Zulip Chat Archive

Stream: maths

Topic: matrix inverse

Chris Hughes (Jul 05 2019 at 14:19):

Is there a total matrix inverse function, defined on all matrices over a field that has the property that it returns left inverses when they exist, and right inverses when they exist, and (AB)−1=B−1A−1(AB)^{-1} = B^{-1}A^{-1} for all matrices?

Kevin Buzzard (Jul 05 2019 at 14:20):

all matrices of any size? Or some fixed square size?

Chris Hughes (Jul 05 2019 at 14:20):

Any size

Kevin Buzzard (Jul 05 2019 at 14:21):

I can't even make this question typecheck

Kevin Buzzard (Jul 05 2019 at 14:22):

If A is an a * b matrix, what size is A^{-1} supposed to be?

Kevin Buzzard (Jul 05 2019 at 14:22):

b * a

Kevin Buzzard (Jul 05 2019 at 14:22):

and whether or not it's a left inverse or right inverse depends on whether a<b or b>a?

Chris Hughes (Jul 05 2019 at 14:23):

b * a

Yes.

Chris Hughes (Jul 05 2019 at 14:24):

and whether or not it's a left inverse or right inverse depends on whether a<b or b>a?

Yes, but maybe neither inverse exists.

Johan Commelin (Jul 05 2019 at 14:25):

Does this work: https://en.wikipedia.org/wiki/Invertible_matrix#In_relation_to_its_adjugate

Chris Hughes (Jul 05 2019 at 14:26):

That's only for square matrices.

Kevin Buzzard (Jul 05 2019 at 14:26):

The problem is with the non-square matrices.

Alexander Bentkamp (Jul 05 2019 at 14:26):

How about this: https://en.wikipedia.org/wiki/Generalized_inverse

Kevin Buzzard (Jul 05 2019 at 14:28):

I am concerned that they talk about the real numbers here. Actually @Abhimanyu Pallavi Sudhir has some paper about inverse matrices in more generality than is usually used by mathematicians, but he definitely needed the ground field to be the reals.

Chris Hughes (Jul 05 2019 at 14:28):

I need it for rationals.

Johan Commelin (Jul 05 2019 at 14:29):

I've never really thought about this... can't the adjugate be generalised? What if you take cofactors where you don't delete a row and a column but "at least one row, and at least one column" so that you end up with squares of size k-1 × k-1 where k = min (m,n).

Kevin Buzzard (Jul 05 2019 at 14:30):

Will that give you a left inverse where one exists?

Johan Commelin (Jul 05 2019 at 14:30):

I have no idea if that would work, but it gives the correct result for square matrices.

Kevin Buzzard (Jul 05 2019 at 14:30):

Wait -- where are you putting these cofactors?

Kevin Buzzard (Jul 05 2019 at 14:30):

They need to be arranged in a nice box

Johan Commelin (Jul 05 2019 at 14:30):

Aah, good point

Mario Carneiro (Jul 05 2019 at 14:31):

I think the pseudoinverse is what you want

Johan Commelin (Jul 05 2019 at 14:31):

/me goes back to commutative algebra

Kevin Buzzard (Jul 05 2019 at 15:08):

/me decides that if Mario is wrong then Chris will just ask again

Mario Carneiro (Jul 05 2019 at 15:13):

the moore-penrose inverse looks right, but also looks complicated -> https://en.wikipedia.org/wiki/Proofs_involving_the_Moore%E2%80%93Penrose_inverse

Mario Carneiro (Jul 05 2019 at 15:14):

it also needs singular value decomposition...

Mario Carneiro (Jul 05 2019 at 15:16):

and apparently both the SVD and the M-P inverse require the base field to be R or C

Johan Commelin (Jul 05 2019 at 15:17):

That's pretty ugly

Johan Commelin (Jul 05 2019 at 15:17):

How can such a thing work for R and C but not for general fields?

Mario Carneiro (Jul 05 2019 at 15:18):

I highly suspect that's not actually true, but you know mathematicians

Johan Commelin (Jul 05 2019 at 15:18):

I understand that certain things only work for alg.closed fields, or only for ordered fields...

Mario Carneiro (Jul 05 2019 at 15:18):

There is quite a lot that is done on fields that are R or C

Kevin Buzzard (Jul 05 2019 at 15:18):

I think that some of these definitions are quite analysis-y

Mario Carneiro (Jul 05 2019 at 15:18):

hilbert spaces and banach spaces come to mind

Kevin Buzzard (Jul 05 2019 at 15:18):

Abhi told me that his matrix inverse stuff didn't work over a general field.

Kevin Buzzard (Jul 05 2019 at 15:18):

hilbert spaces and banach spaces come to mind

That's analysis

Johan Commelin (Jul 05 2019 at 15:19):

Hmm..., maybe it is because they want to mention A^*?

Mario Carneiro (Jul 05 2019 at 15:19):

They are analysis because it says "R or C"

Johan Commelin (Jul 05 2019 at 15:19):

Also, the section "Definition" on wiki, is really just a theorem (-;

Mario Carneiro (Jul 05 2019 at 15:20):

The definition of a pesudoinverse, as a predicate, isn't hard

Mario Carneiro (Jul 05 2019 at 15:20):

but proving existence is hard and requires these analysis-y assumptions

Mario Carneiro (Jul 05 2019 at 15:21):

By the way @Chris Hughes these certainly don't necessarily exist over Q

Johan Commelin (Jul 05 2019 at 15:23):

I expect that you can do similar things over CM fields and totally real fields

Mario Carneiro (Jul 05 2019 at 15:23):

wait, now I'm not sure if that's true. It's not clear to me if eigenvalue decomposition of some kind is required

Kenny Lau (Jul 05 2019 at 15:23):

but Q is totally real

Johan Commelin (Jul 05 2019 at 15:25):

It sure is...

Johan Commelin (Jul 05 2019 at 15:25):

So it works over the algebraic reals

Johan Commelin (Jul 05 2019 at 15:25):

That existence claim is a first order formula, right?

Johan Commelin (Jul 05 2019 at 15:25):

Now I hope that you can do some Galois descent to go to subfields

Mario Carneiro (Jul 05 2019 at 15:26):

The inverse of an invertible matrix exists without any eigenvalue decomp

Mario Carneiro (Jul 05 2019 at 15:26):

So maybe the pseudoinverse also exists

Mario Carneiro (Jul 05 2019 at 15:26):

Doing SVD definitely requires eigenvalues though

Mario Carneiro (Jul 05 2019 at 15:27):

so that restricts application to algebraically closed fields and maybe real closed fields

Johan Commelin (Jul 05 2019 at 15:28):

I would like yo deduce after the fact that it is defined over subfields

Kevin Buzzard (Jul 05 2019 at 15:28):

I would like yo deduce after the fact that it is defined over subfields

You've been doing too many Shimura varieties.

Kevin Buzzard (Jul 05 2019 at 15:29):

I don't think it always works out so easily :-)

Johan Commelin (Jul 05 2019 at 15:31):

@Kevin Buzzard Look at https://en.wikipedia.org/wiki/Proofs_involving_the_Moore%E2%80%93Penrose_inverse#Definition

Johan Commelin (Jul 05 2019 at 15:32):

That characterisation is pretty algebraic..., so I would hope that it lends itself to Galois descent.

Kevin Buzzard (Jul 05 2019 at 15:32):

remember we call it Galois lift here

Johan Commelin (Jul 05 2019 at 15:33):

@Chris Hughes Do you want something that computes?

Johan Commelin (Jul 05 2019 at 15:33):

Because this seems extremely slow...

Abhimanyu Pallavi Sudhir (Jul 05 2019 at 15:42):

I am concerned that they talk about the real numbers here. Actually Abhimanyu Pallavi Sudhir has some paper about inverse matrices in more generality than is usually used by mathematicians, but he definitely needed the ground field to be the reals.

That's determinants, not inverses. And yeah, it's related to the SVD.

Kevin Buzzard (Jul 05 2019 at 15:43):

well, I guess they're related here. So you can take the determinant of a 2x3 matrix with real coefficients, right?

Kevin Buzzard (Jul 05 2019 at 15:43):

But if that 2x3 matrix had rational coefficients, would the determinant be rational?

Abhimanyu Pallavi Sudhir (Jul 05 2019 at 15:43):

No.

Abhimanyu Pallavi Sudhir (Jul 05 2019 at 15:43):

|det(1, 2)| = sqrt(5)

Abhimanyu Pallavi Sudhir (Jul 05 2019 at 15:44):

Or whatever 2^2+1 is.

Kevin Buzzard (Jul 05 2019 at 15:44):

it'll depend on what field you're in

Abhimanyu Pallavi Sudhir (Jul 05 2019 at 15:44):

Oh wait -- but it can be a vector of rational entries.

Abhimanyu Pallavi Sudhir (Jul 05 2019 at 15:44):

If you use the determinant in (Pyle, 1962).

Abhimanyu Pallavi Sudhir (Jul 05 2019 at 15:45):

Or an m-vector with rational entries.

Kevin Buzzard (Jul 05 2019 at 15:46):

what is m?

Abhimanyu Pallavi Sudhir (Jul 05 2019 at 15:46):

The dimension of the domain space.

Abhimanyu Pallavi Sudhir (Jul 05 2019 at 15:46):

So for n by m matrices the "determinant" can just be the wedge product of the m columns.

Abhimanyu Pallavi Sudhir (Jul 05 2019 at 15:47):

The Pyle determinant just flattens it out into an nCm dimensional vector.

Abhimanyu Pallavi Sudhir (Jul 05 2019 at 15:51):

and apparently both the SVD and the M-P inverse require the base field to be R or C

Presumably because the SVD involves the notion of scaling and rotations. If you only allowed scaling, you would need algebraic closure. But allowing rotations allows the complex behavior to be mimicked by real numbers.

Kevin Buzzard (Jul 08 2019 at 17:17):

@Chris Hughes were any of these attempts useful? This seems like a reasonable MO question to me.

Chris Hughes (Jul 08 2019 at 17:45):

I know that the answer is no now. None of the pseidoinverses are particularly useful for what I'm doing.

Last updated: Dec 20 2023 at 11:08 UTC