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Part 1

There are at least two highly rated questions and answers on the relationship between commuting operators and the commonality of eigenvectors [1,2]. In [1], @Algebraic Pavel writes that

Commuting matrices do not necessarily share all eigenvector, but generally do share a common eigenvector.

Frankly, the details of this answer are beyond my comprehension. What I understand from it is that there is some subset of one more eigenvectors common to both operators.

Question 1:

What are the necessary and sufficient conditions such that commuting matrices necessarily share all eigenvectors?

My summarily false (see comment below) initial conjecture was that a necessary and sufficient condition is that each and every eigenvalue is mutually distinct from all the others.

Part 2

There are an abundance of questions that limit the scope only to non-degenerate eigenvectors (cf, [3] and [4] for example). Yet, it comes up that the spectrum might include degenerate eigenvectors. In my reading on quantum mechanics, Landau and Lifshitz [5] propose that:

If two operators commute with each other, they have their entire set of eigenfunctions in common.

I am able to follow along closely to their arguments regarding non-degenerate eigenvectors. However, their outline of a proof for the degenerate case is hand wavy. In addition, it is contrary to the Algebraic Pavel's claim that I rewrite in Part 1 above. There is an inconsistency here.

Question 2:

Given a spectrum of eigenvalues some of which are distinct from all the others and some are not distinct from all the others, can it be proved that all the eigenvectors can be be taken as common to both? What, if anything, is the necessary and sufficient conditions for this to be so?

Bibliography

[1] Do commuting matrices share the same eigenvectors?

[2] Matrices commute if and only if they share a common basis of eigenvectors?

[3] Commuting Operators Have the Same Eigenvectors, but not Eigenvalues.

[4] Eigenvectors of commuting operators on finite complex hilbert spaces.

[5] Landau and Lifshitz, Quantum Mechanics, Third Edition, page 34.

Michael Levy
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    Your conjecture in the first question is clearly false. E.g. every square matrix commutes with itself, but a matrix can have repeated eigenvalues. – user1551 Sep 21 '24 at 19:52
  • If the matrices are diagonalizable and commute, then they can be simultaneously diagonalized; i.e., there is a set of vectors that are eigenvectors of both matrices and which form a basis. Matrices with non-degenerate eigenvalues are diagonalizable, and some matrices with degenerate eigenvalues are as well. But in the case of non-diagonalizable matrices (yuck), it's more complicated... is that the case you're primarily confused about? – mjqxxxx Oct 30 '24 at 22:58
  • I am interested in a case with degenerate eigenvalues. Let us say that yes, a case that is not strictly diagonizable. – Michael Levy Oct 31 '24 at 12:30
  • So, a non-diagonalizable matrix doesn't have a complete set of eigenvectors; i.e., if the matrix is $n\times n$, then there are strictly fewer than $n$ linearly independent eigenvectors. (The remaining dimensions are spanned by generalized eigenvectors, each associated with one of the actual eigenvectors.) Such a matrix can certainly commute with a fully diagonalizable matrix (e.g., the identity matrix), or with a different non-diagonalizable matrix. – mjqxxxx Nov 01 '24 at 17:47
  • In the most extreme case, an $n\times n$ matrix can have just a single eigenvector. An example (which captures the idea) is a lowering operator, which acts on a set of basis vectors $e_1, e_2, \ldots, e_n$, taking each $e_{j}$ to $e_{j-1}$ (and $e_1$ to $0$). Only $e_1$ is an eigenvector (with eigenvalue $0$); the rest are generalized eigenvectors. Call this matrix $A$. Matrices that commute with this matrix are all matrices of the form $P(A)$, where $P$ is a polynomial. (Since $A^n=0$, only polynomials up to that order need to be considered.) – mjqxxxx Nov 01 '24 at 17:53
  • But the point is that $P(A)$ can have more eigenvectors than $A$. Indeed, $A^k$ is going to have $k$ linearly independent eigenvectors. One of them will be $e_1$, certainly. That one must be shared between the two matrices. The rest don't have to be. – mjqxxxx Nov 01 '24 at 17:59
  • Note that the degeneracy of eigenvalues is orthogonal to the "missing eigenvector" problem of non-diagonalizable matrices. If either or both of your commuting matrices just have degenerate eigenvalues, that's fine, as long as they have a complete set of eigenvectors... you can choose a complete set of eigenvectors that are simultaneously eigenvectors of both matrices. As you mention quantum-mechanical references, remember that Hermitian matrices are always diagonalizable. So the nuances associated with non-diagonalizable matrices might not be relevant to your problem. – mjqxxxx Nov 01 '24 at 19:52

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