Commutation and Eigenfunctions

[CanIExplore]

My first question is, does any operator commute with itself? If this is the case, is there a simple proof to show so? If not, what would be a counter-example or a "counter-proof"?

My second question has to do with the properties of an eigenvalue problem. If you have an operator Q such that $$\hat{Q}$$$$\psi$$=$$\lambda\psi$$, then is the following always true?

$$\hat{Q}^{2}$$$$\psi$$=$$\lambda^{2}\psi$$

Since $$\hat{Q}^{2}$$$$\psi$$=$$\hat{Q}\hat{Q}$$$$\psi$$=$$\hat{Q}\lambda\psi$$=$$\lambda\hat{Q}\psi$$=$$\lambda\lambda\psi$$=$$\lambda^{2}\psi$$.

And is there a more general way of saying this? Like,

$$\hat{Q}$$$$_{i}$$$$\hat{Q}$$$$_{j}$$$$\psi$$=$$\hat{Q}$$$$_{i}$$$$\lambda$$$$_{j}$$$$\psi$$=$$\lambda$$$$_{j}$$$$\hat{Q}$$$$_{i}$$$$\psi$$=$$\lambda$$$$_{i}$$$$\lambda$$$$_{j}$$$$\psi$$

Which should always be true even if the operators don't commute because the lambdas are just scalars right?

 

[sgd37]

Of course an operator commutes with itself [Q,Q] is always 0 from the definition of the commutator

For the last part you wouldn't be able to write it like that if the operators don't commute. Commuting operators have simultaneous eigenfunctions, non-commuting ones do not