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

• zohapmkoftid
A*B = AB exists.This is true, but irrelevant.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. But we do not know that A is nonsingular. That is what we are supposed to prove. If you knew that, you wouldn't have to do the proof.Same goes if you if reversed then you will arrive that A and B are both invertible. Hence both AB = I and BA=I.No, that is not "the same". If you assume that both AB= I and BA= I,
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

## The Attempt at a Solution

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

Can you compute ABA?

cristo said:
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

jmegsalazar
fzero said:
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

zohapmkoftid said:
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.

It can poove by taking an example

fzero said:
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?

zohapmkoftid said:
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?

fzero said:
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 ?

zohapmkoftid said:
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 :)

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

HallsofIvy said:
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.

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

Susanne217 said:
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.

Susanne217 said:
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.
Susanne217 said:
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.

uvc29 said:
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.

zohapmkoftid said:
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

xTA = cT

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.

Susanne217 said:
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.

fzero said:
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

jmegsalazar

## 1. What does the statement "AB = I" mean in regards to matrices?

The statement "AB = I" means that the product of matrices A and B is equal to the identity matrix, I. In other words, when matrix A is multiplied by matrix B, the resulting matrix is the identity matrix. This is also known as the inverse property of matrices.

## 2. How does the inverse property of matrices relate to the statement "AB = I"?

The inverse property of matrices states that when a matrix is multiplied by its inverse, the resulting matrix is the identity matrix. Therefore, if AB = I, it follows that the inverse of matrix A is matrix B, and vice versa. This is why the statement "AB = I" is often referred to as the inverse property of matrices.

## 3. Can you provide an example to illustrate the statement "AB = I"?

Yes, for example, consider the following matrices:

A = [2 0; 0 1] and B = [1/2 0; 0 1]

When A is multiplied by B, the resulting matrix is the identity matrix I = [1 0; 0 1]. Therefore, AB = I, proving the statement.

## 4. How does the statement "AB = I" differ from "BA = I"?

The statement "AB = I" means that when matrix A is multiplied by matrix B, the resulting matrix is the identity matrix. On the other hand, the statement "BA = I" means that when matrix B is multiplied by matrix A, the resulting matrix is the identity matrix. While they may seem similar, the order of multiplication is different and can result in different outcomes.

## 5. Why is it important to prove the statement "AB = I, then BA = I"?

Proving the statement "AB = I, then BA = I" is important because it helps to establish the inverse property of matrices. This property is a fundamental concept in linear algebra and is used in various applications, such as solving systems of equations and finding inverses of matrices. Additionally, proving this statement helps to strengthen our understanding of matrix operations and their properties.

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