Matrix Factorization - Numberphile
Why Mathematics Stack Exchange recommends it
Mathematics Stack Exchange askers cite it as the starting point when they want to explore or extend polynomial matrix factorizations in their own questions.
Recommended by 2 people across 2 different months, 2020 to 2022.
What it is
A Numberphile video in which David Eisenbud of MSRI explains matrix factorization of polynomials, starting from Dirac's trick of finding a matrix 'square root' of a sum of squares. It shows how a polynomial like xy - uv can be written as a product of matrices equal to the polynomial times the identity.
What people said
Recommends PickupSticks · 2021-04 · Mathematics Stack Exchange · on “Algorithm for Dirac-like "matrix factorization" of polynomials”
In this [this video] Numberphile video, they discuss Dirac changing to matrix equations to get something like a "square root" of $t^2 + x^2 + y^2 + x^2$ . In particular, they discuss the following multivariate polynomial, which is transformed to a matrix equation to get a kind of square root for this polynomial. $$(xy - uv) \longrightarrow \begin{bmatrix} 0 & 0 & x & u \\ 0 & 0 & v & y \\ y & -u & 0 & 0 \\ -v & x & 0 & 0 \\ \end{bmatrix}^2 = (xy-uv)I$$ They then mention the following paper, which apparently proves the following theorem: any polynomial with no constant or linear terms can be factored as a product of matrices with no constant terms. David Eisenbud, Homological Algebra on a Complete
Recommends Ivan Burbano · 2022-03 · Mathematics Stack Exchange · on “The certainty behind the spanning set of solution for a second order linear homogeneous ODE with constant coefficients”
There is a very neat way of approaching this that leaves absolutely no doubt of all of the solutions. I don't have a lot of time so I will be schematic about this. In any case, I think everything is found in this lecture [LINK] First, as shown in the lecture, you can use an integrating factor to reduce to the case $a=0$ . In that case, the remaining differential equation can be factored as $$\left(\frac{d}{dx}+i\sqrt{b}\right)\left(\frac{d}{dx}-i\sqrt{b}\right)y=0.$$ Setting $$Y=\left(\frac{d}{dx}-i\sqrt{b}\right)y,$$ you are left with a simple first order equation $$\left(\frac{d}{dx}+i\sqrt{b}\right)Y=0,$$ which can be rewritten as $$\frac{d\ln(Y)}{dx}=-i\sqrt{b}.$$ At which point is clear
Cites Axel Kemper · 2020-05 · Mathematics Stack Exchange · on “Matrix factorization of sums of products”
Inspired by David Eisenbud [LINK] 's Matrix Factorization Numberphile video [this video] , I try to factor sums of products of three variables: $$s_k = a_1b_1c_1 + a_2b_2c_2 + a_3b_3c_3 + \dots + a_kb_kc_k$$ A related paper [LINK] of Crisler and Diveris demonstrates how to factor sums of products of two variables: $$f_k = g_1h_1 + g_2h_2 + g_3h_3 + \dots + g_kh_k$$ Their method is inductive: In the case $k=1$ , we have $f_1 = gh$ and $[g][h] = [gh] = [f_1]$ is a $1 \times 1$ factorization. Assume now that $A$ and $B$ are matrix factorizations of $f_{k-1}$ , that $AB = If_{k-1}$ . The following is a matrix factorization of $f_k$ : $$ \huge{ \left[ \begin{array}{c|c} A & -g_k I \\ \hline h_k I & B