Let $R$ be a field, and let $M$ and $N$ be $d$-dimensional $R$-vectorspaces. Suppose we have a pairing: \begin{align*} M \times N \rightarrow R\\ m \times n \mapsto \langle m,n \rangle \end{align*} which is a perfect pairing, meaning in this case that $\Phi_1$ is an isomorphism. \begin{align*} \Phi_1: M \rightarrow N^* \\ m \mapsto (n \mapsto \langle m,n \rangle) \end{align*} This also means that $M \simeq N^* \simeq N$. Suppose we now have a basis for $M$, say $\{m_1,..,m_d\}$, then this must correspond to a basis $\{n_1,..,n_d\}$ under these isomorphisms. What I am now wondering is what we can see about the pairing over two corresponding basis elements, so what can we say about $\langle n_i,m_i \rangle$.
More specifically, my problem is about $R$-linear operators $T_N$ on $N$ and $T_M$ on $M$ that work on both $M$ and $N$ and that 'respect the duality' or are 'compatible', in the sense that: \begin{align*} \langle T_M(m),n \rangle = \langle m,T_N(n) \rangle \end{align*} Suppose now the basis elements $m_i$ is are eigenvectors for the operator $T_M$, so $T(m_i)=\lambda_i m_i$ for all $1 \leq i \leq d$, can we show that the corresponding basis for $N$ is also an eigenbasis for $T_N$, i.e. $T_N(n_i)=\lambda_i(n_i)$? I have been trying to figure this out, but I keep coming back to the first question, what do we know about $\langle m_i,n_i \rangle$?
I will answer the second part of my own question, following from the answer of @57Jimmy. Suppose we have a basis $\{m_1,...,m_d\}$, then this corresponds to a basis $\{n_1,..,n_d\}$ such that $\langle m_i,n_j \rangle = \delta_{ij}$. So for any $n \in N$ such that $n=\sum a_j n_j$ we see that $\langle m_i,n \rangle = a_i$. Suppose now we have $R$-linear operators $T_N$ on $N$, and $T_M$ on $M$ that are compatible with the pairing as defined in the question, and the basis elements $\{m_1,..,m_d\}$ are eigenvectors for $T_M$, so that $T_M(m_i)=\lambda_i m_i$. Then for any basis element $n_i$, such that $T_N(n_i)=\sum a_j n_j$, we see that: \begin{align*} \lambda_i = \lambda_i \langle m_i,n_i \rangle = \langle T_M (m_i),n_i\rangle = \langle m_i,T_N(n_i) \rangle = a_i \end{align*} and \begin{align*} 0 = \lambda_j \langle m_j,n_i \rangle = \langle T_M (m_j),n_i\rangle = \langle m_j,T_N(n_i) \rangle = a_j \end{align*} therefore $T_N(n_i)=\lambda_i n_i$, so $n_i$ is an eigenvector.