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Suppose that I have $n$ matrices $A_1, \ldots, A_n \in \mathbb{R}^{m \times m}$ with $m \gg n$. Can I find $n$ new matrices $B_1, \ldots, B_n \in \mathbb{R}^{n \times n}$ that have the same 3-way cyclic traces: $$ \forall i, j, k : \mathrm{Tr}(A_i A_j A_k) = \mathrm{Tr}(B_i B_j B_k)?$$ By analogy, if I had $n$ vectors $v_1, \ldots, v_n \in \mathbb{R}^m$, it would be easy to construct new vectors $u_1, \ldots, u_n \in \mathbb{R}^n$ that have the same inner products (by choosing an orthonormal basis for the span of the $v_i$ and then writing each $v_i$ in that basis). Parameter counting suggests there should be matrices $B$ that match a given set of cyclic traces (we have $n^3$ parameters to pick and only $n^3/3$ constraints), but I have no idea how you could pick them "naturally" and don't have any reason beyond parameter counting to think they exist.

Context: I'm interested in succinctly summarizing the interactions between the matrices $A$ in a way that lets me make reasonable guesses about arbitrary cyclic traces like $\mathrm{Tr}(A_i A_j A_k A_l)$. I've found no leads on how to guess higher-order traces given lower-order traces, finding a small set of matrices with a given set of low-order traces seems like a natural first step on the problem if it's possible. (I considered instead assuming that higher free cumulants are zero, but this will tend to produce unreasonable estimates including $\mathrm{Tr}(X^2) < 0$.)

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  • $\begingroup$ if all the A_i are nilpotent and commute with one another (for example if they are powers of the shift operator in some basis), then all of these products are nilpotent as well and the traces are zero. So we could set all the B_i to zero and your condition would hold. This principle should allow you to build lots of examples of families of matrices where these traces agree. $\endgroup$
    – pupshaw
    Commented May 27, 2023 at 14:32
  • $\begingroup$ I'm imagining m >> n, I'll clarify the question. If n = 1, then I have only a single matrix A_1, and I can take B_1 to be the cube root of the trace of the cube (i.e. the l3 norm of the eigenvalues of A). $\endgroup$
    – Paul Christiano
    Commented May 28, 2023 at 16:46
  • $\begingroup$ From this question mathoverflow.net/q/391669, we conclude that the functions $(i,j,k)\mapsto\text{Tr}(A_iA_jA_k)$ are precisely the cyclic invariant functions as long as $m$ is large enough. $\endgroup$
    – Joseph Van Name
    Commented May 28, 2023 at 22:08
  • $\begingroup$ With the field of complex numbers, we can factorize $B_1,\dots,B_n$ as upper triangular block matrices where the diagonal blocks have no common invariant subspace. We may then apply Burnside's theorem to conclude that each of the diagonal blocks generates the entire matrix algebra. $\endgroup$
    – Joseph Van Name
    Commented May 28, 2023 at 22:33
  • $\begingroup$ For any particular $n$, the problem is decidable for either the real or the complex numbers. This follows from the reformulation of the question as to whether every cyclic function can be written as $(i,j,k)\mapsto\text{Tr}(A_iA_jA_k)$ which is a first order statement in the theory of real closed fields or respectively algebraically closed fields of characteristic zero. And both the theory of real closed field and algebraically closed fields of characteristic zero are complete and decidable. $\endgroup$
    – Joseph Van Name
    Commented May 28, 2023 at 22:55

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