> Full-text translation from Chinese. The original remains authoritative for its research claims; this translation is not independent mathematical verification. # ACTM Round 16 ## An All-Rank Proof of BADT Linear Support Visibility ### An All-Rank Proof of BADT Linear Support Visibility **Research programme:** Algebraic Combinatorial Transcendence Methodology (ACTM) **Authors:** Neo.K; Aletheia (GPT-5.6 Sol) **Institution:** EveMissLab / 一言諾科技有限公司 **Date:** 2026-10-02 **Version:** Round 16 / v1.0 **Document type:** Formal research note / analytic theorem **Status statement:** This round does not claim to prove the Four Exponentials Conjecture, the impossibility of BADT$_2$, the Matrix Coefficient Conjecture, or the Structural Rank Conjecture. It proves a restricted but all-rank polynomial-matrix support theorem: for BADT$_2$ and its linear-coordinate support $W=\operatorname{span}_{\mathbb Q}\{1,v,v^2,v^4\}$, every polynomial matrix entering the Waldschmidt–Masser strict rank region must be rationally visible. Thus the entire rational-linear GCDA / Waldschmidt–Masser amplification strategy is closed in one stroke. --- ## Abstract The toric chart of BADT$_2$: $$ (X_0,X_1,X_2,X_3) = u(1,v,v^2,v^4) $$ transforms every rational-linear-entry matrix: $$ M(X) = X_0A_0 + X_1A_1 + X_2A_2 + X_3A_3 $$ into: $$ M(u,v) = uN(v), $$ where: $$ N(v) = A_0+vA_1+v^2A_2+v^4A_3. $$ Thus all entries lie in: $$ \boxed{ W = \operatorname{span}_{\mathbb Q} \{1,v,v^2,v^4\}. } $$ Using polynomial minimal bases, the Predictable Degree Property, the missing-cubic map, and exact finite verification, Rounds 11–15 progressively proved that when the normal rank is: $$ 1\le r\le15 $$ all strict Waldschmidt–Masser degeneracies must be rationally visible. Round 16 extracts an analytic proof independent of $r$. Let: $$ N(v) \in M_{m\times n}(W) $$ have normal rank: $$ r. $$ Take a column-reduced minimal basis: $$ U(v) = \begin{pmatrix} u_1(v)&\cdots&u_r(v) \end{pmatrix} $$ such that: $$ N=UV. $$ Let the minimal-basis degrees be: $$ \delta_i \in \{0,1,2,3,4\}. $$ Write: $$ x = \#\{i:\delta_i=0\}, $$ $$ y = \#\{i:\delta_i=4\}. $$ Define: - $F$: multiplier-domain dimension; - $C$: minimal-basis coefficient-vector count; - $I$: missing-$v^3$ inactive coefficient-vector count; - $P$: active coefficient-vector count; - $a=\operatorname{rank}T_3$: active coefficient-span rank. The degree-by-degree resource table is: | $d$ | $F$ contribution | coefficient count | active count | inactive count | |---:|---:|---:|---:|---:| | $0$ | $5$ | $1$ | $1$ | $0$ | | $1$ | $4$ | $2$ | $2$ | $0$ | | $2$ | $3$ | $3$ | $2$ | $1$ | | $3$ | $2$ | $4$ | $2$ | $2$ | | $4$ | $1$ | $5$ | $1$ | $4$ | From this: $$ C+F=6r, $$ $$ P+I=C, $$ and, most importantly: $$ \boxed{ F+I = 4r+x+y. } $$ If the matrix already has a constant rational left/right kernel, a zero row / zero column exceeding the Waldschmidt–Masser visibility threshold is obtained directly. Thus only the case without a constant rational kernel needs to be studied. Define: $$ L=a+I, $$ $$ R=F-a. $$ Then: $$ m\le L, $$ $$ n\le R. $$ Define the resource slacks: $$ \ell=L-m, $$ $$ h=R-n. $$ The Waldschmidt–Masser strict rank condition: $$ \boxed{ r<\frac{mn}{m+n} } $$ on the one hand forces: $$ \boxed{ m+n>4r, } $$ so: $$ \boxed{ \ell+h1, $$ $$ 4-\frac1\theta>1. $$ The strict region also forces: $$ x+y>0. $$ Thus, if both canonical subsets are invisible: $$ \ell+h \ge (4-\theta)x + \left( 4-\frac1\theta \right)y > x+y, $$ which contradicts: $$ \ell+h1. } $$ Thus no hidden linear GCDA of any normal rank exists in the linear-coordinate support of BADT$_2$. --- # 2. Constant-kernel cases If there exists: $$ 0\neq w\in\mathbb Q^m $$ such that: $$ w^TN=0, $$ a rational change of row basis gives a whole zero row: $$ 1\times n. $$ Its visibility is: $$ \frac1m+\frac nn = 1+\frac1m > 1. $$ If there exists: $$ 0\neq c\in\mathbb Q^n $$ such that: $$ Nc=0, $$ the same reasoning gives a: $$ m\times1 $$ zero column, with: $$ \frac mm+\frac1n = 1+\frac1n > 1. $$ Thus from now on assume: $$ \boxed{ \text{no constant rational left/right kernel}. } $$ --- # 3. Polynomial minimal basis Let: $$ \mathcal C $$ be the module generated by the columns of $N$ over: $$ \mathbb Q[v] $$ viewed as a column module. Since: $$ \mathbb Q[v] $$ is a PID, $\mathcal C$ is a free module of rank $r$. Take a column-reduced basis: $$ U(v) = \begin{pmatrix} u_1(v)&\cdots&u_r(v) \end{pmatrix} $$ and: $$ V(v) $$ such that: $$ \boxed{ N=UV. } $$ Write: $$ \delta_i = \deg u_i. $$ By the Predictable Degree Property and the fact that the column degrees of $N$ are at most $4$: $$ \boxed{ 0\le\delta_i\le4. } $$ --- # 4. Resource totals For each degree $d$: $$ f_d=5-d $$ is the multiplier-domain dimension, $$ c_d=d+1 $$ is the coefficient-vector count. The missing cubic: $$ v^3 $$ gives the active coefficient counts: $$ p_d = (1,2,2,2,1), $$ and inactive counts: $$ q_d = (0,0,1,2,4). $$ Define: $$ F = \sum_i f_{\delta_i}, $$ $$ C = \sum_i c_{\delta_i}, $$ $$ P = \sum_i p_{\delta_i}, $$ $$ I = \sum_i q_{\delta_i}. $$ Then: $$ \boxed{ F+C=6r, } $$ $$ \boxed{ P+I=C. } $$ --- # 5. Extreme-degree counts Define: $$ x = \#\{i:\delta_i=0\}, $$ $$ y = \#\{i:\delta_i=4\}. $$ Calculating degree by degree: $$ f_d+q_d = (5,4,4,4,5). $$ Thus: $$ \boxed{ F+I = 4r+x+y. } $$ This is the core support identity of the all-rank proof. --- # 6. Active rank Define the missing-cubic coefficient map: $$ T_3: \mathcal F \to \mathbb Q^m, $$ where: $$ \mathcal F = \bigoplus_i \mathbb Q[v]_{\le4-\delta_i}. $$ Let: $$ a = \operatorname{rank}T_3. $$ Since there is no constant left kernel, all minimal-basis coefficient vectors span: $$ \mathbb Q^m. $$ The active vectors span at most $a$ dimensions and the inactive vectors supply at most $I$ dimensions, so: $$ \boxed{ m\le a+I. } $$ Also, clearly: $$ m\le C. $$ --- # 7. Active leading coefficients Column-reducedness guarantees that: $$ u_1,\ldots,u_r $$ have leading coefficient vectors that are Q-linearly independent. All leading coefficient vectors for: $$ \delta_i\le3 $$ are active. Their number is: $$ r-y. $$ Thus: $$ \boxed{ a\ge r-y. } $$ --- # 8. Right-side capacity The full admissible coefficient space: $$ \ker T_3 $$ has dimension: $$ F-a. $$ The absence of a constant right kernel means the columns of $V$, numbering $n$, are Q-linearly independent, so: $$ \boxed{ n\le F-a. } $$ In addition: $$ a\le P $$ since the active-span rank cannot exceed the number of active vectors. --- # 9. Left/right capacities Define: $$ \boxed{ L=a+I, } $$ $$ \boxed{ R=F-a. } $$ Then: $$ m\le L, $$ $$ n\le R. $$ Define the slacks: $$ \boxed{ \ell=L-m\ge0, } $$ $$ \boxed{ h=R-n\ge0. } $$ Moreover: $$ L+R = F+I = 4r+x+y. $$ --- # 10. Strict rank implies $m,n>r$ Since: $$ \frac{mn}{m+n} < \min(m,n), $$ if: $$ r < \frac{mn}{m+n}, $$ then: $$ \boxed{ m>r, \qquad n>r. } $$ --- # 11. Upper bound for $n$ From: $$ n\le F-a $$ and: $$ a\ge r-y, $$ we have: $$ n \le F-r+y. $$ Expanding degree by degree: $$ F-r+y = 4x + 3c_1 + 2c_2 + c_3 + y, $$ where: $$ c_d = \#\{i:\delta_i=d\}. $$ Each coefficient is at most $4$, so: $$ \boxed{ n\le4r. } $$ --- # 12. Upper bound for $m$ Since: $$ n>r $$ and: $$ n\le F-a, $$ we have: $$ a < F-r. $$ Thus: $$ m \le a+I < F-r+I. $$ Using: $$ F+I=4r+x+y, $$ we obtain: $$ m < 3r+x+y. $$ Since: $$ x+y\le r, $$ we have: $$ \boxed{ m<4r. } $$ --- # 13. Aspect-Ratio Lemma Define: $$ \theta = \frac nm. $$ The strict rank condition: $$ r(m+n)rn. $$ Thus: $$ m > \frac{rn}{n-r}. $$ Since: $$ n\le4r, $$ we have: $$ n-r\le3r. $$ Therefore: $$ m>\frac n3. $$ that is: $$ \boxed{ \theta<3. } $$ Likewise: $$ n(m-r)>rm. $$ From: $$ m<4r $$ we obtain: $$ n>\frac m3. $$ hence: $$ \boxed{ \theta>\frac13. } $$ Thus: $$ \boxed{ \frac13<\theta<3. } $$ --- # 14. Slack Deficit Lemma AM–GM gives: $$ \frac{mn}{m+n} \le \frac{m+n}{4}. $$ The strict rank condition is: $$ r < \frac{mn}{m+n} $$ therefore: $$ \boxed{ m+n>4r. } $$ On the other hand: $$ \ell+h = (L+R)-(m+n). $$ and: $$ L+R = 4r+x+y. $$ so: $$ \boxed{ \ell+h = 4r+x+y-(m+n) < x+y. } $$ In particular: $$ \boxed{ x+y>0. } $$ That is, the strict region must contain at least one extreme minimal-basis degree: $$ 0 \quad\text{or}\quad 4. $$ --- # 15. Canonical subset $J_0$ Let: $$ J_0 = \{i:\delta_i=0\}. $$ Its cardinality is: $$ x. $$ Each degree-$0$ basis column has just one coefficient vector, and that vector is both active and leading. Column-reducedness makes these $x$ vectors Q-linearly independent. Thus: $$ \dim\operatorname{im}(T_3|_{J_0})=x. $$ Its multiplier-domain dimension is: $$ 5x. $$ Therefore: $$ \boxed{ \dim\mathcal A_{J_0} = 4x. } $$ --- # 16. $J_0$ column-combination count The full admissible-space dimension is: $$ F-a = R. $$ The actual coefficient-column span has dimension: $$ n. $$ The Grassmann inequality gives: $$ k_0 \ge n+4x-R. $$ Since: $$ R=n+h, $$ we obtain: $$ \boxed{ k_0 \ge \max(0,4x-h). } $$ The image row-space dimension of $J_0$ is exactly: $$ x. $$ Thus the rational zero block is at least: $$ \boxed{ (m-x) \times \max(0,4x-h). } $$ --- # 17. If $J_0$ Is Invisible If the guaranteed zero block above does not reach: $$ \frac{m'}m+\frac{n'}n>1, $$ then: $$ \frac{m-x}{m} + \frac{\max(0,4x-h)}n \le1. $$ Equivalently: $$ m\max(0,4x-h) \le nx. $$ Thus, regardless of whether: $$ 4x-h $$ is positive, we have: $$ \boxed{ h \ge (4-\theta)x. } $$ where: $$ \theta=\frac nm. $$ --- # 18. Canonical subset $J_3$ Let: $$ J_3 = \{i:\delta_i\le3\}, $$ that is, excluding all: $$ y $$ minimal-basis columns of degree $4$. Thus: $$ F_{J_3}=F-y, $$ $$ P_{J_3}=P-y, $$ $$ I_{J_3}=I-4y, $$ $$ C_{J_3}=C-5y. $$ --- # 19. $J_3$ admissible dimension The restricted active rank is at most: $$ \min(a,P-y). $$ so: $$ \dim\mathcal A_{J_3} \ge F-y-\min(a,P-y). $$ Taking a Grassmann intersection with the actual $n$-dimensional column span: $$ k_3 \ge n + F-y-\min(a,P-y) - (F-a). $$ Put: $$ z=P-a\ge0. $$ Simplifying by cases gives: $$ \boxed{ k_3 \ge n-d, } $$ where: $$ \boxed{ d=\min(y,z). } $$ Thus: $$ d\le y. $$ --- # 20. $J_3$ row-space deficit Let the coefficient-count slack be: $$ c=C-m\ge0. $$ and recall the earlier: $$ \ell=(a+I)-m. $$ The coefficient row-space dimension of $J_3$ is at most: $$ \min \left( m, C-5y, a+I-4y \right). $$ Thus the rational row-annihilator dimension is at least: $$ \boxed{ D = \max \left( 0, 5y-c, 4y-\ell \right). } $$ This gives the guaranteed zero block: $$ \boxed{ D\times(n-d). } $$ --- # 21. If $J_3$ Is Invisible If: $$ \frac Dm + \frac{n-d}{n} \le1, $$ then: $$ nD \le md. $$ Since: $$ D \ge 4y-\ell, $$ and: $$ d\le y, $$ we have: $$ n(4y-\ell) \le my. $$ Therefore: $$ \boxed{ \ell \ge \left( 4-\frac1\theta \right)y. } $$ --- # 22. Final contradiction Suppose both canonical subsets are invisible. Then: $$ h \ge (4-\theta)x, $$ and: $$ \ell \ge \left( 4-\frac1\theta \right)y. $$ By the Aspect-Ratio Lemma: $$ \frac13<\theta<3. $$ so: $$ 4-\theta>1, $$ and: $$ 4-\frac1\theta>1. $$ By the Slack Deficit Lemma: $$ x+y>0. $$ Therefore: $$ \ell+h \ge (4-\theta)x + \left( 4-\frac1\theta \right)y > x+y. $$ But it was proved above that: $$ \ell+h However large the matrix and however high its normal rank, as long as its entries are merely rational linear combinations of the four linear coordinates of BADT$_2$ and one seeks a contradiction by entering: > > $$ > r<\frac{mn}{m+n} > $$ > > the matrix will first automatically produce the rational zero block permitted by the Waldschmidt–Masser theorem. Thus: $$ \boxed{ \text{linear GCDA strategy is completely closed}. } $$ --- # 26. Why This Does Not Duplicate the Matrix Coefficient Conjecture The Matrix Coefficient Conjecture proposes a stronger and more general rational-visibility prediction for general square matrices of algebraic logarithms. R16.1 handles only: - the specific toric support of BADT$_2$; - rational-linear-entry polynomial matrices; - the Waldschmidt–Masser strict rank regime. Thus it is not a general proof of the Matrix Coefficient Conjecture. But it is an: $$ \boxed{ \text{all-rank support-specific visibility theorem}. } $$ --- # 27. The Actual Role of Support Geometry Ultimately the proof needs only two extreme degree counts: $$ x=c_0, $$ $$ y=c_4. $$ The reason is that the five types of minimal-basis degree resources for the support: $$ \{0,1,2,4\} $$ satisfy: $$ f_d+q_d = (5,4,4,4,5). $$ All interior degrees: $$ 1,2,3 $$ contribute exactly: $$ 4, $$ Only the extremes: $$ 0,4 $$ supply an additional unit. Thus: $$ \boxed{ F+I = 4r+x+y. } $$ Strict Waldschmidt–Masser geometry must consume these extreme surpluses. $J_0$ and $J_3$ detect the surpluses at the left and right endpoints respectively. This is the geometric core of the all-rank proof. --- # 28. Extreme-Surplus Principle This round abstracts the following: > When the resource table of a polynomial support has a baseline, only the extreme degree types supply a surplus, and the strict rank region forces total slack below the extreme surplus, two endpoint canonical subsets may suffice to prove visibility at all ranks. This round calls this the: **Extreme-Surplus Principle**. For now, a concrete proof has been completed only for the BADT support. --- # 29. Reconnecting to ACTM's Original "Over-Compression" Problem BADT$_2$ itself remains a hypothetical over-compression configuration. But R16.1 shows: > Such over-compression cannot be refuted by "projecting it into a larger low-rank rational-linear matrix and then applying the Waldschmidt–Masser theorem". This is because the support itself makes all such projections: $$ \boxed{ \text{low-rank} \Rightarrow \text{rationally visible}. } $$ Thus, if a genuine contradiction exists, it must retain more nonlinear information than a rational-linear projection. --- # 30. The Formal Entry Point for Round 17 The linear GCDA hierarchy has ended. The next round should formally turn to nonlinear / support-sensitive routes. Three routes are proposed: ## R17-A: The BADT-Specific Matrix Coefficient Route Directly handle: $$ H_1 = \begin{pmatrix} z_0&z_1\\ z_1&z_2 \end{pmatrix}, $$ $$ H_2 = \begin{pmatrix} z_0&z_2\\ z_2&z_3 \end{pmatrix} $$ and the structure in which they share: $$ z_0,z_2 $$ as displayed above. Rather than amplifying rank further, study whether shared support can directly generate a Matrix Coefficient rational witness. --- ## R17-B: The Dasgupta–Kakde Sparse Auxiliary Polynomial The BADT toric ideal: $$ q_1=X_1^2-X_0X_2, $$ $$ q_2=X_2^2-X_0X_3 $$ itself has only four monomial-support directions. Try to prove directly within the DK framework a BADT-specific: $$ (w)\Rightarrow(m) $$ or the weaker: $$ (w)\Rightarrow(m'). $$ --- ## R17-C: The Roy Geometric Route Study the BADT$_2$ toric surface: $$ X_1^2=X_0X_2, $$ $$ X_2^2=X_0X_3 $$ and the algebraic-logarithm point on it: $$ (z_0,z_1,z_2,z_3). $$ Seek a restricted theorem of the form: $$ \boxed{ \text{nonlinear toric geometry} \Rightarrow \text{rational linear visibility} } $$ displayed above. --- # 31. Round 16 Assessment **PASS — BADT Linear Support Visibility proved for all normal ranks; the entire rational-linear Waldschmidt–Masser amplifier strategy is closed** This round completes: 1. Extraction of two canonical subsets from the finite-rank EARVL pattern. 2. Definition of extreme degree counts: $$ x=c_0, \qquad y=c_4. $$ 3. Construction of: $$ F+I=4r+x+y. $$ 4. Definition of the left/right capacities: $$ L=a+I, \qquad R=F-a. $$ 5. Definition of the slacks: $$ \ell=L-m, \qquad h=R-n. $$ 6. A proof from strict WM that: $$ \ell+h **Linear visibility of the BADT$_2$ support does not require rank-by-rank search: strict Waldschmidt–Masser geometry forces the matrix aspect ratio into $(1/3,3)$ while making the total left/right resource slack smaller than the degree-$0/4$ extreme surplus; if the degree-$0$ subset and the non-degree-$4$ subset are both invisible, they instead each demand more than one unit of extreme slack, and adding the two gives an immediate contradiction.** Thus: $$ \boxed{ R_{\mathrm{SV}} \left( \operatorname{span}_{\mathbb Q} \{1,v,v^2,v^4\} \right) = \infty. } $$ The next round formally leaves the linear rank-amplifier route. --- # References 1. Samit Dasgupta, **Ranks of matrices of logarithms of algebraic numbers I: the theorems of Baker and Waldschmidt–Masser**, *Essential Number Theory* 2 (2023), 93–138. https://arxiv.org/abs/2303.02037 2. W. H. L. Neven and C. Praagman, **Column reduction of polynomial matrices**, *Linear Algebra and its Applications* 188–189 (1993), 569–589. DOI: 10.1016/0024-3795(93)90480-C 3. Andrii Dmytryshyn, Froilán M. Dopico, Paul Van Dooren, **Minimal rank factorizations of polynomial matrices**, *Linear Algebra and its Applications* 721 (2025), 736–774. DOI: 10.1016/j.laa.2025.01.009 4. Samit Dasgupta and Mahesh Kakde, **Ranks of Matrices of Logarithms of Algebraic Numbers II: The Matrix Coefficient Conjecture**, *Advances in Mathematics* 487 (2026), 110753. DOI: 10.1016/j.aim.2025.110753 --- **End of ACTM Round 16**