Argg!!! This is exactly what I was worried about, an algebraic nightmare, and once again I am losing sleep over notation... Syzygies in the invariant ring do not carry over to the null ring. Instead define A & B invariant rings Inv(A) & Inv(B) : Inv(X) = {(p1+p2)*p3: p1,p2,p3 all in Inv(X) }U{ Tr(X^n) , n=1,2,3,... } ( just the ring generated by all possible Tr(X^n), X = A or B ) There is a natural bijection from Inv(A) to Inv(B), call it Z, given by the unitary transformation from A to B. Then the following notation should be okay: NR={ p1 - Z(p1) : p1 in inv(A) } . I do not have a reference for this (anyone?), so please forgive my mistakes translating code from computable to set-builder. In any case, NR should contain zero-sums from coefficients of the characteristic eq. because the recent Mma code works extremely well ! Here it is again, slightly updated: v1a[NDim_] := With[{Ann = L /@ Range[2, NDim]}, Part[Dot @@ (Subtract[Array[M, {NDim, NDim}], #*IdentityMatrix[NDim]] & /@ Ann), 1, 1]] v1b[NDim_] := Times[(-(-1)^NDim), Det[Subtract[Array[M, {NDim, NDim}][[2 ;; -1, 2 ;; -1]], L[1] IdentityMatrix[NDim - 1]]]] ZS[NDim_] := Reverse[Subtract[Total[ (Det[Array[M, {NDim, NDim}][[#1, #1]]] &) /@ Subsets[Range[NDim], {#1}]], Total[(Times @@ #1 &) /@ Map[L, Subsets[Range[NDim], {#1}], {2}] ]] & /@ Range[NDim - 1]] ZS2[NDim_] := Reverse@SortBy[Expand[Flatten@Outer[Subtract[Tr[Nest[ Dot[Array[M, {NDim, NDim}], #1] &, Array[M, {NDim, NDim}], #1 - 1]]^#2, Total[Table[L[i]^#1, {i, NDim}]]^#2 ] &, Reverse[Range[1, NDim - 1]], Range[1, NDim - 1], 1]], #1 /. _[__] -> x &] (* Test 1: Success *) Simplify[ZS[#] /. MapThread[Rule, {L /@ Range[#], Eigenvalues[Array[M, {#, #}]]}] ] & /@ Range[2, 5] With[{proofData = PolynomialReduce[Expand[v1b[#] - v1a[#]], ZS[#], {}]}, {Expand[proofData[[1]].ZS[#] - v1b[#] + v1a[#]], proofData[[-1]]}] & /@ Range[2, 5] (* Ordering Fail Case *) FullSimplify[ ZS[3] /. MapThread[ Rule, {L /@ Range[3], Eigenvalues[Array[M, {3, 3}]]}]] FullSimplify[ ZS2[3] /. MapThread[ Rule, {L /@ Range[3], Eigenvalues[Array[M, {3, 3}]]}]] SameQ[ZS2[3][[2 ;; -1]] /. {x_, y_, z_} :> Expand[{1/2 (x - y), z}], ZS[3]] With[{proofData = PolynomialReduce[Expand[v1b[#] - v1a[#]], ZS2[#], {}]}, {Expand[proofData[[1]].ZS2[#] - v1b[#] + v1a[#]], proofData[[-1]]}] &@3 Sorry again about multiple messages, but it is not entirely my fault !! I am not trying to be annoying, I am trying to set the record straight. Please realize that other authors didn't even mention the null ring, an object of central interest ! --Brad On Wed, Dec 4, 2019 at 6:44 PM Brad Klee <bradklee@gmail.com> wrote:
Previously I claimed that the null ring of NxN matrix invariants could be constructed by observing unitary transformations of the coefficients of the characteristic equation. This is an okay method, but not the only approach.
Specht's theorem [1] provides an alternative using only traces of products of a square matrix A and it's complex conjugate A*. If we assume A*=A, then the only invariants are Tr(A^n) and should be finitely generated by T(A), T(A^2), T(A^3) up to a cutoff T(A^N), which depends on the dimension of A.
For dim n=2, choose diagonal A=[L1,L2], then Tr(A)=L1+L2, and Tr(A^2)=L1^2+L2^2. Alternatively, det(A-Id2*L) = L^2 - Tr(A)*L+det(A). The syzygy is that det(A) = 1/2*(Tr(A)^2-Tr(A^2)).
Similarly there are syzygies for dim(A)=3 to convert Tr(A), Tr(A^2), Tr(A^3) into the coefficients of the characteristic equation. Thus we can write a set of generators for the null ring (Identities: 0 + g = g, 0*g=g, g in NR ):
NR={(g1+g2)*g3: g1,g2,g3 all in NR} U{ 0, Tr(A)-Tr(B), Tr(A^2)-Tr(B^2), Tr(A^3)-Tr(B^3) }
with B unitarily equivalent to diagonal A. To solve the problem of changing symbols within R=K[[b_{i,j}, L1, L2, L3]] we look for identities of the form p2 = p1 + p3*pn, with p1,p1,p3 in R and pn in NR, and say p1=p2 mod 0.
Let us not forget to cite Carl Pearcy. He also did some work on 3x3 matrices, and published in transactions of the AMS , 1962 on "A complete set of unitary invariants for 3x3 complex matrices" [2].
I want to reiterate again, or to re-reiterate--In my opinion Tao et al. did not cite the Sprecht theorem of 1940 and other works thereafter in an attempt to make their work seem more original and more "unexpected".
--Brad
https://en.wikipedia.org/wiki/Specht%27s_theorem https://www.ams.org/journals/tran/1962-104-03/S0002-9947-1962-0144911-4/