Of course it depends on what basis you are using if the matrix representation is diagonal or not. I assumed that you are working in an eigen-basis of the unperturbed Hamiltonian. In this basis the matrix representation of the unperturbed Hamiltonian is diagonal with entries corresponding to the eigen-energies, right? As far as the degenerate subspace is concerned the hamiltonian is diagonal with the same eigen-energies, hence it is proportional to the unit matrix.
The point of degenerate perturbation theory is that can choose a basis of the degenerate subspace arbitrarily and obviously it is most convenient to choose this basis such that it diagonalizes the perturbation. Then once we have made this choice of basis we can perform regular (non-degenerate) perturbation.
The reason why we do not care about diagonalizing the unperturbed part is because any linear combination of states within the degenerate subspace will also be an eigenstate of the unperturbed part with the degenerate energy. This is just another way of saying that the unperturbed Hamiltonian in this subspace is represented by a matrix proportional to the unit matrix.