Radial basis function interpolation

Remarks

By construction $s(x_i) = (A\lambda)_i = y_i$, so $s$ is a multivariate interpolant of the data. For the Gaussian $\varphi(r) = e^{-(\varepsilon r)^2}$ the matrix $A$ is a positive definite matrix, hence invertible, for all distinct points in every dimension. For conditionally positive definite functions such as the thin plate spline, a low-degree polynomial is added to $s$, with matching side conditions on $\lambda$, to get a uniquely solvable system.

Example: $d = 1$, $\varphi(r) = r$, nodes $0, 1$, values $y_1 = 0$, $y_2 = 1$. Then $A = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ and $\lambda = (1, 0)$, so $s(x) = \abs{x}$, and indeed $s(0) = 0$, $s(1) = 1$.

Non-example: take $\varphi(r) = r^2$ on $\R$ and the four nodes $0, 1, 2, 3$. Each translate $(x - x_j)^2$ lies in the space of polynomials of degree at most $2$, which has dimension $3$. So the four columns of $A$, the values of these translates at the nodes, are linearly dependent, and $A$ is not invertible. Then the interpolant is not defined.