Sylvester's law of inertia

Law of inertia. Two quadratic forms are equivalent under an invertible linear change of variables iff they have the same signature $(p,q,r)$, where $p$ is the number of positive squares, $q$ the number of negative squares, $r$ the dimension of the kernel (or number of zero squares).

Reduction 1. $Q=x^2-y^2$ is already diagonal: $p=1$, $q=1$, $r=0$.

Reduction 2. Let $x=u+v$, $y=u-v$. Then
$$Q=(u+v)^2-(u-v)^2=4uv.$$
This is not diagonal. To diagonalize, use $u=(s+t)/2$, $v=(s-t)/2$: $4uv=(s+t)(s-t)=s^2-t^2$. Signature $(1,1,0)$, same as before.

Conclusion. The signature is invariant: $(1,1,0)$ in both reductions.