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Are their other functions of a complex square matrix, not trivially related to trace, which also posses the cyclic property?

Furthermore, do all such functions $f(A)$ depend only on the spectrum of $A$ like the trace and are any of them linear like the trace?

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    $\begingroup$ So $f(A)=f(VAV^{-1})$, and $f$ depends on the JNF only. $\endgroup$ Jul 4, 2014 at 21:54
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    $\begingroup$ Other semi-trivial examples of functions with the cyclic property, but without the other ones that you ask for, are the $\ell^p$-norms of the eigenvalues. $\endgroup$ Jul 4, 2014 at 22:06
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    $\begingroup$ I think it's also pretty clear that $c\,\textrm{tr }A$ is the only linear $f$ because $f$ also needs to be invariant under permutation of Jordan blocks, and the only other candidate, codimension of the span of the eigenvectors (adding up the $1$'s in the JNF) is of course not linear on all matrices. $\endgroup$ Jul 4, 2014 at 22:15
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    $\begingroup$ By the way, Peter Semrl has characterized the linear maps from $C^{n\times n}$ to itself that preserve many different properties, including those that [preserve similarity](www.fmf.uni-lj.si/~semrl/preprints/similarity.pdf). This problem looks related. $\endgroup$ Jul 4, 2014 at 22:18
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    $\begingroup$ In the former case, assume that $C$ is nonsingular; then for any $AB = C$, $A$ is nonsingular, so $BA$ is conjugate to $C$, so any function that depends only on the JNF works (under this restriction to nonsingular matrices). $\endgroup$
    – user44191
    Jul 4, 2014 at 23:22

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To make my comment an explicit answer:

Given $C, D$, there are $A_i, B_i$ with $C = A_0B_0$, $D_i = B_iA_i$, $D = D_n$ if and only if the Jordan normal form of both are the same, up to deleting the block corresponding to the generalized 0-eigenspace. As such, any "cyclic" function depends only on that "reduced" JNF. If you further require continuity, then you restrict yourself to functions which are generated by the coefficients of the characteristic polynomial.

Proof: for one direction, proving it for $C = AB, D = BA$ is enough. Let $v$ be a generalized eigenvector of $C$ with nonzero eigenvalue $\lambda$ of "type" $i$ (I'm not sure of the proper word here; "type" 0 refers to actual eigenvectors, while $(C - \lambda I) v$ is of type $i - 1$ if $v$ is of type $i$). Then $Bv$ is nonzero, so by induction, we can see that it is also a generalized eigenvector of eigenvalue $\lambda$ of type $i$. We therefore have a bijection between general eigenvectors of $C$ and $D$ with nonzero eigenvalues $\lambda$ of type $i$; this means directly that the "reduced" JNF must be the same. Note that this also limits how "quickly" you can get from one to the other (i.e. a minimal n): you can only change the length of any one block by 1 for each "cycle".

The reverse is obvious when dealing only with the Jordan normal forms, and the rest can be gotten by conjugating the relevant matrices. To explain further: David Speyer has shown how to make the relevant generalized 0-eigenspace blocks equivalent; simply adding on the other blocks gives the normal form equivalences. Then if we have that $C' = XCX^{-1}, D' = YDY^{-1}$ are the JNF of $C, D$ and $C' = AB, D' = BA$, then $C = (X^{-1}AY)(Y^{-1}BX), D = (Y^{-1}BX)(X^{-1}AY)$ gives a cycling for C and D.

Therefore, any cyclic function depends entirely on the "reduced" JNF, and any function that does depend on that is cyclic.

As a note, this proof works just as well for rational normal form, so you can even restrict the $A_i, B_i$ to the same field as $C, D$.

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    $\begingroup$ Your result is correct but I don't follow the proof. Where do you prove that $AB$ and $BA$ have the same Jordan structure, except for the nilpotent part? I only see that you have proved this in the case that $AB$ or $BA$ is invertible. $\endgroup$ Jul 5, 2014 at 1:55
  • $\begingroup$ Proof sketch of the things that seem to be missing to me: For any polynomial $f(x)$, we have $B f(AB) A = f(BA) BA$. This shows that $\mathrm{rank} f(AB) \geq \mathrm{rank} f(BA) BA \geq \mathrm{rank} f(AB) (AB)^2 \geq \cdots$. So $\lim_{n \to \infty} \mathrm{rank} f(AB) (AB)^n = \lim_{n \to \infty} \mathrm{rank} f(BA) (BA)^n$. Applying this fact with $f(x)=(x-\lambda)^k$ for all nonzero $\lambda$ and all $k$ shows that the nonzero Jordan blocks match. $\endgroup$ Jul 5, 2014 at 1:58
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    $\begingroup$ Use identities like $\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}$ and $\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$ to show that any nilpotent block can be changed into any other. $\endgroup$ Jul 5, 2014 at 1:58
  • $\begingroup$ Hrm. You're right, I didn't fully explain that part. I'll edit that in. $\endgroup$
    – user44191
    Jul 5, 2014 at 1:59

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