Suppose we have an $n \times m$ grid. The coefficient matrix for the rotated 5-point stencil is \(A = I_{m-1} \otimes (-4I_{n-1}) + D_{m-1} \otimes D_{n-1}\) where $D_n$ is an $n \times n$ matrix with $1$s on the sub and super diagonals and $0$s everywhere else.
Normally if we have an evenly spaced grid such that $h=k$, the coefficient matrix for the standard 5-point stencil is \(A=S \otimes I_{n-1} + I_{n-1} \otimes S\) where $S$ is an $(n-1) \times (n-1)$ toeplitz matrix with $-2$s on the main diagonal, and $1$s on the sub and super diagonals:
\[S = \begin{pmatrix} -2 & 1 & 0 & 0 & \cdots & 0 & 0\\ 1 & -2 & 1 & 0 & \cdots & 0 & 0 \\ 0 & 1 & -2 & 1 & \cdots & 0 & 0 \\ \vdots & \ddots & \ddots & \ddots & \ddots & \ddots & \vdots \\ 0 & 0 & 0 & 1 & -2 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 & -2 & 1 \\ 0 & 0 & 0 & 0 & 0 & 1 & -2 \\ \end{pmatrix}\]If however $h \neq k$, the above method won’t work because it will have the wrong dimensions and incorrect sparsity. To use an unequal grid spacing we can construct it like
\[A= I_{m-1} \otimes M + D_{m-1} \otimes (\lambda I_{n-1})\]where $\lambda = \left(\frac{h}{k} \right)^2$ and $M$ is an $(m-1) \times (m-1)$ matrix with $-(2+2\lambda)$ on the main diagonal, and $1$s on the sub and super diagonals.
\[M = \begin{pmatrix} -(2+2\lambda) & 1 & 0 & 0 & \cdots & 0 & 0\\ 1 & -(2+2\lambda) & 1 & 0 & \cdots & 0 & 0 \\ 0 & 1 & -(2+2\lambda) & 1 & \cdots & 0 & 0 \\ \vdots & \ddots & \ddots & \ddots & \ddots & \ddots & \vdots \\ 0 & 0 & 0 & 1 & -(2+2\lambda) & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 & -(2+2\lambda) & 1 \\ 0 & 0 & 0 & 0 & 0 & 1 & -(2+2\lambda) \\ \end{pmatrix}\]