Of course. Thanks for clearing that up.

Out of curiosity though; obviously

is still a basis for this subspace, no?

If you write "obviously" then why the question "no?" But the answer is no: that is not a basis for that subspace, whose elements are characterized for having their free coefficient equal to the sum of the linear and the quadratic coefficients. The element does not share this characteristic. Tonio
So then, in finding an orthonormal basis for V (via Gram-Schmidt procedure), would there not be multiple, equally valid solutions which arise from starting with either basis?