A robust dimension witness for arbitrary four-record approximations

and let σ2, σ5 > 0 be its second and fifth singular values. If C has a complex Hermitian PSD factorization
of size at most three, then
27L
μ
≥
σ2
2σ2
5
36(σ2 + 18
√
6)2 .
(24)

Proof.
Since σ5 > 0, a factorization cannot be supported on dimension two. Congruence normalization
of its row factors gives a POVM P
a,b Aab = I3 and density matrices bcd with Cab,cd = Tr(Aabbcd).
The row sums imply Tr Aab ≥μ/9, hence a largest eigenvalue at least μ/27. Put Ra = P
d bad and
Qb = P
c bcb. Both have trace three, and the leakage implies

The bound ∥Qb∥≤Tr Qb = 3 and the triangle inequality after applying √Qb give

Every preparation obeys bcd ⪯Qd, hence Tr[(I −P)bcd] ≤lfor this common plane. The projection
bound ∥ρ −PρP∥1 ≤2
p

Tr[(I −P)ρ] therefore bounds every preparation-compression error by 2
√

l.
For completeness, purify ρ, apply P ⊗I, and compute the trace norm of the difference between the
original rank-one projector and the projected, unnormalized rank-one operator: for discarded mass x,
its norm is
√

4x −3x2 ≤2√x. Partial trace contracts this Hermitian trace norm, giving the stated
bound also for mixed ρ.
Replace each bcd by PbcdP and apply the same POVM. The resulting real matrix C(4) has rank
at most four, since Hermitian operators on a two-dimensional plane have real dimension four. A
POVM contracts the trace norm of a Hermitian difference: if D = D+ −D−is its positive/negative
decomposition, then P
ab | Tr(AabD)| ≤Tr D+ + Tr D−= ∥D∥1. Consequently each of the nine column
errors has Euclidean norm at most 2
√
l. Thus

The first inequality follows from the singular-value lower bound for rank-four approximation, which
applies to every real approximant.
Case 2: all Ra, Qb have second eigenvalue below s. Let Pa and Q′
b be their top-eigenvector projectors.
The mass outside each such line is below 2s. Since bab ⪯Ra, Qb,

where D denotes half trace distance. Here D(ρ, |v⟩⟨v|) ≤
p

1 −⟨v, ρv⟩: it is equality for pure states
and follows for mixed states by convexity and Jensen’s inequality. Fix P0. Comparing through b0b gives
D(P0, Q′
b) ≤2
√
2s, whence D(bab, P0) ≤3
√

2s for all nine preparations. Replace every preparation by
this same pure state. The resulting matrix C(1) has identical normalized columns and rank at most
one. Each column difference is zero-sum and has classical total variation at most 3
√

2s by POVM
contraction. For a zero-sum real vector v, its positive and negative parts each have mass ∥v∥1/2, so
∥v∥2
2 ≤2(∥v∥1/2)2. Each column thus has Euclidean norm at most 6√s, and

These two cases exhaust the possible normalized factors.
If the right side of (24) exceeded e, choose s < σ2
2/324 sufficiently close to that upper endpoint.
Then both displayed inequalities would fail, since the limiting first bound is 6√e(1 + 18
√

6/σ2) < σ5.
This choice also satisfies s > e: column stochasticity gives σ2, σ5 ≤∥C∥F ≤3, and the assumed bound
on e is less than σ2
2/(24 · 324). This contradiction proves the necessary inequality.
□

Suppose another complete four-record law P ′ is generated through a quantum causal cut of dimension
at most three. All history-correlated systems used again in the future belong to that cut; independent
fresh ports and an untouched retained archive obey the same convention as the accounting chapter.
The competing law may have a different update rule, lack stationarity, or lie outside the trine family.
Extra null/loss outcomes can remain in its complete alphabet, with zero target probability.
Put δ = TV(P ′, Pλ,4) < 1/9. Restrict its joint table H′ to the target trit pairs and let h′
cd =
P
ab H′
ab,cd ≥1/9 −δ. The column-normalized matrix C′
ab,cd = H′
ab,cd/h′
cd still has PSD rank at most
three. Define

The noiseless growing lower bound is sensitive to small forbidden-event probabilities. A quantitative
replacement is possible, with a poor large-grid margin that should be stated explicitly. Suppose a
row-stochastic matrix eT factors through r classical states as AB, with both A and B row stochastic.
Such a normalization is available for every nonnegative factorization of a stochastic matrix after zero
factors are removed. Set e = maxi eTii and μ = mini̸=j eTij. For τ > 0 put Si = {l: Ail≥τ}. If Si ⊆Sj,
then
eTij ≤e/τ + rτ.
(28)

Indeed, on Si the sum of Blj is at most e/τ because Ajl≥τ, and outside Si the contribution is at
most rτ. Choosing τ =
p

e/r shows that μ > 2√re forces an antichain; the case e = 0 follows by a
limit or the exact support argument.
For an entrywise perturbation of the phase-circle transition by at most δ, write