A set of questions was handed out for you to think about. 
The material discussed in this lecture is also discussed in the textbook,
pages 92-98
The diagonalization of the matrix {{1,0,1},{0,2,0},{1,0,1}} was finished. This is a Hermitian matrix that has two eigenvalues: 0, 2, 2. The value 2 is repeated, requiring one to go through the Gram-Schmidt orthogonalization process to obtain an orthonormal basis. There are many ways of doing this. When finished, one writes the eigenvector components in columns and thereby generates the matrix S-1 which diagonalizes the matrix. The order in which the eigenvectors are written is arbitrary. This was illustrated with the solutions for the above matrix. Also, note that if we pick one of the eigenvectors and apply S, the inverse of S-1, to it we get the vector with only one nonzero component.
The realization of an inner product with wavefunctions was discussed. Gram-Schmidt orthogonalization was discussed with the example of polynomials in x on the interval -1 less than or equal to x less than or equal to +1. The Gram-Schmidt process then results in a series of alternately even and odd functions which are proportional to the Legendre polynomials. (Here we just about did Problem 3.25 in the text.
The conditions for an operator such as the differential operator p, to be Hermitian involve very careful consideration of the Boundary Conditions. Boundary terms have to vanish when you integrate by parts to transfer the differential operator from one wavefunction to the other inside the integral. It works for periodic functions, and it works for wavefunctions which go to zero at infinity.
The matrix representation of the differential operator p was constructed using the polynomials |e2) = xn-1. iThe matrix does not look like a Hermitian matrix, even though the operator is Hermitian, or is it?? Look again at the boundary conditions at =1 and -1.