Vtyjkyuk

I'm trying to read Undergraduate Algebraic Geometry by Miles Reid and I am having some difficulty understanding some of his statements. Specifically I'm trying to solve problem 1.3 which is to prove thay every conic in $mathbbR^2$ can be put to one of the following forms by means of an affine transformation:

Ellipse;
parabola;
hyperbola;
the empty set;
a line ( $x=0$ );
a pair of lines ($xy=0$ );
parallel lines ( $x(x-1)=0$ );
'double line' ( $x^2=0$)

Now, he gives a definition of an affine transformation $T$ in this space in terms of a 2x2 invertible matrix A, and a translation vector B, i.e. an affine transformation is one that has the form $T(x)=Ax+B$, where $ x in mathbbR^2$

I thought maybe I could somehow apply this to the general equation for a conic, i.e. $ax^2+bxy+cy^2+dx+ey+f$, but I'm not sure how to do that or if it even makes any sense.

Any suggestions, hints or clarifications are appreciated.

This is classic "analytic geometry". An affine transformation has the form
beginalign
x&=rx'+sy'+t\
y&=ux'+vy'+w
endalign
where $rv-sune0$. So in the new coordinates, the conic becomes
$$a(rx'+sy'+t)^2+b(rx'+sy'+t)(ux'+vy'+w
)+c(ux'+vy'+w)^2+d(rx'+sy'+t)+e(ux'+vy'+w)+f=0$$
or
$$a(rx'+sy')^2+b(rx'+sy')(ux'+vy')+c(ux'+vy)^2+
textlinear and constant terms=0.$$

Can you show that there are suitable $r$, $s$, $u$, $v$ such that
$$a(rx'+sy')^2+b(rx'+sy')(ux'+vy')+c(ux'+vy)^2$$
is one of $x'^2+y'^2$, $x'y'$, $x'^2$ or zero?
If so you can examine each case in turn. For instance
$$x'^2+y'^2+textlinear and constant terms=0$$
will give either an ellipse (a circle even) or a point or the empty set
(depending on the lower terms) etc.

I find it a bit easier to work with homogeneous coordinates and matrices for this. In the following, I use lower-case bold letters to represent homogenous vectors and a tilde to indicate their corresponding inhomogeneous coordinate tuples.

Your general conic equation can be written as $$mathbf x^TCmathbf x = beginbmatrixx&y&1endbmatrix beginbmatrix a&frac b2&frac d2 \ frac b2 & c & frac e2 \ frac d2&frac e2&f endbmatrix beginbmatrixx\y\1endbmatrix = 0.$$ If you have an invertible point transformation $mathbf x' = Mmathbf x$, then $$mathbf x^TCmathbf x = (M^-1mathbf x')^TC(M^-1mathbf x') = mathbf x'^T(M^-TCM^-1)mathbf x',$$ so the conic’s matrix transforms as $C' = M^-TCM^-1$.

The invertible affine transformation $tildemathbf x = Atildemathbf x'+tildemathbf t$ (note that I’m inverting your transformation for simplicity) can be represented by the $3times 3$ matrix $$M^-1 = left[beginarrayc A & tildemathbf t \ hline mathbf 0^T & 1endarrayright]$$ so that $mathbf x = M^-1mathbf x'$. Writing $C$ in block form as $$C = left[beginarrayc Q & tildemathbf b \ hline tildemathbf b^T & f endarrayright],$$ which corresponds to the form $tildemathbf x^TQtildemathbf x+2tildemathbf b^Ttildemathbf x+f = 0$ of the general conic equation, we then have $$M^-TCM^-1 = left[beginarrayc A^T & mathbf 0 \ hline tildemathbf t^T & 1endarrayright] left[beginarrayc Q & tildemathbf b \ hline tildemathbf b^T & f endarrayright] left[beginarrayc A & tildemathbf t \ hline mathbf 0^T & 1endarrayright]
= left[beginarrayc A^TQA & A^T(Qtildemathbf t+tildemathbf b) \ hline (Qtildemathbf t+tildemathbf b)^T A & mathbf t^TCmathbf t endarrayright].$$ There are several things to note here: the quadratic part of the equation represented by $Q$ is only affected by the linear part of the affine transformation; the new constant term is the left-hand side of the untransformed equation evaluated at $mathbf t$, the translation part of the transformation; and the signs of $det C$ and $det Q$ are preserved by affine transformations. If you examine the signs of these determinants for a canonical example of each the types of conics that you’ve listed, you’ll find that the combination of signs determines the type of conic—opposite signs for an ellipse, $det Q=0$ and $det Cne0$ for a parabola, $det C=0$ and $det Qlt0$ for a pair of intersecting lines, and so on. In fact, $det Q$ is a multiple of the discriminant of the conic equation. It’s a bit tedious, but not very difficult, to work out appropriate transformations to bring an arbitrary conic into one of these canonical forms. (You might need to use the fact that a real symmetric matrix is orthogonally diagonalizable to prove existence of $A$ in all cases.) For example, $det Qne0$ for any central conic, so the linear terms in the equation of a central conic can be eliminated by choosing $tildemathbf t = -Q^-1tildemathbf b$.