Consider a translation-invariant matrix product state \begin{equation} |\psi_L\rangle= \mathrm{Tr}[A(s_1)A(s_2)\ldots A(s_L)]|s_1 s_2\ldots s_L\rangle. \end{equation} I'm interested in the expectation value of a local observable $\hat{O}$~(say a single site operator) in this state $\langle\hat{O}\rangle_{\psi_L}=\langle\psi_L|\hat{O}|\psi_L\rangle/\langle\psi_L|\psi_L\rangle$. To calculate this, define the transfer operator \begin{eqnarray} T^{im}_{jn}&=&\sum_s A_{ij}(s)A^*_{mn}(s),\\ W^{im}_{jn}&=&\sum_{s,s'} A_{ij}(s)A^*_{mn}(s') \langle s'|\hat{O}_l|s\rangle.\nonumber \end{eqnarray} Let the eigenvalues of $T$ be $\{\lambda_j\}^{d^2}_{j=1}$, where $d$ is the bond dimension. Let the left, right eigenvectors be $L_j, R_j$, i.e. $T=\sum^{d^2}_{j=1}\lambda_j R_jL_j^T$. Then \begin{equation}\label{eq:initialstateassumption} \langle\hat{O}\rangle_{\psi_L}=\frac{\mathrm{Tr}[T^{L-1}W]}{\mathrm{Tr}[T^L]}=\frac{\sum^{d^2}_{j=1}\lambda_j^{L-1} (L_j^T W R_j)}{\sum^{d^2}_{j=1}\lambda_j^L}. \tag{1} \end{equation} My question: is the limit $L\to \infty$ is guaranteed to exist?
When $L\to \infty$, only the largest (in absolute value) eigenvalues of $T$ contribute, all other eigenvalues can be neglected. Let the largest eigenvalues be $\lambda_1,\lambda_2,\ldots,\lambda_m$, with $|\lambda_1|=|\lambda_2|=\ldots=|\lambda_m|>|\lambda_{m+1}|>\ldots$. If $\lambda_1=\lambda_2=\ldots=\lambda_m>0$, then the limit (1) clearly exists. But if $\lambda_1,\lambda_2,\ldots,\lambda_m$ are not all equal, and contains some negative/complex values, then $\langle\hat{O}\rangle_{\psi_L}$ in (1) would be oscillating at large $L$. Is this at all possible?