= Solution
For fixed finite matrices, the <matrix exponential> and a <Taylor expansion> show
$$
S(h)=e^{hA/2}e^{hB}e^{hA/2}
=I+h(A+B)+\frac{h^2}{2}(A^2+AB+BA+B^2)+O(h^3).
$$
This agrees through degree two with $e^{h(A+B)}$, even when $A$ and $B$ do not commute. On a fixed time interval, $\|S(h)\|\leq e^{h(\|A\|+\|B\|)}$ and the telescoping identity
$$
S(h)^N-E(h)^N=\sum_{j=0}^{N-1}S(h)^{N-1-j}\bigl(S(h)-E(h)\bigr)E(h)^j,
\qquad E(h)=e^{h(A+B)},
$$
turns the $O(h^3)$ one-step defect into $O(Nh^3)=O(h^2)$ <global error> for $Nh\leq T$. Hence \b[Strang splitting is second order] in general; for commuting matrices it is exact.
For the spatial discretization, use $d=2/(M+1)$, $x_m=-1+md$ and the interior vector $U=(u_1,\ldots,u_M)^T$, with $u_0=u_{M+1}=0$. The <central finite differences> give
$$
\boxed{A=\frac1{d^2}\operatorname{tridiag}(1,-2,1),\qquad
B=\frac{\alpha}{2d}\operatorname{tridiag}(-1,0,1).}
$$
Here $\operatorname{tridiag}$ lists lower diagonal, main diagonal and upper diagonal, in that order. Thus $A^T=A$ and $B^T=-B$. Discrete <summation by parts> gives
$$
U^TAU=-\frac1{d^2}\sum_{m=0}^{M}(u_{m+1}-u_m)^2\leq0.
$$
The <matrix exponential> $e^{hA/2}$ is consequently a contraction in the <Euclidean norm>, while $e^{hB}$ is an <orthogonal matrix> for real $\alpha$. Therefore
$$
\boxed{\|e^{hA/2}e^{hB}e^{hA/2}\|_2\leq1\quad(h\geq0).}
$$
Repeated split steps are contractive in the mesh-weighted norm $d\sum|u_m|^2$, uniformly in the spatial mesh and without a <Courant–Friedrichs–Lewy condition>. \b[The split semidiscretization is unconditionally stable.] No commutativity or shared <eigenvectors> are needed for this <Strang splitting contraction for symmetric diffusion and skew advection>. Its time-order proof is for fixed spatial matrices; as $d\to0$, commutators can grow, so this <stability> result alone is not a uniform-in-mesh second-order error estimate.
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