Are shifted squared terms linearly independent?

22 Views Asked by At

I suspect that my intiuition for this is very wrong, but I feel like polynomials of the form $(x-a)^2$ and $(x-b)^2$ should not be linearly independent (since you can relate them by a shift), and yet if you try to find a non-zero $(\lambda, \mu)$ such that $\lambda (x-a)^2 + \mu (x-b)^2 \neq 0$, it is not possible.

I suppose the latter is defining for linear independence, but I cannot but feel uncomfortable with this. I would appreciate if someone had some intuition for linear independence, that would help me feel more comfortable with these polynmoials being linearly independent.