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

[Mr Davis 97]

Homework Statement

Let A and B be square matrices. Suppose that A is invertible and A = BAB. Show that B is invertible.

Homework Equations

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...

 

[fresh_42]

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.

 

[Mr Davis 97]

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.

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

 

[fresh_42]

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

And it is not necessary. (Probably it isn't true either, but I haven't thought about a counterexample.)
I read your question as you have found C=A^-1BA.

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.

 

[Mr Davis 97]

And it is not necessary. (Probably it isn't true either, but I haven't thought about a counterexample.)
I read your question as you have found C=A^-1BA.

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.

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,

 

[fresh_42]

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,

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?

 

[DrClaude]

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,

To quote for another of your threads:

There is a standard theorem which states that if a square matrix A has a right (left) inverse B, then it also has a left (right) inverse, and that is equal to B as well. You should expand your understanding by trying to prove that theorem.

 

[PeroK]

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

 

[Mr Davis 97]

To quote for another of your threads:

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?

 

[PeroK]

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?

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

 

[Mr Davis 97]

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

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

 

[PeroK]

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

The determinant is related to the existence of an inverse.

 

[Mr Davis 97]

The determinant is related to the existence of an inverse.

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?

 

[PeroK]

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?

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

 

[Mr Davis 97]

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

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?

 

[PeroK]

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?

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.

 

[Mr Davis 97]

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.

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

 

[PeroK]

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

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

 

[fresh_42]

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

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.

 

[Mr Davis 97]

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.

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.

 

[fresh_42]

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.

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.

 

[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.

 

[Mr Davis 97]

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.

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?

 

[fresh_42]

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?

[Emphasis by me.] "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."

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.