Searching in net, I found For an Hilbert space $H$, and an a commuting set $\cal F\subset Hom(H,H)$ of self-adjoint operators. Then under suitable conditions, can be simultaneously (unitarily) diagonalized.( https://en.wikipedia.org/wiki/Compact_operator_on_Hilbert_space#Simultaneous_diagonalisation )
After reading it, I thought about under which conditions is a family of compact positive or normal operators simultaneous diagonalization?
Would you please regard me to answer this question?
For two operators: If selfadjoint operators $A$ and $B$ commute, then $B$ maps $\mathcal{N}(A-\lambda I)$ into itself because $$ (A-\lambda I)x = 0 \implies (A-\lambda I)Bx=B(A-\lambda I)x=0. $$ For $\lambda\ne 0$, the restriction of $B$ to $\mathcal{N}(A-\lambda I)$ is selfadjoint and finite-dimensional; therefore, the restriction of $B$ has an orthonormal basis of eigenvectors. And these eigenvectors are also eigenvectors of $A$ with eigenvalue $\lambda$. The null space with $\lambda=0$ takes a little extra attention.