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I'm trying to reduce this conic : $x^2+y^2+2xy+x+y=0$ to a canonical form.

I started with finding the eigenvalues of the matrix associated to the quadratic form $x^2+y^2+2xy$

I found $z_1=2 , z_2=0$ and a basis for the diagonalized matrix $e_1= ({1\over \sqrt2},{1\over \sqrt2})$ $e_2=({1\over \sqrt2},{-1\over \sqrt2})$

The quadratic form can be rewritten in this base as : $2x^2$ but what should I do next? How can I rewrite the other terms? I know that the solution is a couple of parallel lines...

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Following the method blindly $x^2+y^2+2xy+x+y={\bf x}^T A{\bf x}+K{\bf x}=\begin{pmatrix}x&y\end{pmatrix}\begin{pmatrix}1&1\\1&1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}+\begin{pmatrix}1&1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=0$,

we use $P=\begin{pmatrix}\frac1{\sqrt{2}}&\frac1{\sqrt{2}}\\\frac1{\sqrt{2}}&-\frac1{\sqrt{2}}\end{pmatrix}$.

Then ${\bf x}=P{\bf x}'$ gives ${\bf x}'^T P^TAP{\bf x}'+KP {\bf x}'=2x'^2+\sqrt{2}x'=\sqrt{2}x'(\sqrt{2}x'+1)=0$, two parallel lines, which we could also see directly since: $x^2+y^2+2xy+x+y=(x+y)^2+(x+y)=(x+y)(x+y+1)=0$.

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  • $\begingroup$ In $R^3$ (quadrics) the method remains unchanged? $\endgroup$ – Jonathan Baram May 14 '15 at 17:41
  • $\begingroup$ @JonathanBaram Yes. A general quadric ${\bf x}^T A{\bf x}+K{\bf x}+\text{number}$. Then substitute ${\bf x}=P{\bf x}'$, where P is the eigenvector matrix. $\endgroup$ – Jan-Magnus Økland May 14 '15 at 18:11

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