A nine-point finite-difference stencil on a square grid couples a central value to its four axial neighbours and four diagonal neighbours. Ordering the grid by columns produces a block tridiagonal matrix whose diagonal and off-diagonal blocks are <symmetric tridiagonal Toeplitz matrices>.

Nine-point finite-difference stencil

Let $\theta_k=k\pi h=k\pi/(m+1)$ and define $q^{(k)}_j=\sin(j\theta_k)$. The artificial boundary values are $q^{(k)}_0=q^{(k)}_{m+1}=0$. For the <symmetric tridiagonal Toeplitz matrix> $H=[\beta,\alpha,\beta]$, the <sine addition formula> gives
$$
(Hq^{(k)})_j
=\alpha\sin(j\theta_k)
+\beta\bigl[\sin((j-1)\theta_k)+\sin((j+1)\theta_k)\bigr]
=\bigl(\alpha+2\beta\cos\theta_k\bigr)q^{(k)}_j.
$$
Thus all such matrices have the same <discrete sine transform> eigenvectors, with eigenvalues
$$
\boxed{\lambda_k=\alpha+2\beta\cos(k\pi h),
\qquad k=1,\ldots,m.}
$$
The sine vectors are mutually <orthogonal vectors> and nonzero, so the matrix $Q=(\sin(ij\pi h))_{i,j=1}^m$ is invertible. If $D=\operatorname{diag}(\lambda_1,\ldots,\lambda_m)$, then $HQ=QD$ and therefore
$$
\boxed{H=QDQ^{-1}.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 2 / 39C / a / Solution

An $m\times m$ symmetric tridiagonal Toeplitz matrix has one constant diagonal $\alpha$ and equal constant subdiagonal and superdiagonal $\beta$. Its eigenvectors are the <discrete sine transform> vectors $q^{(k)}_j=\sin(jk\pi/(m+1))$, with eigenvalues $\alpha+2\beta\cos(k\pi/(m+1))$.

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