Does AB = I Imply BA = I for Square Matrices?

[zohapmkoftid]

Homework Statement

Let A and B be n × n matrices. Show that if AB = I, then also BA = I, so A and B
are invertible, A = B⁻¹ and B = A⁻¹.

How can I prove this?
Thanks

Homework Equations

The Attempt at a Solution

 

[cristo]

How do you think you should go about solving the problem? Show some work!

 

[fzero]

Can you compute ABA?

 

[zohapmkoftid]

How do you think you should go about solving the problem? Show some work!

Sorry, I have no clue about this question. I found some solutions on the web but they are related to vector space which I haven't learned yet

 

[zohapmkoftid]

Can you compute ABA?

The thing confuses me is the definition of inverse

B is the inverse of A if AB = BA = I
B is the inverse of A if AB = I

Which one is the correct definition?

ABA = IA
A^-1ABA = A^-1A
BA = I

 

[fzero]

The thing confuses me is the definition of inverse

B is the inverse of A if AB = BA = I
B is the inverse of A if AB = I

Which one is the correct definition?

You're supposed to show that if AB = I, then BA=I as well, so the two definitions are equivalent.

ABA = IA
A^-1ABA = A^-1A
BA = I

This isn't a proof since you assumed the existence of the inverse of A, which is what you're trying to prove.

 

[uvc29]

It can poove by taking an example

 

[zohapmkoftid]

You're supposed to show that if AB = I, then BA=I as well, so the two definitions are equivalent.



This isn't a proof since you assumed the existence of the inverse of A, which is what you're trying to prove.

Could you give me some hints?

 

[fzero]

Could you give me some hints?

You're given AB = I. You can therefore show that ABA = A. What about BAB? What can you conclude about BA given what happens when you multiply with it on the left or right of these matrices?

 

[zohapmkoftid]

You're given AB = I. You can therefore show that ABA = A. What about BAB? What can you conclude about BA given what happens when you multiply with it on the left or right of these matrices?

ABA = A
BAB = B

We can conclude that BA = I ?

 

[Susanne217]

1. Homework Statement

Let A and B be n × n matrices. Show that if AB = I, then also BA = I, so A and B
are invertible, A = B⁻¹ and B = A⁻¹.

How can I prove this?
Thanks

As Scotty from Star Trek would say, Damn it man :)

Well You know the property of the invertible n \times n matrix which states

AB = I which by inturn mean that A \cdot A^{-1}=I

If B = A^{-1} then also implies BA = I

Back then I went to High School during the Clinton years I learned

Let a be a real number then if you say a \cdot a^{-1}= 1

e.g. 5 \cdot 5^{-1} = 1

What you need to realize, man.

Let A and B be two square matrices of equal size which means that

[a_1 \ a_2 \ \cdots \ a_n] \cdot B = A \cdot B exists (remember that the Matrix-Matrix product only exists if you can use the row column rule from lin-alg).

We know that the basic property of the invertible square matrices is
A \cdot A^{-1}=I

then by letting B = A^{-1} then by the row-column rule

[a_1 \ a_2 \ \cdots \ a_n] \cdot A^{-1} = A \cdot A^{-1} = I Thus

A\cdot B = I

Then you can easily show the other way BA = I yourself :)

 

[HallsofIvy]

But the problem did not say, initially, that A and B are invertible. That was one of the things to be shown.

 

[Susanne217]

But the problem did not say, initially, that A and B are invertible. That was one of the things to be shown.

HallsofIvy,

I know you have the ability to use the force at a higher level than the rest of us.

But the assigment does say

"Show that if AB = I, then also BA = I, so A and B
are invertible."

In my opinion it all comes down to how you read the problem at hand.

1) For A and B to be invertible then they must live up to AB = I, which implies that either
AA^-1 = I if B = A^-1. Or if BA = I which implies that A = B^-1.

2) Hence then for the matrix product to exist then it has to live up to the row column rule. Then I choose A and B to be square matrices, then A*B = AB exists.

3) For A to be invertible then A has to be non-singular. Then if A is non singlar and I replace B with A^-1 and since we know that AB = I, then A is invertible.

Same goes if you if reversed then you will arrive that A and B are both invertible. Hence both AB = I and BA=I.

 

[zohapmkoftid]

Can I prove like this?

AB = I
det(AB) = detI
(detA)(detB) = 1
detA != 0 and detB != 0

Therefore, A^-1 and B^-1 exist

AB = I
A^-1AB = A^-1
B = A^-1
BA = A^-1A
BA = I

 

[Mark44]

HallsofIvy,

I know you have the ability to use the force at a higher level than the rest of us.

But the assigment does say

"Show that if AB = I, then also BA = I, so A and B
are invertible."

In my opinion it all comes down to how you read the problem at hand.

The problem description is clear, with little room for interpretation. We are given that AB = I, with A and B n x n matrices.

1) For A and B to be invertible then they must live up to AB = I, which implies that either
AA^-1 = I if B = A^-1. Or if BA = I which implies that A = B^-1.

You are starting by assuming that which is to be proved.

2) Hence then for the matrix product to exist then it has to live up to the row column rule. Then I choose A and B to be square matrices, then A*B = AB exists.

3) For A to be invertible then A has to be non-singular. Then if A is non singlar and I replace B with A^-1 and since we know that AB = I, then A is invertible.

Same goes if you if reversed then you will arrive that A and B are both invertible. Hence both AB = I and BA=I.

 

[D H]

It can poove by taking an example

No, you can't. A single counterexample works just fine for disproving some conjecture but an example does not suffice for proving some conjecture.

 

[vela]

Can I prove like this?

AB = I
det(AB) = detI
(detA)(detB) = 1
detA != 0 and detB != 0

Therefore, A^-1 and B^-1 exist

AB = I
A^-1AB = A^-1
B = A^-1
BA = A^-1A
BA = I

While there's nothing wrong with what you wrote, I don't think it was the intent of the problem. The idea is to show if A has a right inverse B, B is also its left inverse. Then by definition A and B are invertible and are inverses for each other.

It there's a more straightforward proof than what I'll suggest, I don't see it, but then I've never been particularly good at figuring out clever manipulations. Start by showing that if you have a vector c, you can always find a solution x to the equation

x^TA = c^T

In particular, you can find a solution when c is equal the i-th column of the identity matrix. Combine all n equations into one matrix equation and use the fact that the identity matrix is equal to its transpose to show BA=I.

 

[HallsofIvy]

HallsofIvy,

I know you have the ability to use the force at a higher level than the rest of us.

But the assigment does say

"Show that if AB = I, then also BA = I, so A and B
are invertible."

In my opinion it all comes down to how you read the problem at hand.

1) For A and B to be invertible then they must live up to AB = I, which implies that either
AA^-1 = I if B = A^-1. Or if BA = I which implies that A = B^-1.

No, that's wrong. In order for A and B to be invertible, both AB= I and BA= I must be true.

2) Hence then for the matrix product to exist then it has to live up to the row column rule. Then I choose A and B to be square matrices, then A*B = AB exists.

3) For A to be invertible then A has to be non-singular. Then if A is non singlar and I replace B with A^-1 and since we know that AB = I, then A is invertible.

Same goes if you if reversed then you will arrive that A and B are both invertible. Hence both AB = I and BA=I.

 

[WastedGunner]

Can you compute ABA?

I don't think this is enough.

You can't prove this only using the group properties.

Prove instead that (BA-I)b=0 for any nx1 matrix b

Then show that the components of BA-I are 0 by choosing specific vectors b