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I understand that in ZF set theory without the Axiom of Choice (AC), it is consistent to have models in which there exist vector spaces over some (unspecified) field $k$ without a basis.

So in particular, in linear algebra over $k$, we would find matrices without an eigenbasis.

My question lives at the other end of the spectrum: could we find a matrix over some field $\ell$ in some model of ZF set theory without AC which has different eigenbases of different cardinalities ?

THC
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    It seems to me this reduces to a question about whether the base vector space can have two bases of differing cardinalities. If yes, then take the identity matrix or zero matrix, and use the two different bases; if no, then you can't have bases of different cardinalities, and therefore can't have eigenbases of different cardinalities. – user44191 Aug 02 '21 at 19:47
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    @user44191: Yes, a vector space can have two bases of different cardinalities. – Asaf Karagila Aug 02 '21 at 19:48
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    With matrix do you mean finite-dimensional ones? Linear algebra for finite-dimensional spaces works just fine in ZF. Also, there are matrices without eigenbases even in ZFC - any non-diagonalizable matrix works. – Wojowu Aug 02 '21 at 20:11
  • There is a misunderstanding in this question. As @Wojowu says: in infinite dimensions, there are plenty of linear operators which do have any eigenvectors (over any field extension). For example the operator $\partial/\partial x$ acting on $\mathbb{C}[x]$. – Theo Johnson-Freyd Aug 02 '21 at 20:35
  • @AsafKaragila I didn't know that! Could you post a "construction" of a vector space with bases of different cardinalities (and what axiom contradicting Choice it will require) as an answer? I encourage OP to accept such a discussion! – Theo Johnson-Freyd Aug 02 '21 at 20:37
  • @Theo: This is pretty much the one structural construction that I know of, but not the intimate details of. Nevertheless, this appears as a problem (in the sense of an exercise) in Jech's "Axiom of Choice" in Chapter 10 and it is originally due to Läuchli in his PhD thesis and his seminal paper "Auswahlaxiom in der Algebra", Comment. Math. Helv. 37 (1962-1963), 1–18. (Also relevant, https://mathoverflow.net/questions/93242/sizes-of-bases-of-vector-spaces-without-the-axiom-of-choice) – Asaf Karagila Aug 02 '21 at 20:57
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    @TheoJohnson-Freyd Poor choice of an example, constants are eigenvectors of the derivative operator (with eigenvalue zero). Multiplication by $x$ has no eigenvectors though. Also, for lack of eigenbasis we don't even need to go to infinite dimensions - $\begin{pmatrix}1&1\0&1\end{pmatrix}$ has eigenvectors, but no basis consisting of such. – Wojowu Aug 02 '21 at 21:02
  • @TheoJohnson-Freyd I took the question to be asking whether there was some operator that had a basis of eigenvectors; if a linear operator doesn't have an eigenvector, it clearly can't have a basis made of eigenvectors (excluding when the space is 0-dimensional). – user44191 Aug 03 '21 at 00:41
  • @user44191 : can you think of other interesting matrices (except the identity matrix or 0-matrix) with eigenbases of different cardinalities ? – THC Oct 18 '21 at 12:01

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