Commuting operators Let's consider a number of linear operators, defined on a finite dimensional complex vector space, which two by two commutes with each other. (the amount of them can be infinite). How to prove that that will have a common eigenvector?
The finite case can be done by induction:
1) $n=2$, $AB=BA$, then let $x$ be an eigenvector of $A$ (it does exist, because we are working over a $\mathbb{C}$) and $\alpha$ - an eigenvalue. Then, $A(x)=\alpha \cdot x, B(A(x))=A(B(x))=B(\alpha x)=\alpha B(x)$, so $B(x)$ is also an eigenvector of $A$, associated with $\alpha$ eigenvalue.
Analogically, we do it for $n>2$.
But, what can i do, while working with an infinite number of operators( induction doesn't work here, actually).
Any help would be appreciated. 
 A: Let's say your vector space is $\mathbb{C^n}$. Then $M_n(\mathbb{C})$, i.e. all $n$ by $n$ matrices are the bounded operators on $\mathbb{C^n}$.
Now, $M_n(\mathbb{C})$ is finite dimensional. So, even if you have infinitely many operators, say $A_1,A_2,\cdots$, there will exist $i_1,i_2,\cdots,i_k$ such that $A_{i_1},\cdots,A_{i_k}$ will span the rest of the matrices. So, your case for finitely many operators will work.
A: Here is an inductive proof:
Let ${\cal C}$ be the commuting family of operators.
Let me call an operator $A$ on a subspace $S$ a multiplier operator on $S$ iff there exists some $\lambda$ such that $Ax=\lambda x$ for all $x \in S$.
Let me call a subspace $S$ invariant iff $AS \subset S$ for all $A \in {\cal C}$.
Note that if a one dimensional space $S$ is invariant, then it is an eigenspace for all $A \in {\cal C}$.
Pick $A \in {\cal C}$, and suppose $\lambda $ is an eigenvalue of $A$, then
let $S_1 = \ker (A-\lambda I)$. It is straightforward to show that $S_1$ is invariant.
If every $B \in {\cal C}$ is a multiplier operator on $S_1$ then we are finished.
Otherwise pick $B \in {\cal C}$ that is not a multiplier operator on $S_1$ and suppose $\mu$ is an eigenvalue of $B$ on $S_1$. Then
$S_2 = \ker (B-\mu I) \cap S_1$ is non empty, invariant and $\dim S_2 < \dim S_1$.
Now repeat the process, noting that it must end because the space is finite dimensional.
