I'm studying quadratic forms: In the book I'm reading, he starts by looking at quadratic forms such as:
$$\varphi (x,y)=Ax^2+2Bxy+Cy^2$$
And that given this quadratic form, one can introduce via axis rotation the new coordinates $(s,t)$ with:
$$x= as-bt \quad \quad \quad y= bs-at$$
And
$$a=\cos \theta \quad \quad \quad \quad b=\sin \theta$$
And then this reduces to:
$$\varphi(x,y)=\varphi(as-bt,\; bs-at)=A's^2+2B'st+C't$$
Where:
$$\begin{eqnarray*} {A'}&=&{Aa^2+2Bab+Cb^2} \\ {B'}&=&{-Aab+B(a^2-b^2)+Cab} \\ {C'}&=&{Ab^2-2Bab-Ca^2} \end{eqnarray*}$$
I don't understand why $A',B',C'$ becomes that. Shouldn't:
$$\varphi(as-bt,\; bs-at)\stackrel{?}{=}A (a s-b t)^2+2 B (a s-b t) (b s-a t)+C (b s-a t)^2$$
In a trial to do something similar to what he did, I took one term and expanded:
$$C (b s - a t)^2=b^2 C s^2 - 2 a b C s t + a^2 C t^2=s^2(b^2 C-2abCs^{-1}+a^2Cts^{-2})$$
And I guess that $(b^2 C-2abCs^{-1}+a^2Cts^{-2})$ would be my $C'$, but it's very different from the $C'$ given by the book.
After you expand all terms, you should combine similar terms.
$$A(as-bt)^2+2B(as-bt)(bs-at)+C(bs-at)^2\\ =Aa^2s^2-2Aabst+Ab^2t^2+2Babs^2-2Ba^2st-2Bb^2st+2Babt^2+Cb^2s^2-2Cabst+Ca^2t^2\\ =(Aa^2+2Bab+Cb^2)s^2+2(-Aab+B(a^2-b^2)+Cab)st+(Ab^2-2Bab-Ca^2)t^2$$
With this you can define the $A',B',C'$ as desired.