Skip to main content

[Community Question] Geometry: Factoring singular conics into linear forms

One of our user asked:

I'm looking for an easy way to factor singular conics into linear forms in order the following exercise.

Which of the following quadratic forms define a singular > conic? Write those as a product of two linear forms.

(a) $x_0^2-2x_0x_1+4x_0x_2-8x_1^2+2x_1x_2+3x_2^2$

(b) $x_0^2-2x_0x_1+x_1^2-2x_0x_2$

(c) $3x_0^2-2x_0x_1$

The matrices of (b) and (c) have full rank and so the quadratic forms are non-degenerate. For (a), we have the matrix

$$ M= \begin{bmatrix} 1 & -1 & 2 \\ -1 & -8 & 1 \\ 2 & 1 & 3 \end{bmatrix} $$

where $det(M)=0$. One way to solve the exercise would be to orthogonally diagonalize M to get rid of the mixed terms. However this involves computing the eigenvalues of M and finding the corresponding eigenvectors. So my question is: Is there a quicker way to do this or do I have to go through the calculations?


Labels:

Comments

Post a Comment

Popular posts from this blog

[Community Question] Calculus: Manifold with boundary - finding the boundary

One of our user asked: I have the manifold with boundary $M:= \lbrace (x_1,x_2,x_3) \in \mathbb R^3 : x_1\geq 0, x_1^2+x_2^2+x_3^2=1\rbrace \cup\lbrace (x_1,x_2,x_3) \in \mathbb R^3 : x_1= 0, x_1^2+x_2^2+x_3^2\leq1\rbrace$ and I need to find the boundary of this manifold. I think it is $\lbrace (x_1,x_2,x_3) \in \mathbb R^n : x_1= 0, x_2^2+x_3^2=1\rbrace$ , the other option is that the boundary is the empty set? I think the first is right? Am I wrong?
Post a Comment
Read more

[Community Question] Linear-algebra: Are linear transformations between infinite dimensional vector spaces always differentiable?

One of our user asked: In class we saw that every linear transformation is differentiable (since there's always a linear approximation for them) and we also saw that a differentiable function must be continuous, so it must be true that all linear operators are continuous, however, I just read that between infinite dimensional vector spaces this is not necessarily true. I would like to know where's the flaw in my reasoning (I suspect that linear transformations between infinite dimensional vector spaces are not always differentiable).
Post a Comment
Read more

Order of elements of the Prüfer groups $\mathbb{Z}(p^{\infty})$

Let $\mathbb{Z}(p^{\infty})$ be defined by $\mathbb{Z}(p^{\infty}) = \{ \overline{a/b} \in \mathbb{Q}/ \mathbb{Z} / a,b \in \mathbb{Z}, b=p^i$ $ with$ $ i \in \mathbb{N} \}$ , I wish show that any element in $\mathbb{Z}(p^{\infty})$ has order $p^n$ with $n \in \mathbb{N}$ . i try several ways but I have not been successful, some help ?? thank you
Post a Comment
Read more