Expanding the trace in the spin basis gives
$$
\operatorname{Tr}(M^N)
=\sum_{s_1,\ldots,s_N}M_{s_1s_2}M_{s_2s_3}\cdots M_{s_Ns_1}
=\sum_{\{s_i\}}e^{-\beta H}=Z,
$$
so the <transfer matrix for the one-dimensional Ising model> reproduces the <partition function>.

In the ordered basis $s=+1,-1$,
$$
M=\begin{pmatrix}
e^{\beta(J+B)}&e^{-\beta J}\\
e^{-\beta J}&e^{\beta(J-B)}
\end{pmatrix}.
$$
Its <eigenvalues> are
$$
\boxed{\lambda_\pm=e^{\beta J}\cosh(\beta B)
\pm\sqrt{e^{2\beta J}\sinh^2(\beta B)+e^{-2\beta J}}}.
$$
Since $\lambda_+>|\lambda_-|$ at every positive <temperature>, the <thermodynamic limit> gives the <Helmholtz free energy> per spin
$$
\boxed{f=-kT\log\lambda_+}.
$$
This is an <analytic function> of $T$ and $B$ for $T>0$, so the one-dimensional short-range <Ising model> has no finite-temperature <phase transition>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 4 / 35A / a / Solution

For nearest-neighbour interactions, a two-by-two transfer matrix assigns one <Boltzmann factor> to every adjacent spin pair. With periodic boundary conditions, summing all spin configurations gives the <partition function> as the trace of a power of this matrix.

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 Incoming links: Transfer matrix for the one-dimensional Ising model