I think it is better to think in terms of linear transformations- any linear transformation can be represented as a matrix in a given basis: Apply the linear tranformation to each of the basis vectors in turn, the write the result as a linear combination of the basis vectors- the coefficients are the columns of the matrix.
The three distinct eigenvalues must have 3 independent eigenvectors. Using those eigenvectors as a basis for the vector space, the linear operator is represented by a diagonal matrix with the eigenvalues on the diagonal.
Equivalently, if A is a matrix with three distinct eigenvalues, B is the matrix having those three eigenvectors as columns, then B^{-1}AB is the diagonal matrix having the eigenvalues on the diagonal.
By the way, having three independent eigenvectors is a necessary condition for a matrix to be diagonalizable. Having three distinct eigenvalues is not necessary.