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

by SrEstroncio
Tags: inverses, matrix, multiplication
 P: 62 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.
 Emeritus Sci Advisor PF Gold P: 4,500 Try using the fact that B=BAB
P: 8
 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

P: 65
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.
HW Helper
Thanks
P: 5,086
 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
P: 65
 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.
 Sci Advisor HW Helper Thanks P: 5,086 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
P: 65
 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.
 Sci Advisor HW Helper Thanks P: 5,086 I guess I interchanged A and B from the original question. RGV
P: 65
 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.
 P: 867 This thread's almost 2 years old... Anyway, what about: AB = I CA = I C(AB) = CI (CA)B = C IB = C B = C
 P: 65 You dont know there exists a C such that CA = I. That is the first part of the problem.
P: 867
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
P: 65
 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.)