533 reputation
49
bio website wallenborn.net
location Bonn, Germany
age
visits member for 2 years, 9 months
seen Nov 24 at 21:48

Jul
11
accepted $\alpha x^2+\beta y^2=\gamma$ solvable over $\mathbb Q$ iff $ax^2+by^2=z^2$ solvable over $\mathbb Z$ with coprime $x,y,z$?
Jul
11
asked $\alpha x^2+\beta y^2=\gamma$ solvable over $\mathbb Q$ iff $ax^2+by^2=z^2$ solvable over $\mathbb Z$ with coprime $x,y,z$?
Jun
27
comment algorithm for solving diagonal quadratic equations over real or complex numbers
Yes I thought that it's not in the spirit of the paper, but I want to be very precise at this point. You say that solving linear equations in exact arithmetic has $O(n^3)$ (by Gaussian elimination of course). And speaking of which, I realize that I forgot a square :-/ (even though I wrote "quadratic equation" in the title) - now corrected. So I'm in the case of a quadratic equation where one cannot apply Gauss' algorithm. The next question is: what do you mean with "exact arithmetic"? Working symbolically wherever necessary? Thanks for all your time btw!
Jun
27
revised algorithm for solving diagonal quadratic equations over real or complex numbers
added 2 characters in body
Jun
27
comment algorithm for solving diagonal quadratic equations over real or complex numbers
I added a link to the paper. It is a rather algebraic context. What you say is, concerning my first follow-up question, equivalent to extending the alphabet to $\mathbb F$ and adding special states for adding and multiplying two cells to the set of states of the turing machine, right?
Jun
27
revised algorithm for solving diagonal quadratic equations over real or complex numbers
added 86 characters in body
Jun
27
asked algorithm for solving diagonal quadratic equations over real or complex numbers
Jun
25
awarded  Scholar
Jun
25
accepted Two elements $a$ and $b$ of a quadratic space generate a nondegenerate two dimensional subspace if and only if $(a\cdot a)(b\cdot b)\neq(a\cdot b)^2$
Jun
25
comment Two elements $a$ and $b$ of a quadratic space generate a nondegenerate two dimensional subspace if and only if $(a\cdot a)(b\cdot b)\neq(a\cdot b)^2$
So either $\dim(P)<2$ or ($(b.b)=0$ and $(a.b)=0$). The later implies that $b\in\mathop{\rm rad}$. Now if $b=0$, $\dim(P)<2$ and if $b\neq 0$, $P$ is degenerate. Thank you very much!
Jun
25
comment Two elements $a$ and $b$ of a quadratic space generate a nondegenerate two dimensional subspace if and only if $(a\cdot a)(b\cdot b)\neq(a\cdot b)^2$
looking at this element (let's call it $c=(b.b)a-(a.b)b$) i get $c.a=0$ and $c.b=0$ yielding that $c$ is in the radical of $P$. This implies that $c$ is zero or $P$ is degenerate. In the later case we are done. If $c$ is zero, we have that $(b.b)a-(a.b)b=0$ which is a linear combination of $0$ in $P$. So either $\dim(P)<2$ or $(b.b)=0$ and $(a.b)=0$.
Jun
25
awarded  Editor
Jun
25
comment Two elements $a$ and $b$ of a quadratic space generate a nondegenerate two dimensional subspace if and only if $(a\cdot a)(b\cdot b)\neq(a\cdot b)^2$
of course "is zero"... sorry (edited).
Jun
25
revised Two elements $a$ and $b$ of a quadratic space generate a nondegenerate two dimensional subspace if and only if $(a\cdot a)(b\cdot b)\neq(a\cdot b)^2$
added 2 characters in body
Jun
25
asked Two elements $a$ and $b$ of a quadratic space generate a nondegenerate two dimensional subspace if and only if $(a\cdot a)(b\cdot b)\neq(a\cdot b)^2$
Jun
10
awarded  Supporter
May
28
awarded  Teacher
May
28
answered Is there a rational univariat polynomial of degree 3 with 3 irrational roots?
May
27
awarded  Student
May
27
comment Is there a rational univariat polynomial of degree 3 with 3 irrational roots?
@RagibZaman yes, that's absolutely right and it works out perfectly. Thank you all! I'll post the answer to this question in around 6 hours, when the software here allows me to.