Power Traces Do Not Determine Similarity

Abstract

Two explicit matrices have identical positive-power traces and characteristic polynomial but belong to different similarity classes.

Theorem 1.1 (All power traces can miss the similarity class).

$$\forall K:Type,\mathrm{Field}\left(K\right)\Rightarrow A=\mathrm{zeroMatrix}\left(2,K\right),N=\mathrm{single}\left(2,0,1,1,K\right),(\forall k\in \mathbb{N},1\le k\Rightarrow (\mathrm{tr}\left(A^k\right)=0\wedge \mathrm{tr}\left(N^k\right)=0))\wedge \mathrm{charpoly}\left(A\right)=X^2\wedge \mathrm{charpoly}\left(N\right)=X^2\wedge \mathrm{rank}\left(A\right)=0\wedge \mathrm{rank}\left(N\right)=1\wedge \neg (\exists P\in \mathrm{GL}\left(2,K\right):PA{P}^{-1}=N)\wedge \neg (\forall M,C\in \mathrm{Matrix}\left(2,2,K\right),(\forall k\in \mathbb{N},1\le k\Rightarrow \mathrm{tr}\left(M^k\right)=\mathrm{tr}\left(C^k\right))\Rightarrow \exists P\in \mathrm{GL}\left(2,K\right):PM{P}^{-1}=C).$$

Proof. Machine-checked in Lean as D5/S0/Observation/PowerTraceSimilarityCountermodel.power_traces_do_not_determine_similarity (✓ std3). ∎

Source. Repository-derived.

Commentary.

Over an arbitrary field, take A to be the two-by-two zero matrix and N to have its only nonzero entry, one, in row zero and column one. The matrix N is nonzero and square-zero.

Every positive power of A and N has trace zero, and both characteristic polynomials are X squared. Their ranks are zero and one, so no invertible change of basis conjugates A to N. The same pair directly refutes the universal claim that all positive-power traces determine matrix similarity.

The result is stronger than the source’s characteristic-zero context: the countermodel works over every field. Pinned Mathlib supplies the two-dimensional characteristic-polynomial formula and rank bounds, but no theorem packages this full countermodel.

References

• Truth anchor: D5/S0/Observation/PowerTraceSimilarityCountermodel.power_traces_do_not_determine_similarity