To prove right inverse implies left inverse for square matrices.

In summary: 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
  • #1
SrEstroncio
62
0

Homework Statement


Let A be a square matrix with right inverse B. To prove A has a left inverse C and that B = C.


Homework Equations


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

The Attempt at a Solution



I don't 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 can't get around proving one implies the other.
 
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  • #2
Try using the fact that B=BAB
 
  • #3
Office_Shredder said:
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
 
  • #4
WastedGunner said:
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.
 
  • #5
who_ said:
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
 
  • #6
Ray Vickson said:
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.
 
  • #7
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
 
  • #8
Ray Vickson said:
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.
 
  • #9
I guess I interchanged A and B from the original question.

RGV
 
  • #10
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.
 
  • #11
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
You don't know there exists a C such that CA = I. That is the first part of the problem.
 
  • #13
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
 
  • #14
Bohrok said:
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.)
 
  • #15
who_ said:
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
 
  • #16
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.
 

1. What is a right inverse for a square matrix?

A right inverse for a square matrix is a matrix that, when multiplied by the original matrix on the right, produces the identity matrix. In other words, it "undoes" the original matrix when multiplied on the right. This is denoted as A*A-1 = I.

2. What is a left inverse for a square matrix?

A left inverse for a square matrix is a matrix that, when multiplied by the original matrix on the left, produces the identity matrix. In other words, it "undoes" the original matrix when multiplied on the left. This is denoted as A-1*A = I.

3. Why is it important for a square matrix to have both a left and right inverse?

Having both a left and right inverse for a square matrix is important because it guarantees that the matrix is invertible. This means that the matrix can be "undone" and the original values can be recovered. It also allows for solving systems of linear equations and performing other mathematical operations on the matrix.

4. How can I prove that a right inverse implies a left inverse for a square matrix?

To prove that a right inverse implies a left inverse for a square matrix, you can use the fact that A*A-1 = I and A-1*A = I. By multiplying these two equations together, you can show that A-1 is both a left and right inverse for A.

5. Is the converse also true, that a left inverse implies a right inverse for a square matrix?

No, the converse is not always true. A left inverse does not necessarily imply a right inverse for a square matrix. This is because a matrix can have a left inverse that is not compatible with its right inverse, meaning they cannot be multiplied together to produce the identity matrix.

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