Factorisation using Matrices

We explore the idea of taking a polynomial multiplied by an identity matrix, and factorising it into two matrices.
Further reading:
Crisler, D., & Diveris, K. (2016). Matrix factorizations of sums of squares polynomials. Pi Mu Epsilon Journal, 14(5), 301-306.
Eisenbud, D. (1980). Homological algebra on a complete intersection, with an application to group representations. Transactions of the American Mathematical Society, 260(1), 35-64.
00:00 Intro
01:07 2x2 Matrices
03:42 Polynomials with 2 terms
05:25 Polynomials with 3 terms
09:24 Block matrix multiplication
12:23 A possible 4x4 factorisation
16:06 Ensuring commutativity
18:36 Factorisation for 3 terms
20:00 Generalisation for 3 terms
22:21 Even more terms

[@thedude882]

If you use the standard representation of non-zero complex numbers to real invertible 2 by 2 matrices, then the factorization of x^2+y^2 as (x+iy)(x-iy) corresponds to the matrix factorization you showed.

[@rv706]

Thought the same as soon as I watched the video! :)

[@saj_f0]

Deserve more subscriptions

[@hypnovia]

These matrices are of a similar form to the matrix representation of quaternions! a^2+b^2+c^2+d^2=(a+bi+cj+dk)(a-bi-cj-dk)

[@DrBarker]

Yes, and there is also some shared structure with complex numbers, which can be represented by 2x2 matrices of a certain form.

[@omograbi]

I was a little bit confused because i guess in 1:47 you've missed a negative sign on c in the second matrix.

[@DrBarker]

Well-spotted! Yes, this should be "-c".

[@categorygrp]

i read the title and then scrolled away before realizing this is a really neat approach

[@rockapedra1130]

This is incredible!!!😊

[@aweebthatlovesmath4220]

Wow what a cool way of factoring multivariable polynomials! At3:26 the first matrix is the complex representation of x+iy as a 2 by 2 matrix and second one is the conjugate of it so it's kind of generalisation of factoring using complex numbers?!

[@ConManAU]

Yep! There's an isomorphism between the complex numbers and 2x2 matrices of that form.

[@zathrasyes1287]

Pretty cool stuff!

[@NightWanderer31415]

The 2x2 factorization of x^2+y^2 is actually equivalent to the complex number factorization if we construct complex numbers as 2x2 matrices.

[@tcoren1]

For the first example. Define J= (0 -1) (1 0) We can clearly see J^2=-I. Then we take x^2+y^2=(x+iy)(x-iy), and replace the i's with J, we get the solution you found. This is hardly surprising, if we treat a complex number as a vector of two real numbers, multiplying by J exactly matches with multiplying the complex number by i

[@jmathg]

It's really cool that the factorizations correspond to matrix representations, e.g. [0 -1 1 0] being the representation of i, which helps explain the factorization x^2 + y^2 = (x + iy)(x - iy). It got me thinking, can we express the quadratic formula using 2x2 matrices? Could we do the same with cubic or quartic? What about quintic - will the nonsolvability appear in some linear algebraic way? Or are the jacobi theta functions (that allow a general formula for the roots of a quintic) also representable as nxn matrices for some n?

[@user-xn3js2bw4h]

How intersting! I didn't know you could use matrices like that.

[@pableraspfgpfg468]

Which are the possible uses for this method? Is there some way to take advantage of this factorization for any known problem that cannot be solved as simply without this?

[@DrBarker]

I believe the idea of factorising matrices related to polynomials may be of interest in homological algebra, but I'm not sure of the details. You could check out the further reading in the video description.

[@bjornfeuerbacher5514]

@@DrBarker Isn't this essentially the same idea which was used by Dirac when he constructed the relativistic version of the Schroedinger equation?

[@TranquilSeaOfMath]

Interesting algebra.

[@crimfan]

Using the 2x2 matrix representation of complex numbers. Cool. Is this (or some fancier version) how programs like Mathematica factor polynomials?

[@nathanisbored]

the only thing i dont like about this is that, when factorizing polynomials, theres an implicit restriction that the factors have to both also be polynomials. otherwise we could use for example a rational function times a polynomial, which of course would be a trivial "factorization". it seems like when we translate to matricies, theres no analogous restriction (perhaps on say, the determinants, or the eigenvalues or something). For example, you could just factorize the starting matrix as its square times its inverse, as a trivial factorization. maybe the methods in the video are actually interesting factorizations, but its not clear to me that they should be. the methods themselves at least are interesting though.

[@ivanerofeev1269]

They are still polynomials with matrices as coefficients. (Or if you prefer matrices with polynomial inputs)

[@nathanisbored]

only with respect to the field multiplication, but we're considering a factorization over the matrix multiplication

[@DrBarker]

This is a good point. I believe that this overall type of factorisation/structure is of interest in homological algebra, but that goes way above my level of knowledge of abstract algebra. So I'm not sure exactly which factorisations would be widely considered "interesting" and which would be "trivial", but I personally find this factorisation interesting!

[@japanada11]

@@nathanisbored "we're considering a factorization over the matrix multiplication" having polynomial entries is still a valid and interesting restriction! Given any ring R you can define the ring M_n(R) of nxn matrices with coefficients in R, and ask for factorizations that use only elements of M_n(R). This is an extremely important construction with lots of applications (e.g. in the study of lattices, which occur everywhere from number theory through the study of modular forms, to chemistry and crystallography, to post-quantum cryptography, the change-of-basis matrices we consider must lie in M_n(Z) where Z is the integers: a basis change with non-integer entries would take you outside of the grid!) so it is actually very important to understand how factoring works even when matrix multiplication is restricted to matrices with no denominators in any entry.

[@japanada11]

In particular, the proposal of factoring a matrix into its square times its inverse is not valid if you're working in M_n(R), because most nxn matrices with entries in R don't have an inverse matrix with entries in R. Granted, it is still true that there are still lots of "trivial" factorizations (e.g. take any nxn integer matrix U whose inverse is also an integer matrix, then A = (AU^-1) . U is a factorization), in the same way that we have trivial integer factorizations like 6=1*6 or 6=(-1)*(-6). So it is true that one does have to add extra constraints, such as requiring neither factor to be invertible in M_n(R) (which is a constraint satisfied by all the factorizations in the video: the inverse matrices of all the factorizations have nontrivial denominators). Finding nontrivial factorizations in M_n(R) is not easy (in fact it's an area of active research; look up "integer matrix factorization" on Google scholar for a bunch of recent papers in the subject); this video gives a tiny glimpse of some of the tools that go into this very rich and complicated subject.

[@gibbogle]

What I missed is the point of this. What do we use this for? How do we get the factorisation of the original polynomial?

[@ojas3464]

👍Observing that Half of the Matrix Factors at each stage will have zeroes on their antidiagonals, does it save computational time by starting with zeroes as these elements? If so some rigorous justification maybe needed, beyond my skill-sets.

[@cparks1000000]

Since matrix multiplication has zero divisors, this seems less useful than standard factorization. It's still cool though.

[@marcfreydefont7520]

Not sure how this is ‘factorising’ polynomials. The formula shown here leads to p(x,y,z,s).I = A.B. Isn’t it obvious that A and B are the inverse of each other, subject to dividing that the polynomial p(x,y,z,s)? Then when one takes determinants on both sides, depending on the dimension of the identity matrix (2x2, 4x4, 8x8, etc.), one gets, [p(x,y,z,s)]^8 = q(x,y,z,t).r(x,y,z,t). But it seems to me that by construction, the q and r polynomials are equal and both are [p(x,y,z,s)]^4? The interesting thing would be to have a construction where q and r are different polynomials but I don’t see how this can be… leading to a trivial decomposition of p^(2k)=p^k.p^k? Or am I missing something?

[@japanada11]

The exact same thing happens with complex numbers. In the same way that matrices have determinants that take you back to real numbers, complex numbers have norms (square of absolute value), N(x+iy) = x^2+y^2, that take you back to real numbers. So by your exact argument, the factorization x^2+y^2 = (x+iy)(x-iy) is useless, because when you take norms on both sides, you get the trivial decomposition (x^2+y^2)^2 = (x^2+y^2)(x^2+y^2). So would you argue that complex numbers are useless, because when you take norms of these factorizations you just get trivial identities?

[@japanada11]

The point with both complex numbers and matrices is that you're introducing _new elements_ to your number system to allow for factorizations that didn't exist before. Taking norms/determinants takes you back to the original system and kills off all that new power, so of course you only get trivial stuff. The point is to _not_ throw the new system away, and treat the matrices / complex numbers as new elements in their own right, giving new factorizations that didn't exist before. In fact, complex numbers are a special case of this exact matrix construction: if you represent a+bi by the matrix [[a, b], [-b, a]] then you recover all the same rules of arithmetic (the complex numbers can be embedded into the ring of 2x2 matrices). We extend the reals to the complex numbers to allow for new factorizations; what he's doing in this video is a direct generalization of this.

[@spitsmuis4772]

Any (easy) applications?

[@DrBarker]

Not that I'm aware of. I suppose we could say an application is making an interesting exercise for students learning about linear algebra, but that doesn't really answer your question!

[@theor4343]

@@DrBarker I wonder if something interesting happens with second order PDEs. The resulting matrix factors would be first ordered and solvable by characteristics I think

[@dacianbonta2840]

addled exposition very much in the style of Barnard.Child.Algebra. Not that England cannot produce great educators (see Ferrar) looks like great math minds like Turing, Conway, Wiles, Maynard appear despite of, not because of, english mathematics education. I suppose that if a few years of the above don't kill your love for math, tensorial fibrations of non-riemannian simplicial polytopes do appear a walk in the park

[@mauisstepsis5524]

I have to disagree with the very beginning of this setup. x^2 + y^2 is a 1-by-1 scalar, (x^2 + y^2) * I is a 2-by-2 matrix.

[@koenth2359]

Your writing of right brackets and B, although it may looking a little bit clumsy, is a clear sign that you taught yourself to write before school, clearly a consequence of high intelligence.

[@stanislawpalka9015]

Completelly wrog!. For Polynomial of multiples variables with complex coefficients one can always states if it is factorisable. If is coefficients of factorisation can be written. No matter how coefficients are.