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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 ?

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    $\begingroup$ 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. $\endgroup$
    – user44191
    Commented Aug 2, 2021 at 19:47
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    $\begingroup$ @user44191: Yes, a vector space can have two bases of different cardinalities. $\endgroup$
    – Asaf Karagila ♦
    Commented Aug 2, 2021 at 19:48
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    $\begingroup$ 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. $\endgroup$
    – Wojowu
    Commented Aug 2, 2021 at 20:11
  • $\begingroup$ 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 $\mathbb{C}[x]$. $\endgroup$
    – Theo Johnson-Freyd
    Commented Aug 2, 2021 at 20:35
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    $\begingroup$ @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. $\endgroup$
    – Wojowu
    Commented Aug 2, 2021 at 21:02

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