My question concerns the use of a linear change of coordinates applied to an algebraic variety.
I would like to illustrate my question with an example:
If we consider the curve $x^2+y^2-1=0$ over $\mathbf{R}$, and then apply a linear change of coordinates given by $(x, y) \mapsto (x+y, x-y)$ one can check that this is clearly a linear isomorphism, and applying it to the variables in our equation we obtain $2x^2 +2y^2-1=0$.
Now clearly this does not send points of one curve to points of the image curve as one can clearly check that $(1, 0) \mapsto (1, 1)$ does not lie on the image curve. Geometrically we can immediately see that the image curve is the circle of radius $\frac{1}{\sqrt{2}}$ and in this case there is a clear correspondence between the image curve and the original curve, but we do not always have a nice geometric interpretation for varieties in general.
Now I have seen results concerning the classification of varieties up to a linear change of coordinates, but I had naively assumed that these linear maps took the solution set of the original variety to the solution set of the variety obtained after applying the change of coordinates. Now the example above illustrates that this is clearly not the case, which leads me to the question:
How does the solution set of the variety after a linear change of coordinates correspond to the variety described by the original polynomial?
Any sources would be greatly appreciated in addition to an answer.
I think you confused yourself a bit with this example.
You start with the variety $x^2+y^2-1=0$. The change of coordinates you propose seems to be $x=u+v,y=u-v$. (It's better not to use the same variable names for the original and transformed coordinates). With that, we do indeed have
$(u+v)^2+(u-v)^2-1 = 0$
$2u^2+2v^2-1=0$
and of course every solution in the $x,y$ plane is equivalent to a solution in the $u,v$ plane, because the transformation is invertible. Indeed, the solution $(x,y)=(1,0)$ corresponds to $u+v=1,u-v=0$ which means $(u,v)=(\frac{1}{2},\frac{1}{2})$, and not $(1,0)$. The point $(\frac{1}{2},\frac{1}{2}$) certainly does lie on the curve $2u^2+2v^2-1=0$.
In algebraic geometry, we usually study varieties up to birational equivalence, which is more flexible than merely linear changes of coordinates. You can read about birational geometry in any introductory text on algebraic geometry.
Separately, the situation you describe here is a bit more reminiscent of the theory of binary quadratic forms, where linear invertible transformations play a central role. This is covered in many Number Theory texts, starting with the classics by Dirichlet and Gauss through modern treatments (Hardy & Wright, Cox, and the somewhat unusual "The Sensual Quadratic Form" by Conway).