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3h
comment Using resultants to check if multivariate polynomials have a common factor - is my proof correct?
That will work, but for some reason I prefer not to look at conjugates, so I look at the all the square ("$6 \times 6$") submatrices and compare their determinants to zero. I'm okay with having several simultaneous polynomial conditions and not just one.
3h
comment Determinant-like expression for non-square matrices
Thank you, this is informative. In my use, I cannot use conjugation. However, I found that I can work with several polynomial expressions being $0$ simultaneously and not just one (I didn't write it in the original post because I didn't think about it when I posted it). So I ended up comparing the determinants of all $n \times n$ submatrices to zero, which works in any field, as detailed by Joonas's answer.
3h
accepted Notation for vector space of polynomials of bounded degree
22h
asked Notation for vector space of polynomials of bounded degree
23h
accepted Using resultants to check if multivariate polynomials have a common factor - is my proof correct?
1d
comment Using resultants to check if multivariate polynomials have a common factor - is my proof correct?
Follow up: I figured out that since I'm looking at unions of algebraic conditions anyway, then the condition "one of the $6\times 6$ sub-matrices has determinant zero" is an algebraic condition. So, we're good!
1d
comment Using resultants to check if multivariate polynomials have a common factor - is my proof correct?
Thank you for your answer! I tried to follow the approach you suggested, with a big resultant. However I encountered an issue: Consider $\deg f = 2, \deg g = 2, n = 2$. Then we need to check that the $6$ polynomials $f, xf, yf, g, xg, yg$ are dependent, but the space of polynomials of degree $3$ or less is $10$-dimensional. So the Sylvester matrix is not a square but of size $6 \times 10$ - and it's unclear how to define the resultant. See also my follow up question about it: math.stackexchange.com/questions/903028/…
1d
accepted Determinant-like expression for non-square matrices
1d
comment Determinant-like expression for non-square matrices
Answering myself: Consider complex $2 \times 1$ matrices. Only the zero matrix isn't full rank. So we need a polynomial $P(z, w) : \mathbb{C}^2 \to \mathbb{C}$ whose only zero is at $z=w=0$. This is impossible.
2d
comment Determinant-like expression for non-square matrices
This is an interesting approach! Can it be generalized to the complexes, or a general field?
2d
comment Determinant-like expression for non-square matrices
@DanielV, it is a multivariate polynomial of degree $n$ in $n^2$ variables (the matrix entries). The monomials are the products of "generalized diagonals" in the matrix and the coefficients are $\pm 1$.
2d
asked Determinant-like expression for non-square matrices
Aug
17
comment An infinite group $G$ and $\forall x\in G, x^n=e$
In fact... Let $G$ be the group generated by $a, b$ with only the relations $a^2 = b^2 = e$. Then $a, b \in H$ but $ab$ is not in $H$ because it has infinite order, so in the non-Abelian case, $H$ might not be a subgroup.
Aug
17
revised An infinite group $G$ and $\forall x\in G, x^n=e$
Oops...
Aug
17
comment An infinite group $G$ and $\forall x\in G, x^n=e$
Oops - I don't know whether it's a subgroup if the group is not Abelian.
Aug
17
answered An infinite group $G$ and $\forall x\in G, x^n=e$
Aug
17
comment Solve $5^{2x+2}-5^{x+2}+6=0 $
To a certain degree of approximation, yes.
Aug
16
comment Do the ratios of successive primes converge to a value less than 1?
Could you please explain the reason for the downvote?
Aug
16
answered Do the ratios of successive primes converge to a value less than 1?
Aug
14
comment Representing 2D objects in 4D
I don't know if it has a special name. The idea is the following: Let's say we're at 3D space. Then a line has two perpendicular directions. (If you place a person in the space so the line goes through his stretched-out arms, then the perpendicular directions are "forward" and "up" from the person's point of view. Note this is arbitrary and more choices may be made...) The two equations represent these perpendicular directions, since a linear equation is $(x,y,z) \cdot v = c$ where $v$ is a direction vector. $(x,y,z) \cdot v$ is constant if and only if the line is perpendicular to $v$.