Question regarding quadratic form exercise in Hoffman Kunze In the book the quadratic form associated with a bilinear form f is defined as $q(\alpha)=f(\alpha,\alpha)$. Then, if $U$ is a linear operator on $\mathbb R^2$ an operator $U^\dagger$ on the space of quadratic forms on $\mathbb R^2$ is defined by the equation $U^\dagger q(\alpha)=q(U(\alpha))$. Then, an exercise asks to find an operator $U$ such that if the quadratic form $q$ on $\mathbb R^2$ is $q=ax_1^2+2bx_1x_2+cx_2^2$, then $U^\dagger q=ax_1^2+\left(c- \frac {b^2}a\right)x_2^2$
A hint is given that recommends to complete the square in other to find $U^{-1}$, but I have no idea what to do.
 A: I've already tried several things but it is sadly clear I'm not understanding the hint given at all. This is what I've managed to do so far (too long for a comment):
Completing the square in the unique place where it makes some sense to me:
$$q(x,y):=ax^2+2bxy+cy^2=a\left(x+\frac bay\right)^2+\left(c-\frac{b^2}a\right)y^2$$
Now, any linear operator on the real plane is of the form
$$U\binom xy=\binom{\alpha x+\beta y}{\gamma x+\delta y}\;,\;\;\alpha,\beta,\gamma,\delta\in\Bbb R$$
(This is a good exercise: a map $\;T:\Bbb F^n\to\Bbb F^m\;,\;\;\Bbb F\;$ a field,  is a linear map iff it is of the form
$$T(x_1,\ldots,x_n):=\left(t_1(x_1,...,x_n),\ldots,t_m(x_1,..,x_n)\right)$$
with every $\;t_i\;$ a homogeneous polynomial of degree one in $\;x_1,...,x_n\;$) .
So we then have
$$U^\dagger(q(x,y))=q\left(U\binom xy\right):=a(\alpha x+\beta y)^2+2b(\alpha x+\beta y)(\gamma x+\delta y)+c(\gamma x+\delta y)^2$$
Somehow then, it seems to be we should be able to compare (equality?) the last xpression with the first one above and get what the coefficients $\;\alpha,\beta,\gamma,\delta\;$ are...
Where and how does $\;U^{-1}\;$ kick in with the "complete the square" thing is, so far, beyond my comprehension...
