This matrix is already in diagonal form. You find the roots of the characteristic polynomial det(A-Ix). In this case, they are just the things on the diagonal.

To find the "characteristic vectors" I assume you are talking about the eigenvectors. Simply put the eigenvalue in for x in A-Ix (just subtract the value from the diagonal). Then you set that equal to 0 and find a basis for this solution space (the null space). This is the basis for the kernel of .