having an inverse with a having non-zero determinant.

23. Oct 30, 2016

### Mr Davis 97

Wait, so if I have $AC = I$, then this does not guarantee that A has a left inverse, right? However, if I take the determinant, we see that $det(A) det(B) = 1$,so A and B both must be invertible. But I thought that $AC = I$ doesn't guarantee a left-inverse?

24. Oct 31, 2016

### Staff: Mentor

Beside this, how did you find $B\,$? $AC=I$ gives you $\det(A) \det(C) = 1$, so both, $A$ and $C$ are invertible and $C$ is a right inverse of $A$ which is a left inverse of $C$. How come $B$ as a left inverse of $A$ into play without any additional conditions or deviations? Will $B=C$ always do the job? How can we guarantee that $\det (C) \det(A) = 1$ implies $CA = I\,$? And how, that $\det (C) \det(A) = 1$ even holds? Where is it stated, that our matrix elements have a commutative multiplication?

I have no good counterexamples at hand, for in most common cases, a right inverse matrix is also the left inverse matrix, and as shown above, together with associativity they are the same then. I know of non associative examples, but they also lack of the classical unity matrix.

Nevertheless, one has to be careful about what is explicitly given and what is somehow automatically assumed for the reasons: "How can it be not the case?" or "It has always been so!" Both are no valuable mathematical arguments. The fact, that counterexamples might be difficult to find, does not mean there are none.