Left and right inverses of a non-square matrix

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In summary, the conversation discusses a problem involving a system of equations with a matrix, and the conditions for A′(AA′)⁻¹y and A⁻¹y to be solutions. The given solution shows that A′(AA′)⁻¹y is a solution, while the source from StackExchange suggests that A′(AA′)⁻¹y is not a solution. There is also a discussion about the use of right and left inverses in solving the system. Overall, the conversation delves into the nuances and complexities of solving a system of equations with a matrix.
  • #1
PainterGuy
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Homework Statement
I was working with right inverse and left inverse of a matrix.
Relevant Equations
Please check my work below.
Hi,

It's actually not a homework problem but I still decided to post it here.

Problem:
Consider Ax=y, where A is mxn and has rank m. Is (A′A)⁻¹A′y a solution? If not, under what condition will it be a solution? Is A′(AA′)⁻¹y a solution?

The given solution is:
Consider Ax=y with A mxn and rank(A)=m, which implies n⋝m. If n>m then A′A is nxn and singular . Thus (A′A)⁻¹ is not defined and (A′A)⁻¹A′y is not a solution. Because AA′ is mxm and nonsingular, substituting A′(AA′)⁻¹y into Ax yields y. Thus A′(AA′)⁻¹y is a solution. If n=m then both reduce to A⁻¹y and are solutions of Ax=y.
Question 1:
A′(AA′)⁻¹ is right inverse but the system in question statement is written as "Ax=y". The yellow highlighted text below says that a right inverse, A′(AA′)⁻¹, is useful for solving XA=Y system. You notice the contradiction. Which source is correct? Given solution or the yellow highlighted text? Could you please help me?
1616399245942.png

Source: https://math.stackexchange.com/a/1335707/775285

Question 2:
Mostly it's written AX=B where B is a constant matrix. In the case above the system is written instead as Ax=y. Is B=y? This link is relevant here . Could you please help me with it?
 
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  • #2
I cannot interpret this:
PainterGuy said:
substituting A′(AA′)⁻¹y into Ax yields y.
If they mean premultiplying by A′(AA′)⁻¹ it yields A′(AA′)⁻¹Ax=A′(AA′)⁻¹y. If A were non singular we could expand that inverse and get x=A′(AA′)⁻¹y, but not otherwise.
The veracity of the statement in yellow is easily checked.

For q2, the usual is to use uppercase for a matrix and lowercase for a vector, but of course the latter is just a special case of the former.
 
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Thank you.

I'm sorry but I still don't get it.

Of course, a right inverse is useful for solving an equation of the form XA=Y and a left inverse is useful for solving an equation of the form AX=Y.
Source: https://math.stackexchange.com/a/1335707/775285

In the given solution right inverse is used but the given system has the form AX=Y. The choice of using right inverse is correct, in my opinion, which means that the StackExchange source is wrong.

Consider Ax=y with A mxn and rank(A)=m, which implies n⋝m. If n>m then A′A is nxn and singular . Thus (A′A)⁻¹ is not defined and (A′A)⁻¹A′y is not a solution. Because AA′ is mxm and nonsingular, substituting A′(AA′)⁻¹y into Ax yields y. Thus A′(AA′)⁻¹y is a solution. If n=m then both reduce to A⁻¹y and are solutions of Ax=y.

haruspex said:
For q2, the usual is to use uppercase for a matrix and lowercase for a vector, but of course the latter is just a special case of the former.

Once again sorry but I think my question wasn't understood. Mostly, it's AX=B and not AX=Y. To me, AX=Y doesn't make much sense.

Question 2:
Mostly it's written AX=B where B is a constant matrix. In the case above the system is written instead as Ax=y. Is B=y?
Note to self:
1616489810622.png

...
1616489827889.png

Source: https://en.wikipedia.org/wiki/Generalized_inverse
 
  • #4
these formulas you give for left and right inverses work over ##\mathbb R## but do not work over other fields -- so the thread is a bit misleading; I assume we are working over ##\mathbb R## from here forward. I also wouldn't read too much into the stackexchange linked post.

If ##A## is surjective and you are trying to solve for an ##\mathbf x## such that ##A\mathbf x = \mathbf b## then you know a solution exists (by surjectivity) and so you actually want to use the right inverse formula. The 'left inverse formula' actually corresponds to the case of ##A## being injective, so if a solution exists, it will be unique and you will find it, but a solution may not exist-- in such a case you actually have ##A\mathbf x\neq \mathbf b## for all ##\mathbf x##, and working over ##\mathbb R## you cannot 'solve' such an equation but you can find a unique candidate ##\mathbf x## that minimizes ##\Big \Vert A\mathbf x- \mathbf b\Big \Vert_2##
 
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  • #5
Thank you!

StoneTemplePython said:
If ##A## is surjective and you are trying to solve for an ##\mathbf x## such that ##A\mathbf x = \mathbf b## then you know a solution exists (by surjectivity) and so you actually want to use the right inverse formula.

Possibly, this is something dumb but we are writing the right inverse on the left of A, isn't right inverse of A supposed to be written on the right of A?

1616548987242.png
Why is the original system written as AX=Y and not as AX=B? Could you please guide me?
 
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  • #6
you've written AX = Y vs AX = B many times and I see a distinction without a difference -- i.e. it seems irrelevant to me so I don't know what/why you are asking. It is a bit concerning though.

As for the other part of your question:
Suppose ##A\in \mathbb R^{m\times n}## is surjective and you want to find some ##\mathbf x## such that ##A\mathbf x = \mathbf b##, where of course ##A## and ##\mathbf b## are specified beforehand.

Then one valid solution is ##\mathbf x := (AA^T)^{-1}\mathbf b## because ##(AA^T)^{-1}## is a right inverse so
##A(AA^T)^{-1}= I_m## (which is the definition of a right inverse)
##\implies A\mathbf x = A\big((AA^T)^{-1}\mathbf b\big)= \big(A(AA^T)^{-1}\big)\mathbf b = I_m \mathbf b=\mathbf b##

it turns out in reals this is the minimum length (2 norm) solution for ##\mathbf x## that satisfies ##A\mathbf x = \mathbf b##. But based on the level of questioning I think this is way out side the scope.
 
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  • #7
PainterGuy said:
Why is the original system written as AX=Y and not as AX=B? Could you please guide me?
I mentioned before that a common convention is to use uppercase for matrices and lowercase for vectors. Another is to use A, B.. for matrices and X, Y.. for vectors; but, confusingly, also using X for an unknown matrix. So in AX=B suggests matrices whereas in AX=Y X and Y may well be vectors.
But these are only conventions. In any given usage the meanings should be made clear by the context,
 
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  • #8
Thank you very much!

I understand it now that why the right inverse was written on the left for the given case.

1616657757111.png
 

Related to Left and right inverses of a non-square matrix

1. What is a left inverse of a non-square matrix?

A left inverse of a non-square matrix is a matrix that, when multiplied on the left by the original matrix, results in the identity matrix. This means that the left inverse "undoes" the original matrix operation.

2. Can a non-square matrix have both a left and right inverse?

No, a non-square matrix can only have either a left or a right inverse. If a non-square matrix has a left inverse, it cannot have a right inverse, and vice versa.

3. How do you find the left and right inverses of a non-square matrix?

The left and right inverses of a non-square matrix can be found using the Moore-Penrose pseudoinverse. This is a special type of inverse that can be calculated for any matrix, regardless of its size or shape.

4. What is the significance of left and right inverses of a non-square matrix?

The existence of a left or right inverse for a non-square matrix can be used to solve systems of linear equations. It can also be used to determine if a matrix is invertible or not.

5. Can a non-square matrix have more than one left or right inverse?

Yes, a non-square matrix can have multiple left and/or right inverses. However, all of these inverses will have the same number of columns as rows in the original matrix.

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