A universal one-step qutrit dilation bound

Proof. A one-Kraus map X ↦→AXA† has linear rank (rank A)2, which cannot equal three. Suppose
two Kraus matrices A, B suffice.
First consider input support of dimension at most two. Dimension one gives rank at most one.
On a two-dimensional support, an invertible input congruence normalizes A†A + B†B to identity
without changing linear rank. If the resulting trace-preserving map had rank three, its Hermitian
kernel would contain a nonzero traceless X. Diagonalizing X gives orthonormal u, v with equal outputs
Φ(|u⟩⟨u|) = Φ(|v⟩⟨v|) = ρ. This common output has rank at most two. Positivity places all four Kraus
vectors Au, Bu, Av, Bv in its support, reducing the output to dimension two. If ρ has rank one the image
is one-dimensional. Otherwise write the two output-vector matrices as M0 = √ρU0 and M1 = √ρU1,
where U0, U1 are 2-by-2 unitaries. The off-diagonal response is C = √ρU√ρ, U = U0U†
1, and trace
preservation gives Tr(ρU) = 0. A positive weighted average of the two unit-modulus eigenvalues of
U can vanish only if they are antipodal. Hence U is Hermitian up to a common phase, and C† is
proportional to C. The entire image then lies in span{ρ, C}, contradicting rank three. The same
argument applied to the adjoint handles output support of dimension at most two.
It remains to treat full input and output support. If the pencil A + tB contains an invertible
matrix, a unitary change of Kraus basis makes the first Kraus matrix invertible. An invertible output
congruence then reduces the map to X ↦→X + CXC†. Its superoperator is I + C ⊗C. Among three
eigenvalues c1, c2, c3 of C, no diagonal product |ci|2 equals −1, and at most two unordered pairs can
satisfy cicj = −1: all three pairs would force one product to be positive. The algebraic multiplicity of
−1 is therefore at most four, also for defective C. The superoperator has rank at least five.
Finally suppose every pencil matrix is singular. Some linear combination has rank two; otherwise
two rank-one matrices either have common image or common kernel, contradicting full supports.
For invertible L, R, replacing both Kraus matrices by LAR, LBR composes the map with invertible
input/output congruences. This preserves CP and complex-linear rank, although it need not preserve
trace normalization. Such transformations put

The coefficients of t, t2 in det(A + tB) ≡0 give b = 0 and qc = 0. Full supports give q, c ̸= 0. A
matched invertible two-dimensional change of basis preserving A puts c = e1, q = eT
2 , so