## To prove right inverse implies left inverse for square matrices.

1. The problem statement, all variables and given/known data
Let A be a square matrix with right inverse B. To prove A has a left inverse C and that B = C.

2. Relevant equations
Matrix multiplication is asociative (AB)C=A(BC).
A has a right inverse B such that AB = I

3. The attempt at a solution

I dont really know where to start, I mean, proving that if both B and C exist then B = C is not that hard, but I really cant get around proving one implies the other.
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 Blog Entries: 1 Recognitions: Homework Help Try using the fact that B=BAB

 Quote by Office_Shredder Try using the fact that B=BAB
I am pretty sure that is not going to get you the answer.

I believe you will be better off showing this lemma:

Let A and B be nxn matrices such that AB=I,
then (BA-I)b=0 for every b nx1 matrix (a vector)

Then you can prove the result easily by considering the components of (BA-I) by clevorly choosing specific 'b's

## To prove right inverse implies left inverse for square matrices.

 Quote by WastedGunner Let A and B be nxn matrices such that AB=I, then (BA-I)b=0 for every b nx1 matrix (a vector)
How does that follow? (Well, that is true, but you're missing some steps here :) But that is the right idea (kind of).

First, you want to work of the definition of inverses: If B is the right inverse of A, then for every vector b, there is an x such that Ax = b (simply choose x = Bb). What does that tell you about the number of pivot columns for A?

Then what does that tell you about A's relationship with I? Think back to Echelon reductions and elementary matrices.

Recognitions:
Homework Help
 Quote by who_ How does that follow? (Well, that is true, but you're missing some steps here :) But that is the right idea (kind of). First, you want to work of the definition of inverses: If B is the right inverse of A, then for every vector b, there is an x such that Ax = b (simply choose x = Bb). What does that tell you about the number of pivot columns for A? Then what does that tell you about A's relationship with I? Think back to Echelon reductions and elementary matrices.
Can you prove the following? If a square matrix has a right inverse, then it also has a left inverse.

If BA = I, then A is a right inverse of B. Let D be a left inverse of B and then do the obvious.

RGV

 Quote by Ray Vickson Can you prove the following? If a square matrix has a right inverse, then it also has a left inverse. If BA = I, then A is a right inverse of B. Let D be a left inverse of B and then do the obvious. RGV
What are you doing here? You can't just assume that BA = I, and you can't use the statement you are proving to prove that statement.

The proof of the OP's problem will require specifics involving elementary matrices or something similar. That kind of detail is necessary; otherwise, one would be saying that in any algebraic group, the existence of a right inverse implies the existence of a left inverse, which is definitely not true.
 Recognitions: Homework Help I never claimed that the left inverse of B was A---in fact, that is what we are trying to prove. I just claimed that a left inverse of B EXISTS, nothing more. RGV

 Quote by Ray Vickson I never claimed that the left inverse of B was A---in fact, that is what we are trying to prove. I just claimed that a left inverse of B EXISTS, nothing more. RGV
Yes, but where did you get the fact that BA = I from? I don't believe that is something trivially true.
 Recognitions: Homework Help I guess I interchanged A and B from the original question. RGV

 I guess I interchanged A and B from the original question.
Well, that does change the question quite a bit. As far as I know, the OP's question is not trivial in the sense that you can just move the matrices around and hope to get somewhere. You must delve deeper and understand what it means for a matrix to have inverses, and go on from there.
 This thread's almost 2 years old... Anyway, what about: AB = I CA = I C(AB) = CI (CA)B = C IB = C B = C
 You dont know there exists a C such that CA = I. That is the first part of the problem.

I did a Google search on the first sentence in the OP; two of the results wrote the problem as follows (which I believe is the actual problem):
 Let A be a square matrix with right inverse B and left inverse C. Show that B=C

 Quote by Bohrok I did a Google search on the first sentence in the OP; two of the results wrote the problem as follows (which I believe is the actual problem):
Umm... there is no reason that the OP's question is incorrect. I believe the OP meant what s/he said. The question clearly says:

Given that a square matrix A has right inverse B, show that that A has a left inverse C and that C = B.

So, you can NOT assume that C exists. (Just because a Google search reveals variations of this problem doesn't mean that those variations are this problem.)

Recognitions:
Homework Help
 Quote by who_ Well, that does change the question quite a bit. As far as I know, the OP's question is not trivial in the sense that you can just move the matrices around and hope to get somewhere. You must delve deeper and understand what it means for a matrix to have inverses, and go on from there.
It does not change anything at all: my A = your B and my B = your A. What we really have is that UV=I for two square matrices U and V (so V is a right inverse of U and U is a left inverse of V). We need to prove that U is also a right inverse of V (and V is a left inverse of U). I asked the OP: can you prove that if a matrix has a right inverse it also has a left inverse? (That was, in fact, part of the original question.) The OP could answer "yes", in which case my hint could be useful, or could answer "no", in which case my hint would be useless. So much depends on what the OP already knows and is allowed to use. For example, can he/she use the fact that the row and column ranks are equal? etc., etc.

RGV
 Well, that was my point - that one could not use your hint until they knew the answer to the existence of the left inverse, which is the harder part of the question.

 Tags inverses, matrix, multiplication