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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.

hyperkahler
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    Hint: Use induction on the dimension. It they are all scalar opertors, there is nothing to do. Otherwise, one of them has an eigenspace of strictly smaller dimension than the whole space and that eigenspace is invariant under all the operators since they commute with each other. – Geoff Robinson Jan 23 '15 at 00:12

2 Answers2

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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.

voldemort
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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.

copper.hat
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