Well...as I learned earlier today from a weirdly coincidental Mathematica.StackExchange thread, one can get results using random numeric values instead of symbolic mu and lambda. See comments in the thread at the link below.

http://mathematica.stackexchange.com/questions/46949/is-there-a-built-in-procedure-for-simultaneous-diagonalization-of-a-set-of-commu

As you observed in one note, eigenvectors corresponding to a multiple eigenvalue are not in general uniquely defined. So it might require combinations of eigenvectors from a given eigenspace from the first matrix in order to make an eigenvector for the second one, and vice versa.

If instead you use the pencil approach that I indicated, it all works just fine. I started with your definitions for matrices H and TotSz.

Check the eigenvalues and eigenvectors for each matrix individually.

I'm not sure if this is what you are seeking but you could look at the eigensystem for (what I believe is called the matrix pencil). If your matrices are m1 and m2, this is mu*m1+lambda*m2.


For your notebook example I believe this gives common eigenvectors.

Dear Prof. Blinder,
Thanke you for your response.
You are completely right, if two operators do not have duplications in their eigenvalues and they commute.
On the  contrary if they have duplications in their eigenvalues then the linear combination of the degenerated states corresponding to the same eigenvalues still is an eigenvector for the first operator, but not necessarily for the second (actually the corresponding theorem states that in case of commutation one can always find a common set of eigenvectors, and in case of absence of duplications in eigenvlues the set is the onley possible one, but in case of degenaration  not every set is common,
see as a reference Landau Lifshitz, "Quantum Mechanics"  
https://archive.org/details/QuantumMechanics_104
page 34).
For example here, in the file I uploaded,  I present a particular hamiltonian of three interacting spin-1 particles, which commutes with the total angular momentum in the z direction, but some eigenvectors of the hamiltonian are not eigenvectors for the total angular momentum.