Background
Consider 1-D Ising model of n lattice points with periodic boundary condition,
$\beta H(\sigma_1,\sigma_2,...,\sigma_N) = -\sum_{i=1}^nk(\sigma_i\sigma_{i+1})-\sum_{i=1}^n\sigma_i$
$k=\beta J$ and $h=\beta B$ where $J$ is the coupling constant and $B$ is the applied magnetic field.
Partition function, $Z=\sum_{\sigma_1,\sigma_2,...,\sigma_N}e^{-\beta H(\sigma_1,\sigma_2,...,\sigma_N)}$
where $\sum_k$ is the sum over all the possible microstates and $H_k$ is the hamiltonian of the $k-th$ microstate.
$Z = \sum_{\sigma_1,\sigma_2,...,\sigma_N}exp\Big(k(\sigma_1\sigma_2+\sigma_2\sigma_3+...+\sigma_N\sigma_1)\Big)exp\Big(h(\sigma_1+\sigma_2+...+\sigma_N)\Big)$
$\implies Z = \sum_{\sigma_1,\sigma_2,...,\sigma_N}exp\Big(k\sigma_1\sigma_2+\frac{(\sigma_1+\sigma_2)}{2}\Big)exp\Big(k\sigma_2\sigma_3+\frac{(\sigma_2+\sigma_3)}{2}\Big)...exp\Big(k\sigma_N\sigma_1+\frac{(\sigma_N+\sigma_1)}{2}\Big)$
The $Z$ can be written in terms of transfer matrix as
$Z=\sum_{\sigma_1,\sigma_2,...,\sigma_N}\langle\sigma_1|T|\sigma_2\rangle\langle\sigma_2|T|\sigma_3\rangle...\langle\sigma_N|T|\sigma_1\rangle$
where $T=\begin{pmatrix}e^{k-h}&e^{-k}\\e^{-k}&e^{k+h}\end{pmatrix}$
$\sigma_1=-1$ corresponds to $|\sigma_1\rangle=\begin{pmatrix}1\\0\end{pmatrix}$ and $\sigma_1=+1$ corresponds to $|\sigma_1\rangle=\begin{pmatrix}0\\1\end{pmatrix}$
So, $Z = \sum_{\sigma_1}\langle\sigma_1|T^N|\sigma_1\rangle=\sum_{\sigma_1}\langle O\sigma_1|OT^NO^T|O\sigma_1\rangle\tag{1}$
where $O$ is the orthogonal matrix that diagonalize $T$ and thus $T^N$.
$OTO^T=\begin{pmatrix}\lambda_1&0\\0&\lambda_2\end{pmatrix}$
where $\lambda_{1,2}=e^k\cosh h\pm\sqrt{e^{2k}\sinh^2 h+e^{-2k}}$
So, $Z=Tr(T^N)=Tr(OT^NO^T)=\lambda_1^N+\lambda_2^N\tag{2}$
Work
Now, I am trying to calculate $\langle\sigma_i\rangle$ when $h=0$ using transfer matrix.
We know that when $h=0$, $\langle\sigma_i\rangle=0$
$\langle\sigma_i\rangle=\sum_{\sigma_1,\sigma_2,...,\sigma_N}\langle\sigma_1|T|\sigma_2\rangle\langle\sigma_2|T|\sigma_3\rangle...\langle\sigma_{i-1}|T|\sigma_i\rangle\sigma_i\langle\sigma_i|T|\sigma_{i+1}\rangle...\langle\sigma_N|T|\sigma_1\rangle$
$\langle\sigma_i\rangle=\sum_{\sigma_1,\sigma_i}\langle\sigma_1|T^{i-1}|\sigma_i\rangle\sigma_i\langle\sigma_i|T^{N-i+1}|\sigma_1\rangle$
As quantities inside $\sum$ are just numbers (matrix elements and $\sigma_i$ are numbers),
So, $\langle\sigma_i\rangle=\sum_{\sigma_1,\sigma_i}\sigma_i\langle\sigma_i|T^{N-i+1}|\sigma_1\rangle\langle\sigma_1|T^{i-1}|\sigma_i\rangle$
$\implies\langle\sigma_i\rangle=\sum_{\sigma_i}\sigma_i\langle\sigma_i|T^{N-i+1}\sum_{\sigma_1}\Big(|\sigma_1\rangle\langle\sigma_1|\Big)T^{i-1}|\sigma_i\rangle$
$\implies\langle\sigma_i\rangle=\sum_{\sigma_i}\sigma_i\langle\sigma_i|T^{N-i+1}T^{i-1}|\sigma_i\rangle$
$\implies\langle\sigma_i\rangle=\sum_{\sigma_i}\sigma_i\langle\sigma_i|T^N|\sigma_i\rangle$
$\implies\langle\sigma_i\rangle=\sum_{\sigma_i}\sigma_i\langle O\sigma_i|OT^NO^T|O\sigma_i\rangle$
$\implies\langle\sigma_i\rangle=\sum_{\sigma_i}\sigma_i\langle O\sigma_i|\begin{pmatrix}\lambda_1^N&0\\0&\lambda_2^N\end{pmatrix}|O\sigma_i\rangle\tag{3}$
Doubt
But now how to show that $\langle\sigma_i\rangle=0$?
One way is to calculate $O$ explicitly and plug this in $(3)$ and show that it is $0$. This I have verified.
But I have seen that in books, they introduce Pauli matrix to show that it is $0$. I am not able to understand how that follows from $(3)$.