Consider a kernel $K$ which satisfies the conditions of Mercer's theorem. We know that $K(x,y) = \sum\limits_{i=1}^{+\infty}\lambda_ie_i(x)e_i(y)$ where $\lambda_i$ and $e_i$ are the eigenvalues/eigenfuncions associated with the integral operator $T_K$ of $K$. Now, consider for example $K(x,y) = e^{-c|x-y|}$, that is the exponential kernel. From Stochastic Finite Elements: A Spectral Approach written by Ghanem and Spanos (Stochastic Finite Elements: A Spectral Approach), we know an expression of the eigenvalues/eigenfunctions (see p.28/29). Consider an approximation of the kernel by $K_{N}(x,y) = \sum\limits_{i=1}^{N}\lambda_ie_i(x)e_i(y)$ and we know from Mercer's theorem that $K_{N}$ converges uniformly to $K(x,y)$ as $N\to+\infty$. Take the first eigenvalue/eigenfunction of $T_K$, that is $K_{1}(x,y) = \lambda_1e_1(x)e_1(y)$. From Ghanem's book, the expression of is $e_1(x) = \frac{cos(\omega x)}{\sqrt{a + \frac{sin(2\omega a)}{2\omega}}}$ where $\omega$ is the first solution of $c - \omega tan(\omega a)= 0 $.
Finally, take 3 points $x_0 = -1, x_1 = 0 ,x_2 = 1$ and form the Gram matrix $K_{ij} = \lambda_1e_1(x_i)e_1(x_j)$ and you want to us this Gram matrix for regression. One knows that the value of at an unobserved point $x_*$ is $f(x_*) = K_1(x_*, X)K_1(X,X)^{-1}Y$ where $X = (x_0, x_1, x_2)$. The issue is that $K_1(X,X)$ is not invertible as the first row is equal to the third row and so the determinant of $K_1(X,X)$ is equal to 0 but $K(X,X)$ is invertible. What am I missing ?