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I've come across a paper that mentions this as a fact...where can I find the proof?

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You can find a good and detailed discussion of this fact in the book "Linear Algebra", by Kenneth Hoffman and Ray Kunze, chapter 6 (Elementary Canonical forms). – Ronaldo Oct 7 '10 at 17:00
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As a physics student, I would like to mention that this simple theorem, is tremendously useful and important in quantum mechanics, and gives rise to the concepts of compatible and incompatible variables, which is essential for the measurements and probabilistic interpretation of quantum mechanics. – ramanujan_dirac Feb 17 at 13:01

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Suppose that $A$ and $B$ are $n\times n$ matrices, with complex entries say, that commute.
Then we decompose $\mathbb C^n$ as a direct sum of eigenspaces of $A$, say $\mathbb C^n = E_{\lambda_1} \oplus \cdots \oplus E_{\lambda_m}$, where $\lambda_1,\ldots, \lambda_m$ are the eigenvalues of $A$, and $E_{\lambda_i}$ is the eigenspace for $\lambda_i$. (Here $m \leq n$, but some eigenspaces could be of dimension bigger than one, so we needn't have $m = n$.)

Now one sees that since $B$ commutes with $A$, $B$ preserves each of the $E_{\lambda_i}$: if $A v = \lambda_i v, $ then $A (B v) = (AB)v = (BA)v = B(Av) = B(\lambda_i v) = \lambda_i Bv.$

Now we considered $B$ restricted to each $E_{\lambda_i}$ separately, and decompose each $E_{\lambda_i}$ into a sum of eigenspaces for $B$. Putting all these decompositions together, we get a decomposition of $\mathbb C^n$ into a direct sum of spaces, each of which is a simultaneous eigenspace for $A$ and $B$.

NB: I am cheating here, in that $A$ and $B$ may not be diagonalizable (and then the statement of your question is not literally true), but in this case, if you replace "eigenspace" by "generalized eigenspace", the above argument goes through just as well.

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I'm a bit unclear on "putting decompositions together" part...does it give you an explicit form for the common set of eigenvectors? – Yaroslav Bulatov Oct 7 '10 at 18:32
@Yaroslav: Dear Yaroslav, Yes, you will be able to get an explicit form for the common set of eigenvectors, but it may be complicated to compute (at least if you are computing by hand). What you need to do is to find a basis for each $E_{\lambda_i}$. You can then write a matrix for $B$ (restricted to $E_{\lambda_i}$) in terms of this basis, and then by the usual diagonalization procedure (applied to this new matrix) find a new basis for $E_{\lambda_i}$ consisting of eigenvectors for $B$. (So these will be simultaneous eigenvectors for $B$ and $A$ --- with $A$-eigenvalue all equal to ... – Matt E Oct 7 '10 at 18:33
... $\lambda_i$.) You do this for each $\lambda_i$, and putting all of the bases you found for each $E_{\lambda_i}$ together, you get a basis for $\mathbb C^n$ consisting of simultaneous eigenvectors. As Qiaochu points out, the situation is easiest when all the $E_{\lambda_i}$ are one-dimensional (i.e. when $A$ has distinct eigenvalues) since then you don't need to do the second step of computing eigenvectors for $B$; each $E_{\lambda_i}$ will automatically consists of simultaneous eigenvectors for $A$ and $B$ (again, this is just in the special case when $A$ has distinct eigenvalues). – Matt E Oct 7 '10 at 18:37
Also, as your probably know, in the quantum mechanics literature, the situation when the $E_{\lambda_i}$ are bigger than one-dimensional is sometimes referred to by saying that the operator $A$ has degeneracies. – Matt E Oct 7 '10 at 18:38
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This is false in a sort of trivial way. The identity matrix $I$ commutes with every matrix and has eigenvector set all of the underlying vector space $V$, but no non-identity matrix has this property.

What is true is that two matrices which commute and are also diagonalizable are simultaneously diagonalizable. The proof is particularly simple if at least one of the two matrices has distinct eigenvalues.

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(And this can be improved to "simultaneously Jordanized", among other things) – Mariano Suárez-Alvarez♦ Oct 7 '10 at 13:56
Dear Qiaochu, I suspect that the question means "admit a common set of eigenvectors", not that every eigenvector for one is an eigenvector fo the other. – Matt E Oct 7 '10 at 15:19
Yes, I meant Matt's version, the precise statement was "Thus, the Hamiltonian and all the translation operators of the crystal commute with each other. They possess, therefore, a common set of eigenstates." which I figured would have an equivalent statement in linear algebra – Yaroslav Bulatov Oct 7 '10 at 17:56
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@MarianoSuárez-Alvarez: That's not true. Take the 3 x 3 matrix nilpotent $A$ with ones just above the diagonal and all other entries zero. Then $A$ and $A^2$ commute, but are not simultaneously Jordanized (because $A$ is a single Jordan block, so any $P$ Jordanizing it must give $PAP^{-1} = A$ and thus $PA^2P^{-1} = A^2$ is not in Jordan form). It is true that any two commuting matrices can be simultaneously put in upper triangular form, though, maybe that's what you were remembering? – Omar Nov 26 '12 at 18:50

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