Let $g(x,y)=ax^2+bxy+cy^2$ for some real $a,b,c$.
Find a basis (or show one exists) $\{v_1,v_2\}$ for $\mathbb{R}^2$ such that $g(xv_1+yv_2)=gx^2+hy^2$ where $g,h=\pm 1$ or $0$.
A hint would be appreciated. Am I allowed to choose $a,b,c$? I know I can write this as a quadratic form using a matrix, but I don't see how that would help. How do I get the middle term $bxy$ to vanish by choosing a fixed basis?
(1) First, let $A = \begin{pmatrix}\alpha&\beta\\\beta&\gamma\end{pmatrix}$ be an arbitrary symmetric matrix. Find its eigenvalues $\lambda_{1,2}$ and corresponding eigenvectors $u_{1,2}$. Then you have $A = UDU^T$ with $U = [u_1,u_2]$ and $D = \operatorname{diag}(\lambda_1,\lambda_2)$, everything in terms of $\alpha,\beta,\gamma$. The matrix $U$ will be unitary.
(2) Then find a symmetric(!) matrix $A$ such that $g(x,y) = z^TAz$, where $z = (x,y)^T$. In the first step you have calculated the decomposition $A = UDU^T$ already. Thus, you have $$ g(z) = z^TAz = z^TUDU^Tz = w^TDw, $$ where $w = U^Tz$. As $z = Uw$, this gives $$ g(w_1u_1 + w_2u_2) = g(Uw) = w^TDw = d_1w_1^2 + d_2w_2^2. $$ That's almost what was asked for. You are only left with scaling everything appropriately.