The non-self-adjoint matrix M has non-degenerate eigenvalues, that is $M \psi_i = e_i \psi_i$, and its adjoint matrix satisfies $M^\dagger \chi_j= e_j^* \chi_j$.
I know that $(\chi_j, \psi_i) = (\psi_i, \chi_j ) = 0$ for $i \neq j$, but why $(\chi_i, \psi_i), \forall i$ cannot be zero?
If $\lambda$ is a non-degenerate eigenvalue of $M$, the matrix $M$ is similar to a Jordan form $$\tag{1} P^{-1}MP=J:=\lambda\oplus\tilde{J}, $$ where $P$ is a nonsingular matrix.
Let $p$ and $s^*$ denote the first column and row, respectively, of $P$ and $P^{-1}$. It is easy to verify that $p$ and $s^*$ are, respectively, right and left eigenvectors associated with $\lambda$ by evaluating the first column of $MP=PJ$ and the first row of $P^{-1}M=JP^{-1}$ taking into account the form (1) of $J$.
Now since $P^{-1}P=I$, $s^*p=1\neq 0$.
The assumption that $\lambda$ is non-degenerate is important. A simple example of its importance is given by $$ M=\pmatrix{0&1\\0&0} $$ with orthogonal left and right eigenvectors. This assumption is not needed, however, to show the biorthogonality of eigenvectors corresponding to distinct eigenvalues.
