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Let $A$ be a diagonalizable transformation on a finite-dimensional vector space. If $B$ commutes with all transformations commuting with $A$, does it follow that $B=p(A)$ for some polynomial $p$?

This is based on an exercise of Halmos (1958, §84 exc. 5). I seem to think I've shown it to be true, but the original exercise has the much more restricted assumption that $A$ is Hermitian. So is it still true for any diagonalizable matrix?

Here is a sketch of my arguments:

  1. Let $C$ be a transformation with unique eigenvalues and with an eigenbasis agreeing with $A$. Then $C$ commutes with $A$.
  2. Therefore, $C$ commutes with $B$ (by assumption). Since $C$ has unique eigenvalues, it follows that $B$ and $C$ are simultaneously diagonalizable, which also means that $A$ and $B$ are simultaneously diagonalizable (since $A$ and $C$ share the unique eigenbasis of $C$).
  3. Let $\sigma_i$ be the eigenvalues of $A$ with corresponding eigenvectors $v_i$, and $\lambda_i$ be the eigenvalues of $B$. Let $E_{ij}$ be the transformation given by $$ E_{ij}v_k = \begin{cases}v_i&k=j\\v_j&k\ne j\end{cases}. $$ (That is, $E_{ij} = (\delta_{ij})_{ij}$ in the eiegenbasis). Then $\sigma_i=\sigma_j$ implies that $E_{ij}$ commutes with $A$, and thus also with $B$, which then implies that $\lambda_i=\lambda$.
  4. Finally, $B=p(A)$, where $p$ is a polynomial such that $p(\sigma_i)=\lambda_i$ (which is well-defined by the point above).

I can't pinpoint any problems with the above, and I can't see in particular why an orthogonal eigenbasis or real eigenvalues should make a difference. Any turnip?

Milten
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  • I didn't read your argument, but you don't even need diagonalizability -- the bicommutant of $A$ (i.e. the transformations commuting with all transformations commuting with $A$) is the set of polynomials in $A$. – blargoner Oct 16 '23 at 18:36
  • @blargoner Thank you, I was missing that term. So is this the “Von Neumann bicommutant theorem” in the finite-dimensional case? I don’t really have the knowledge to parse it properly (and I’m having a hard time finding a dumbed down version of it), but I’m guessing that the (weak and strong) closure of $A$ is ${p(A)}$ in the finite case? – Milten Oct 16 '23 at 19:45
  • I'm closing this myself as a duplicate of https://math.stackexchange.com/q/3330133/620957. Though I'd still appreciate a comment on whether my proof outline works for the simpler diagonalizable case – Milten Oct 16 '23 at 20:44
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    Yes I believe that's the generalization but I've only studied this in the finite-dimensional case. A good reference for the latter (over fields of characteristic 0) is Greub Linear Algebra 4th ed, p. 422, where it is proved using a cyclic decomposition. – blargoner Oct 16 '23 at 20:52

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