= Solution
For an open chain there are $L-1$ bonds, with both endpoint <Ising spins> independently summed. Retain the same <spin-chain transfer matrix>, but replace the trace by endpoint vectors. With $\mathbf1=(1,1)^{\mathsf T}$,
$$
\boxed{Z_L^{\mathrm{open}}=\mathbf1^{\mathsf T}W^{L-1}\mathbf1.}
$$
This follows directly by expanding the product of $L-1$ <Boltzmann factors>. If $\mathbf v_\pm$ are orthonormal <eigenvectors> of the <symmetric matrix> $W$, the equivalent <spectral decomposition> is
$$
Z_L^{\mathrm{open}}=\sum_{\nu=\pm}(\mathbf1^{\mathsf T}\mathbf v_\nu)^2\lambda_\nu^{L-1}.
$$
Fixed endpoint <Ising spins> replace the corresponding $\mathbf1$ by a basis vector; extra endpoint <energies> replace its entries by the corresponding <Boltzmann factors>. The leading bulk <Helmholtz free energy> per spin agrees with the periodic result, because the boundary changes only subextensive terms. In particular, $L=1$ gives $Z_1^{\mathrm{open}}=\mathbf1^{\mathsf T}\mathbf1=2$, as required for one free <Ising spin>.
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