Perturbation Theory in Orbital Interactions in Chemistry In the entire book, perturbation theory is used as a qualitative tool to rationalise some chemical phenomena. The authors write that
$$\psi_i = \sum_\mu{T_{ji}\psi_j^\circ}\tag{1}$$
The proof of the derivation for $T_{ji}$ involves the following statement:
$$T_{ji} = \left(\mathbf{C}_j^\circ\right)^\mathrm{T}\mathbf{S}^\circ\mathbf{C}_i \tag{2}$$
where
$$\left(\mathbf{C}_j^\circ\right)^\mathrm{T} = \left(C_{1j}^\circ \quad C_{2j}^\circ \quad \cdots \quad C_{mj}^\circ\right)\tag{3}$$
(indexing the coefficients for the jth unperturbed MO), and
$$
\mathbf{C}_i =
\begin{pmatrix}
C_{1i} \\
C_{2i} \\
\vdots \\
C_{mi} 
\end{pmatrix}
\tag{4}
$$
(indexing the coefficients for the $i$th perturbed MO); $\mathbf{S},$ the overlap matrix, is defined as usual.
I'm not quite sure how the aforementioned statement is obtained. Could someone please walk me through how
$$T_{ji} = \sum\sum C_{nj}^\circ S_{nm}C_{mi}?$$
Source: Orbital Interactions in Chemistry, 2nd edition, page 794-802
Rather than focusing on the individual matrix entries, it may be more intuitive to look at the matrix $\mathbf{T}$ formed by putting together all entries $T_{ij}$. In this case, $\mathbf{T}=\mathbf{C}^T\mathbf{S}^0\mathbf{C}$, where $\mathbf{C}$ is the coefficient matrix with column $k$ denoting how atomic orbitals mix together into molecular orbital $k$. 

Now the point is that molecular orbitals, or the columns of $\mathbf{C}$, are orthonormal and are eigenfunctions of $H_{eff}$ (see page 34). This orthonormality makes sense if you think of the Hamiltonian matrix, which is certainly self-adjoint, and solving the eigenproblem is equivalent to diagonalizing the matrix with an orthonormal basis of eigenvectors.

Hence, $\mathbf{C}$ is unitary, and $\mathbf{T}=\mathbf{C}^T\mathbf{S}^0\mathbf{C}$ can be simply seen as a change-of-basis transformation done on the overlap matrix from atomic basis to the molecular orbital basis.  

I hope this rather mathematical perspective may be of help to you. Thanks!