To prove right inverse implies left inverse for square matrices.by SrEstroncio Tags: inverses, matrix, multiplication 

#1
Oct1810, 08:58 PM

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. 



#2
Oct1810, 11:19 PM

Mentor
P: 4,499

Try using the fact that B=BAB




#3
Oct2110, 02:01 PM

P: 8

I believe you will be better off showing this lemma: Let A and B be nxn matrices such that AB=I, then (BAI)b=0 for every b nx1 matrix (a vector) Then you can prove the result easily by considering the components of (BAI) by clevorly choosing specific 'b's 



#4
Jul2812, 02:23 AM

P: 65

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



#5
Jul2812, 03:39 PM

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P: 4,677

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 



#6
Jul2812, 05:14 PM

P: 65

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. 



#7
Jul2812, 05:56 PM

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I never claimed that the left inverse of B was Ain fact, that is what we are trying to prove. I just claimed that a left inverse of B EXISTS, nothing more.
RGV 



#8
Jul2812, 05:59 PM

P: 65





#9
Jul2812, 06:11 PM

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P: 4,677

I guess I interchanged A and B from the original question.
RGV 



#10
Jul2812, 06:15 PM

P: 65





#11
Jul2812, 07:18 PM

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 



#12
Jul2812, 07:21 PM

P: 65

You dont know there exists a C such that CA = I. That is the first part of the problem.




#13
Jul2812, 08:12 PM

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):




#14
Jul2812, 08:20 PM

P: 65

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.) 



#15
Jul2812, 10:24 PM

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RGV 



#16
Jul2912, 05:35 AM

P: 65

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.



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