- #1
zohapmkoftid
- 27
- 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
cristo said:How do you think you should go about solving the problem? Show some work!
fzero said:Can you compute ABA?
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?
ABA = IA
A^{-1}ABA = A^{-1}A
BA = I
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.
zohapmkoftid said:Could you give me some hints?
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?
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
HallsofIvy said:But the problem did not say, initially, that A and B are invertible. That was one of the things to be shown.
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: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.
You are starting by assuming that which is to be proved.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.
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.
No, you can't. A single counterexample works just fine for disproving some conjecture but an example does not suffice for proving some conjecture.uvc29 said:It can poove by taking an example
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.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^{-1}AB = A^{-1}
B = A^{-1}
BA = A^{-1}A
BA = I
No, that's wrong. In order for A and B to be invertible, both AB= I and BA= I must be true.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.
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?
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.
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.
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.
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.
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.