Show that there is no affine transformation that takes a circle to a hyperbola in $\mathbb{R}^2$

• In $\mathbb R^2$, a circle is a bounded set and a hyperbola is not.

• An invertible affine transformation of the plane maps bounded sets to bounded sets.

(One can replace «bounded» by «compact» or «connected»...)

The other Mariano is of course cheating: this is algebraic geometry after all...

The gentlemanly way of doing this is to observe how the coefficients of the homogeneous part of top degree changes when you do an affine transformation, and then invoke Sylvester's Law of Inertia.

In more detail... Let $f\in \mathbb R[x,y]$ be a polynomial of degree two, which we can write uniquely as $$f(x,y) = v^tav+b^tv+c$$ with $v$ the vector $\left(\begin{smallmatrix}x\\y\end{smallmatrix}\right)$, $a$ a symmetric $2\times 2$ matrix, $b\in\mathbb R^2$ and $c\in\mathbb R$.

If $[A,\bar s]$ is an affine transformation, then $$[A,\bar s]\cdot f=v^ta'v+b'^tv+c'$$ for some new $a'$, $b'$ and $c'$ which we can compute explicitly. Most interestingly, we have $$a'=A^taA.$$ Sylvester's Law of Inertia then implies that $a$ and $a'$ have the same number of positive eigenvalues, the same number of negative eigenvalues, and the same number of zero eigenvalues.

Now, the matrix $a$ corresponding to your $f$ is $\begin{pmatrix}1&0\\0&1\end{pmatrix}$, while that of $g$ is $\begin{pmatrix}1&0\\0&-1\end{pmatrix}$. Since they have different numbers of negative eigenvalues, there is no affine transformation mapping one to the other.

Why think when you can simply compute?

So $f(x,y)=x^2+y^2-1$. Consider the affine transformation given by $$\left\{\begin{array}{l}x\leftarrow ax+by+s\\y\leftarrow cx+dy+t\end{array}\right.$$ Applying it to $f$ we get $$f(ax+by+s, cx+dy+t)=(a^2+c^2) x^2+2(a b + c d)xy+(b^2 + d^2)y^2+\text{terms of lower degree}.$$ If this is to be equal to $g(x,y)=x^2-y^2-1$, then, in particular, looking at the coefficient of $y^2$ we see that we must have $$b^2+d^2=-1.$$ Of course, this is not going to work...