I Canonical Form for quadratic equations *with* linear terms

arestes
Messages
84
Reaction score
4
TL;DR Summary
Not sure about the accepted "canonical form" for a quadratic equation WITH linear term
Hello:
I'm not sure if there's an accepted canonical form for a quadratic equation in two (or more) variables:
$$ax^2+by^2+cxy+dx+ey+f=0$$

Is it the following form? (using the orthogonal matrix Q that diagonalizes the quadratic part):

$$ w^TDw+[d \ \ e]w+f=0$$
$$w^TDw+Lw+f=0$$
where
$$ w=\begin{pmatrix}
x' \\
y'
\end{pmatrix} = Q^T
\begin{pmatrix}
x \\
y
\end{pmatrix}
$$
and
$$ L=[d \ \ e] $$

Or is it a form with translated coordinates:
$$a(x'-m)^2+b(y'-n)^2+c(x'-m)(y'-n)+d(x'-m)+e(y'-n)+f'=0$$
with some to-be-determined constants m and n such that the linear terms vanish, which can be then used to change variables x'=x+m and y'=y+n.

I tried to find these m and n (expanding the binomials) but the simultaneous equations to satisfy in order to remove the linear terms are restricting and it seems to be impossible when $$c^2-4ab=0$$

Is it enough to leave the linear terms and call it "canonical form" just by diagonalizing the quadratic terms?
 
Physics news on Phys.org
That's a neat condition (I haven't checked your work). ##c^2-4ab=0## means that ##D## is going to be singular, which people often want to exclude when talking about quadratic forms anyway.

In particular, suppose your quadratic form was ##x^2+y=0##. It's pretty clear you're not going to change coordinates to make this look nice, since it's just not quadratic in one direction.
 
My personal preference is the first one (simple scalar).
 
WWGD said:
I think you're referring to Quadratic forms:
https://en.wikipedia.org/wiki/Quadratic_form
Yeah but quadratic forms don't have linear terms.

Thanks for reminding me of wikipedia. I did find the info I needed (almost) here:https://en.wikipedia.org/wiki/Matrix_representation_of_conic_sections

I found that there is a general form to write the equation of a conic (valid for central conics, which excludes parabolas because the K mentioned below doesn't exist):
1623477016440.png


Here they call this equation "standard canonical form" but I'm not sure if it's THE canonical form.
My question still stands regarding the naming convention but I just realized that it's just that: semantics.

So, in the end (after learning stuff about projective geometry and homogeneous coordinates) is the name "canonical form" used for the "standard canonical" form above (although it doesn't work for parabolas for which, I guess it'll be just y^2=ax or x^2=ay)?
 
I asked online questions about Proposition 2.1.1: The answer I got is the following: I have some questions about the answer I got. When the person answering says: ##1.## Is the map ##\mathfrak{q}\mapsto \mathfrak{q} A _\mathfrak{p}## from ##A\setminus \mathfrak{p}\to A_\mathfrak{p}##? But I don't understand what the author meant for the rest of the sentence in mathematical notation: ##2.## In the next statement where the author says: How is ##A\to...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
When decomposing a representation ##\rho## of a finite group ##G## into irreducible representations, we can find the number of times the representation contains a particular irrep ##\rho_0## through the character inner product $$ \langle \chi, \chi_0\rangle = \frac{1}{|G|} \sum_{g\in G} \chi(g) \chi_0(g)^*$$ where ##\chi## and ##\chi_0## are the characters of ##\rho## and ##\rho_0##, respectively. Since all group elements in the same conjugacy class have the same characters, this may be...
Back
Top