{"claim_id": "prop:symmetry", "title": "Covariance and normalization", "environment": "proposition", "component": "area_feedback", "status": "manuscript_proved", "statement_latex": "\\label{prop:symmetry}\nOrthogonal isomorphisms intertwine the legal update. For every nonzero real $c$, the similarity $(x,A,a)\\mapsto(cx,c^2A,ca)$ also intertwines it. Replacing the update by $A'=A+\\gamma x\\wedge a$ for a fixed $\\gamma\\neq0$ produces an equivalent normalized system under $\\widetilde A=A/\\gamma$.", "proof_latex": "The identities $(QAQ^*)(Qa)=Q(Aa)$ and $(Qx)\\wedge(Qa)=Q(x\\wedge a)Q^*$ give the first assertion. The scaled gate is $c^3Aa=0$ and the scaled area increment is $c^2x\\wedge a$, proving the second. Division by $\\gamma$ preserves the kernel and converts the third rule into~\\eqref{eq:update}.", "references_in_statement_or_proof": ["eq:update"], "source_file": "manuscript/Avenyra_Area_Feedback_v1.0.0.tex", "proof_audit_file": "audits/PROOF_AUDIT.txt", "priority_status": "No per-claim priority certification; see audits/NOVELTY_AUDIT.txt.", "verification_limit": "Manuscript proof status is not formal certification; finite code checks are separate."}
{"claim_id": "prop:area", "title": "Exact area accumulation", "environment": "proposition", "component": "area_feedback", "status": "manuscript_proved", "statement_latex": "\\label{prop:area}\nFor any finite word from $(x_0,A_0)$, evaluated by the algebraic update even if legality is ignored,\n\\begin{align}\n x_n&=x_0+\\sum_{j=1}^n a_j,\\\\\n A_n&=A_0+x_0\\wedge\\sum_{j=1}^n a_j+\n              \\sum_{1\\leq i<j\\leq n}a_i\\wedge a_j.\n \\label{eq:area}\n\\end{align}\nOn $V\\oplus\\bigwedge^2V$, the product\n\\[\n (x,A)\\star(y,C)=(x+y,A+C+x\\wedge y)\n\\]\nis associative, with identity $(0,0)$ and inverse $(x,A)^{-1}=(-x,-A)$. A legal update is right multiplication by $(a,0)$ subject to $Aa=0$.", "proof_latex": "The endpoint formula follows by summation. Substituting $x_{j-1}=x_0+\\sum_{i<j}a_i$ into $A_j-A_{j-1}=x_{j-1}\\wedge a_j$ and summing proves~\\eqref{eq:area}. Both associations of three factors have endpoint $x+y+z$ and area $A+C+F+x\\wedge y+x\\wedge z+y\\wedge z$, proving associativity. The remaining assertions follow from $x\\wedge x=0$.", "references_in_statement_or_proof": ["eq:area"], "source_file": "manuscript/Avenyra_Area_Feedback_v1.0.0.tex", "proof_audit_file": "audits/PROOF_AUDIT.txt", "priority_status": "No per-claim priority certification; see audits/NOVELTY_AUDIT.txt.", "verification_limit": "Manuscript proof status is not formal certification; finite code checks are separate."}
{"claim_id": "lem:decomposition", "title": "Orthogonal decomposition", "environment": "lemma", "component": "area_feedback", "status": "manuscript_proved", "statement_latex": "\\label{lem:decomposition}\nFor real skew-adjoint $A$, $R=\\im A=(\\ker A)^\\perp$. The restriction $B=A|_R$ is an invertible skew-adjoint operator on $R$. If $h=B^{-1}u$, then $\\inner{u}{h}=0$.", "proof_latex": "A vector $v$ is perpendicular to $\\im A$ exactly when $\\inner{v}{Az}=0$ for all $z$, equivalently $A^*v=-Av=0$. Thus $V=R\\oplus K$. The subspace $R$ is $A$-invariant since $A$ maps all of $V$ into $R$. Its restricted kernel is $R\\cap K=0$, so $B$ is invertible in finite dimension. Its inverse is skew-adjoint: $(B^{-1})^*=(B^*)^{-1}=-B^{-1}$. Hence $\\inner{u}{B^{-1}u}=0$. When $R=0$, all statements are interpreted on the zero space.", "references_in_statement_or_proof": [], "source_file": "manuscript/Avenyra_Area_Feedback_v1.0.0.tex", "proof_audit_file": "audits/PROOF_AUDIT.txt", "priority_status": "No per-claim priority certification; see audits/NOVELTY_AUDIT.txt.", "verification_limit": "Manuscript proof status is not formal certification; finite code checks are separate."}
{"claim_id": "thm:kernel", "title": "Kernel transport and rank creation", "environment": "theorem", "component": "area_feedback", "status": "manuscript_proved", "statement_latex": "\\label{thm:kernel}\nLet $a\\in K=\\ker A$, write $x=u+w$ with $u\\in R$ and $w\\in K$, and set $h=(A|_R)^{-1}u$. Let $A'=A+x\\wedge a$.\n\nIf $w,a$ are linearly independent, then\n\\begin{equation}\n \\ker A'=K\\cap w^\\perp\\cap a^\\perp,\n \\qquad \\rank A'=\\rank A+2,\n \\qquad P_{\\ker A'}(x+a)=0.\n \\label{eq:birth}\n\\end{equation}\nIf $w,a$ are linearly dependent, including $a=0$, then\n\\begin{equation}\n \\ker A'=\\{z-h\\inner{a}{z}:z\\in K\\},\n \\qquad \\rank A'=\\rank A.\n \\label{eq:graph}\n\\end{equation}\nIn the second case, $P_{\\ker A'}(x+a)=0$ if and only if $w+a=0$.", "proof_latex": "Relative to $V=R\\oplus K$, the new operator is\n\\begin{equation}\n A'=\\begin{pmatrix}\n B&ua^T\\\\\n -au^T&wa^T-aw^T\n \\end{pmatrix}.\n \\label{eq:block}\n\\end{equation}\nLet $r+z\\in R\\oplus K$ be in its kernel. The upper equation gives\n\\[\n Br+u\\inner{a}{z}=0,\n \\qquad r=-h\\inner{a}{z}.\n\\]\nSince $\\inner{u}{h}=0$, substitution into the lower equation leaves\n\\begin{equation}\n w\\inner{a}{z}-a\\inner{w}{z}=0.\n \\label{eq:lower}\n\\end{equation}\nIf $w,a$ are independent, both coefficients vanish. Then $r=0$ and $z\\in K\\cap a^\\perp\\cap w^\\perp$. These are two independent linear conditions on $K$, proving the rank claim. Also $x+a=u+w+a$ is perpendicular to this kernel.\n\nIf $w,a$ are dependent, $w\\wedge a=0$, so~\\eqref{eq:lower} holds for every $z\\in K$. The displayed graph follows. Its projection to $K$ is the identity, so its dimension is $\\dim K$. Finally, for every graph vector,\n\\[\n \\inner{x+a}{z-h\\inner{a}{z}}=\\inner{w+a}{z},\n\\]\nbecause $u\\perp K$ and $u\\perp h$. This vanishes for every $z$ exactly when $w+a=0$.", "references_in_statement_or_proof": ["eq:lower"], "source_file": "manuscript/Avenyra_Area_Feedback_v1.0.0.tex", "proof_audit_file": "audits/PROOF_AUDIT.txt", "priority_status": "No per-claim priority certification; see audits/NOVELTY_AUDIT.txt.", "verification_limit": "Manuscript proof status is not formal certification; finite code checks are separate."}
{"claim_id": "thm:rho", "title": "Exact update of the exclusion radius", "environment": "theorem", "component": "area_feedback", "status": "manuscript_proved", "statement_latex": "\\label{thm:rho}\nUnder the hypotheses and notation of Theorem~\\ref{thm:kernel}, put $\\rho'=\\rho(x+a,A')$.\nIf $w,a$ are independent, then\n\\begin{equation}\n (\\rho')^2=\\rho^2+\\norm{w+a}^2.\n \\label{eq:rho-birth}\n\\end{equation}\nIf they are dependent, then\n\\begin{equation}\n (\\rho')^2=\\rho^2+\n \\frac{\\norm{h}^2\\inner{a}{w+a}^2}{1+\\norm{h}^2\\norm{a}^2}.\n \\label{eq:rho-graph}\n\\end{equation}\nConsequently $\\rho'\\geq\\rho$ for every legal move.", "proof_latex": "In the independent case, Theorem~\\ref{thm:kernel} says $x+a\\in\\im A'$. Therefore $(\\rho')^2=\\norm{x+a}^2=\\norm{u}^2+\\norm{w+a}^2$.\n\nIn the dependent case, define $J:K\\to V$ by $Jz=z-h\\inner{a}{z}$. Its image is the new kernel. Orthogonality of $K,R$ gives\n\\[\n J^*J=I_K+\\norm{h}^2aa^T,\n \\qquad J^*(x+a)=w+a.\n\\]\nWrite $v=w+a$ and $c=\\norm{h}^2$. The squared norm of the projection of $x+a$ onto $\\im J$ is\n\\[\n v^T(I_K+caa^T)^{-1}v.\n\\]\nThe inverse identity\n\\[\n (I_K+caa^T)^{-1}=I_K-\n \\frac{caa^T}{1+c\\norm{a}^2}\n\\]\nis verified by multiplication; its denominator is positive, including when $a=0$. The projection norm squared is thus\n\\[\n \\norm{v}^2-\\frac{c\\inner{a}{v}^2}{1+c\\norm{a}^2}.\n\\]\nSubtract this from $\\norm{x+a}^2=\\norm{u}^2+\\norm{v}^2$ to obtain~\\eqref{eq:rho-graph}. Both increments are nonnegative.", "references_in_statement_or_proof": ["eq:rho-graph", "thm:kernel"], "source_file": "manuscript/Avenyra_Area_Feedback_v1.0.0.tex", "proof_audit_file": "audits/PROOF_AUDIT.txt", "priority_status": "No per-claim priority certification; see audits/NOVELTY_AUDIT.txt.", "verification_limit": "Manuscript proof status is not formal certification; finite code checks are separate."}
{"claim_id": "thm:minimum", "title": "Global optimal endpoint and uniqueness", "environment": "theorem", "component": "area_feedback", "status": "manuscript_proved", "statement_latex": "\\label{thm:minimum}\nFor every state $(x,A)$, with $u=P_{\\im A}x$ and $w=P_{\\ker A}x$, equation~\\eqref{eq:main} holds. The endpoint $u$ is reached by the legal move $a=-w$. If $u\\neq0$, every nonzero move preserving $\\rho$ must be exactly $-w$; after that move every further nonzero legal move strictly raises $\\rho$.", "proof_latex": "For any finite legal trajectory, Theorem~\\ref{thm:rho} gives $\\rho(s_n)\\geq\\rho(s_0)$. Every endpoint satisfies $\\norm{x_n}\\geq\\rho(s_n)$ because orthogonal projection cannot increase norm. Thus no reachable endpoint has norm below $\\rho(s_0)$.\n\nThe vector $a=-w$ belongs to $K$, yields $x'=u$, and is in the dependent case of Theorem~\\ref{thm:kernel}. Since $w+a=0$, it also yields $P_{\\ker A'}u=0$, so $\\rho'=\\norm{u}=\\rho$. This proves attainment.\n\nIf $u=0$, the only vector of minimum norm zero is the zero vector, which has just been reached. Suppose $u\\neq0$. Then $h\\neq0$. An independent move strictly raises $\\rho$ because independent $w,a$ cannot have $w+a=0$. For a nonzero dependent move, $w+a$ is a scalar multiple of $a$, and~\\eqref{eq:rho-graph} vanishes exactly when $w+a=0$. Hence the only nonzero move preserving $\\rho$ is $a=-w$. At its endpoint the new $w$ is zero and the new $u$ remains nonzero. Every later nonzero $a$ then has $\\inner{a}{w+a}=\\norm{a}^2>0$ and strictly raises $\\rho$. A trajectory attaining the initial minimum can therefore contain only zero moves, possibly one projection move, and then zero moves. Its endpoint is necessarily $u$.", "references_in_statement_or_proof": ["eq:main", "eq:rho-graph", "thm:kernel", "thm:rho"], "source_file": "manuscript/Avenyra_Area_Feedback_v1.0.0.tex", "proof_audit_file": "audits/PROOF_AUDIT.txt", "priority_status": "No per-claim priority certification; see audits/NOVELTY_AUDIT.txt.", "verification_limit": "Manuscript proof status is not formal certification; finite code checks are separate."}
{"claim_id": "cor:origin", "title": "Exact origin test", "environment": "corollary", "component": "area_feedback", "status": "manuscript_proved", "statement_latex": "\\label{cor:origin}\nFrom an arbitrary state $(x,A)$, the origin is reachable if and only if $Ax=0$. If it is reachable, one move $a=-x$ reaches it.", "proof_latex": "Theorem~\\ref{thm:minimum} makes origin reachability equivalent to $P_{\\im A}x=0$, which by Lemma~\\ref{lem:decomposition} is $x\\in\\ker A$. The stated move is then legal, and $x\\wedge(-x)=0$.", "references_in_statement_or_proof": ["lem:decomposition", "thm:minimum"], "source_file": "manuscript/Avenyra_Area_Feedback_v1.0.0.tex", "proof_audit_file": "audits/PROOF_AUDIT.txt", "priority_status": "No per-claim priority certification; see audits/NOVELTY_AUDIT.txt.", "verification_limit": "Manuscript proof status is not formal certification; finite code checks are separate."}
{"claim_id": "cor:closed", "title": "Closed-loop rigidity from zero memory", "environment": "corollary", "component": "area_feedback", "status": "manuscript_proved", "statement_latex": "\\label{cor:closed}\nFor a legal history starting at $(0,0)$, if its final endpoint is zero, then every area tensor along the history is zero. More strongly, if $j$ is its first step with $x_{j-1}\\wedge a_j\\neq0$, then\n\\[\n \\norm{x_n}\\geq\\rho(s_n)\\geq\\norm{x_j}>0\n \\qquad\\text{for every }n\\geq j.\n\\]", "proof_latex": "Before the first nonzero area increment, $A_{j-1}=0$. Thus $R=0$, $w=x_{j-1}$, and the nonzero wedge means $w,a_j$ are independent. Formula~\\eqref{eq:rho-birth} gives $\\rho(s_j)=\\norm{x_j}>0$. Monotonicity proves the inequality and rules out a later zero endpoint. If no such first increment exists, every $A_j$ is zero.", "references_in_statement_or_proof": ["eq:rho-birth"], "source_file": "manuscript/Avenyra_Area_Feedback_v1.0.0.tex", "proof_audit_file": "audits/PROOF_AUDIT.txt", "priority_status": "No per-claim priority certification; see audits/NOVELTY_AUDIT.txt.", "verification_limit": "Manuscript proof status is not formal certification; finite code checks are separate."}
{"claim_id": "thm:exterior", "title": "Orthogonal exterior increments", "environment": "theorem", "component": "area_feedback", "status": "manuscript_proved", "statement_latex": "\\label{thm:exterior}\nEvery legal update satisfies, for each integer $j\\geq1$,\n\\begin{equation}\n E_j(A')=E_j(A)+\n \\norm{\\frac{\\alpha^{\\wedge(j-1)}}{(j-1)!}\\wedge x\\wedge a}^2.\n \\label{eq:exterior}\n\\end{equation}\nAll $E_j$ are therefore nondecreasing along legal histories.", "proof_latex": "The bivector $x\\wedge a$ is decomposable, so its square under exterior multiplication is zero. Since degree-two elements commute under exterior multiplication,\n\\[\n \\frac{(\\alpha+x\\wedge a)^{\\wedge j}}{j!}\n =\\frac{\\alpha^{\\wedge j}}{j!}\n +\\frac{\\alpha^{\\wedge(j-1)}}{(j-1)!}\\wedge x\\wedge a.\n\\]\nThe tensor $\\alpha$ is supported on $R=\\im A$. If $a=0$, the second summand vanishes. Otherwise choose an orthonormal basis beginning with $a/\\norm a$ and continuing through $R$ and its complement. The first summand has no $a$ component because $a\\perp R$. Every nonzero term of the second summand contains that component. The two summands are orthogonal. Taking squared norms proves~\\eqref{eq:exterior}, also in degrees exceeding $d$.", "references_in_statement_or_proof": ["eq:exterior"], "source_file": "manuscript/Avenyra_Area_Feedback_v1.0.0.tex", "proof_audit_file": "audits/PROOF_AUDIT.txt", "priority_status": "No per-claim priority certification; see audits/NOVELTY_AUDIT.txt.", "verification_limit": "Manuscript proof status is not formal certification; finite code checks are separate."}
{"claim_id": "prop:scalar", "title": "Two scalar identities", "environment": "proposition", "component": "area_feedback", "status": "manuscript_proved", "statement_latex": "\\label{prop:scalar}\nLet $E=E_1(A)$, $N=\\norm{Ax}^2$, and\n\\[\n D=\\norm{x\\wedge a}^2=\\norm{x}^2\\norm{a}^2-\\inner{x}{a}^2.\n\\]\nFor every legal update,\n\\begin{equation}\n E'=E+D,\n \\qquad N'=N+\\norm{x+a}^2D.\n \\label{eq:scalar}\n\\end{equation}\nIf $E>0$, the weaker radius $\\sqrt{N/E}$ is nondecreasing and satisfies $\\sqrt{N/E}\\leq\\rho$.", "proof_latex": "The first identity is the case $j=1$ of Theorem~\\ref{thm:exterior}. Put $X=\\norm{x}^2$, $Y=\\norm{a}^2$, $c=\\inner{x}{a}$. Since $Aa=0$,\n\\[\n A'(x+a)=Ax+x(c+Y)-a(X+c).\n\\]\nThe vector $Ax$ is perpendicular to both $x$ and $a$. The squared norm of the remaining vector is\n\\[\n X(c+Y)^2+Y(X+c)^2-2c(c+Y)(X+c)\n =(XY-c^2)(X+Y+2c).\n\\]\nThis proves the second identity.\n\nFor completeness, a skew-adjoint operator has an orthonormal decomposition into $2\\times2$ skew blocks with nonnegative magnitudes $\\sigma_i$, plus its kernel. This follows by diagonalizing the symmetric operator $-A^2$ and pairing a unit vector $v$ with $Av/\\sigma_i$ in each positive eigenspace. Hence $E=\\sum_i\\sigma_i^2\\geq\\norm{A}_{\\mathrm{op}}^2$, and $\\norm{Ax}^2=\\norm{Au}^2\\leq E\\norm{u}^2$. Also $A(x+a)=Ax$, so $\\norm{x+a}^2\\geq N/E$. Therefore\n\\[\n \\frac{N'}{E'}=\n \\frac{E(N/E)+D\\norm{x+a}^2}{E+D}\\geq\\frac NE.\n\\]", "references_in_statement_or_proof": ["thm:exterior"], "source_file": "manuscript/Avenyra_Area_Feedback_v1.0.0.tex", "proof_audit_file": "audits/PROOF_AUDIT.txt", "priority_status": "No per-claim priority certification; see audits/NOVELTY_AUDIT.txt.", "verification_limit": "Manuscript proof status is not formal certification; finite code checks are separate."}
{"claim_id": "thm:rank", "title": "Minimum steps to maximal rank", "environment": "theorem", "component": "area_feedback", "status": "manuscript_proved", "statement_latex": "\\label{thm:rank}\nLet $k=\\dim\\ker A$, $m=\\lfloor k/2\\rfloor$, and $w=P_{\\ker A}x$. The maximal rank reachable from $(x,A)$ is\n\\[\n \\rank A+2m=2\\lfloor d/2\\rfloor.\n\\]\nIf $m=0$, that rank is already attained. If $m>0$, the minimum number of legal steps needed to attain it is\n\\begin{equation}\n \\begin{cases}\n 2m,&w=0,\\\\\n 2m-1,&w\\neq0.\n \\end{cases}\n \\label{eq:ranksteps}\n\\end{equation}", "proof_latex": "Theorem~\\ref{thm:kernel} permits rank increases only by two, and at most $m$ such increases fit in dimension $d$. A rank increase resets the new kernel projection of the endpoint to zero. From a state with that projection zero, the next move cannot raise rank because its $w$ is zero. Thus at least one intervening move is required between increases; an initial intervening move is also required when the initial $w=0$. This proves the lower bounds.\n\nFor attainment, if $k\\geq2$ and $w\\neq0$, choose $a\\in K\\setminus\\R w$. It raises rank by two and resets $w$ to zero. If $w=0$, first choose any nonzero $a\\in K$. This is a rank-preserving move and has new kernel projection nonzero, by the last assertion of Theorem~\\ref{thm:kernel}. One can then choose a vector independent of that projection and raise rank. Repeat while at least two kernel dimensions remain. This attains exactly the stated number of steps. The parity identity follows because the rank of a real skew operator is even.", "references_in_statement_or_proof": ["thm:kernel"], "source_file": "manuscript/Avenyra_Area_Feedback_v1.0.0.tex", "proof_audit_file": "audits/PROOF_AUDIT.txt", "priority_status": "No per-claim priority certification; see audits/NOVELTY_AUDIT.txt.", "verification_limit": "Manuscript proof status is not formal certification; finite code checks are separate."}
{"claim_id": "cor:parity", "title": "Even and odd ambient dimensions", "environment": "corollary", "component": "area_feedback", "status": "manuscript_proved", "statement_latex": "\\label{cor:parity}\nIn even dimension, a full-rank state permits only the zero move. In odd dimension, every state permits a nonzero move. From $(0,0)$, maximal rank can be reached in $d$ steps for even $d$ and $d-1$ steps for odd $d\\geq1$.", "proof_latex": "The kernel is zero exactly at full rank. An odd-dimensional real skew matrix always has nonzero kernel. Apply Theorem~\\ref{thm:rank} with $k=d$ and $w=0$; the cases $d=0,1$ already have maximal rank zero.", "references_in_statement_or_proof": ["thm:rank"], "source_file": "manuscript/Avenyra_Area_Feedback_v1.0.0.tex", "proof_audit_file": "audits/PROOF_AUDIT.txt", "priority_status": "No per-claim priority certification; see audits/NOVELTY_AUDIT.txt.", "verification_limit": "Manuscript proof status is not formal certification; finite code checks are separate."}
{"claim_id": "thm:launch", "title": "Orthogonal launch family", "environment": "theorem", "component": "area_feedback", "status": "manuscript_proved", "statement_latex": "\\label{thm:launch}\nLet $e_1,\\ldots,e_d$ be orthonormal. The prefix $e_1,\\ldots,e_n$, $0\\leq n\\leq d$, is legal from zero and yields\n\\[\n x_n=\\sum_{i=1}^n e_i,\\qquad\n A_n=\\sum_{1\\leq i<j\\leq n}e_i\\wedge e_j.\n\\]\nIts rank is $2\\lfloor n/2\\rfloor$. For $n\\geq1$,\n\\begin{equation}\n \\rho_n^2=\\begin{cases}n,&n\\text{ even},\\\\ n-1/n,&n\\text{ odd},\\end{cases}\n \\qquad E_j(A_n)=\\binom{n}{2j}.\n \\label{eq:launch}\n\\end{equation}\nFor $n=0$, $\\rho_0=0$. Unused coordinate directions remain in the kernel. In the occupied $n$ coordinates, the characteristic polynomial is\n\\begin{equation}\n \\det(tI-A_n)=\\sum_{j=0}^{\\lfloor n/2\\rfloor}\\binom{n}{2j}t^{n-2j}\n =\\frac{(t+1)^n+(t-1)^n}{2}.\n \\label{eq:launchpoly}\n\\end{equation}", "proof_latex": "Before step $i$, the area is supported on earlier coordinates, so it annihilates $e_i$. Formula~\\eqref{eq:area} gives the state. In the occupied coordinates the matrix has every upper-triangular entry one. If it annihilates a vector $v$, subtracting successive row equations gives $v_{i+1}=-v_i$. The first row then forces $v_1=0$ when $n$ is even, and imposes no additional condition when $n$ is odd. In the latter case the kernel is spanned by $k_n=(1,-1,\\ldots,1)$, with $\\norm{k_n}^2=n$ and $\\inner{x_n}{k_n}=1$. The rank and projection formulas follow.\n\nThe coefficient of a basis $2j$-wedge in $\\alpha_n^{\\wedge j}/j!$ is the Pfaffian of the corresponding principal submatrix. Every such submatrix is the same upper-one skew matrix of size $2j$. Its Pfaffian is one: expansion along its first row gives the previous Pfaffian multiplied by $\\sum_{i=2}^{2j}(-1)^i=1$, starting with the empty Pfaffian one. There are $\\binom{n}{2j}$ principal subsets. Their squared coefficients sum to that number, proving the exterior formula. Expanding $\\det(tI-A_n)$ by principal minors, whose even determinants are these squared Pfaffians and whose odd determinants vanish, proves the first polynomial identity. The binomial theorem proves the second.", "references_in_statement_or_proof": ["eq:area"], "source_file": "manuscript/Avenyra_Area_Feedback_v1.0.0.tex", "proof_audit_file": "audits/PROOF_AUDIT.txt", "priority_status": "No per-claim priority certification; see audits/NOVELTY_AUDIT.txt.", "verification_limit": "Manuscript proof status is not formal certification; finite code checks are separate."}
{"claim_id": "thm:accessibility", "title": "Three steps are sufficient and necessary for endpoint interior at a specified state", "environment": "theorem", "component": "area_feedback", "status": "manuscript_proved", "statement_latex": "\\label{thm:accessibility}\nStart in $\\R^3$ at $x=(1,1,0)^T$, $k=(0,0,1)^T$, equivalently $A=e_1\\wedge e_2$. The set of endpoints reachable in three legal moves has nonempty interior in $\\R^3$. The union of endpoints reachable in at most two moves has empty interior.", "proof_latex": "Use scalar controls $p,q,r$ in~\\eqref{eq:axial}. All are legal and all intermediate $k$ remain nonzero. Direct substitution gives the polynomial endpoint map\n\\begin{equation}\n F(p,q,r)=\n \\begin{pmatrix}\n 1+pq+pr+qr+p^2qr\\\\\n 1-pq-pr-qr+p^2qr\\\\\n p+q+r-2pqr\n \\end{pmatrix}.\n \\label{eq:endpointpoly}\n\\end{equation}\nFor clarity, after the first two steps the endpoint is\n\\[\n (1+pq,1-pq,p+q)^T,\n\\]\nand the axial vector is\n\\[\n (p+q+p^2q,-p-q+p^2q,1-2pq)^T.\n\\]\nAdding $r$ times the latter vector proves~\\eqref{eq:endpointpoly}. Differentiation and expansion of a $3\\times3$ determinant give\n\\[\n \\det DF=2p(q-r)(2p^2qr+p^2-pq-pr+2qr).\n\\]\nAt $(p,q,r)=(1,1,2)$ the endpoint is $(8,-2,0)^T$ and the determinant is $-12$. The finite-dimensional inverse function theorem therefore gives an open neighborhood of this endpoint contained in the three-step reachable set.\n\nAfter at most two steps, every endpoint satisfies $x_1+x_2=2$, as seen in the displayed two-step formula and its one-step and empty-word cases. This plane has empty interior in $\\R^3$.", "references_in_statement_or_proof": ["eq:axial", "eq:endpointpoly"], "source_file": "manuscript/Avenyra_Area_Feedback_v1.0.0.tex", "proof_audit_file": "audits/PROOF_AUDIT.txt", "priority_status": "No per-claim priority certification; see audits/NOVELTY_AUDIT.txt.", "verification_limit": "Manuscript proof status is not formal certification; finite code checks are separate."}
{"claim_id": "thm:accessibility-general", "title": "Three-dimensional endpoint accessibility classification", "environment": "theorem", "component": "area_feedback", "status": "manuscript_proved", "statement_latex": "\\label{thm:accessibility-general}\nLet $V=\\R^3$ and $A\\neq0$. The finite reachable endpoint set has nonempty interior if and only if $Ax\\neq0$. Whenever $Ax\\neq0$, the exact minimum number of steps needed for an endpoint set with nonempty interior is three.", "proof_latex": "If $Ax=0$, then $x\\in\\ker A$, which is a line. Every legal input lies on that line, writes zero wedge, and leaves the kernel fixed. All endpoints remain on the line, so the reachable set has empty interior.\n\nIf $Ax\\neq0$, let $k=k(A)$. An orientation-preserving orthogonal change of coordinates puts the state in the form\n\\[\n x=(b,0,z)^T,\\qquad k=(0,0,c)^T,\n \\qquad b,c>0,\\ z\\in\\R.\n\\]\nThree scalar controls $p,q,r$ in~\\eqref{eq:axial} give\n\\[\n F_{b,c,z}(p,q,r)=\n \\begin{pmatrix}\n b+bc^2p^2qr+bczpqr\\\\\n -bc(pq+pr+qr)\\\\\n z+c(p+q+r-b^2pqr)\n \\end{pmatrix}.\n\\]\nThis follows by substituting the first two steps, whose endpoint is $(b,-bcpq,z+c(p+q))^T$. At $(1,1,2)$ the Jacobian is\n\\[\n \\begin{pmatrix}\n 2bc(2c+z)&2bc(c+z)&bc(c+z)\\\\\n -3bc&-3bc&-2bc\\\\\n c(1-2b^2)&c(1-2b^2)&c(1-b^2)\n \\end{pmatrix}.\n\\]\nSubtract the second column from the first. The first column becomes $(2bc^2,0,0)^T$, and the remaining $2\\times2$ determinant is $-bc^2(1+b^2)$. Hence\n\\[\n \\det DF_{b,c,z}(1,1,2)=-2b^2c^4(1+b^2)\\neq0.\n\\]\nThe inverse function theorem gives endpoint interior after three steps. After at most two steps, the first coordinate of every endpoint remains $b$, so those endpoints lie in a plane and have empty interior. Orthogonal covariance transports both statements back to the original coordinates.", "references_in_statement_or_proof": ["eq:axial"], "source_file": "manuscript/Avenyra_Area_Feedback_v1.0.0.tex", "proof_audit_file": "audits/PROOF_AUDIT.txt", "priority_status": "No per-claim priority certification; see audits/NOVELTY_AUDIT.txt.", "verification_limit": "Manuscript proof status is not formal certification; finite code checks are separate."}
{"claim_id": "thm:dormant", "title": "Invariant inaccessible summand", "environment": "theorem", "component": "area_feedback", "status": "manuscript_proved", "statement_latex": "\\label{thm:dormant}\nFor a state $(x,A)$, let $K=\\ker A$, $u=P_{\\im A}x$, and define\n\\[\n C=\\Span\\{A^j u:j\\geq0\\},\\qquad S=K\\oplus C,\n \\qquad H=S^\\perp.\n\\]\nThen $S,H$ are $A$-invariant and $A|_H$ is invertible. Every future legal input and every future endpoint lies in $S$; the block $A|_H$ never changes. Replacing it by any invertible skew-adjoint operator on $H$ leaves all future legal input words and endpoints unchanged.\n\nThe subspace $S$ is the smallest $A$-invariant subspace containing $x$ and $K$. This does not assert that it is the smallest behavioral realization.", "proof_latex": "The cyclic subspace $C$ is contained in $\\im A$ and is $A$-invariant. Indeed, the span stabilizes in finite dimension, and multiplication by $A$ shifts each spanning vector to the next. Since $A$ is skew-adjoint, the orthogonal complement of an invariant subspace is invariant. Thus $S$ and $H$ are invariant. Also $H\\subseteq K^\\perp$, so the restriction to $H$ has zero kernel and is invertible.\n\nInitially $x\\in S$ and $\\ker A\\subseteq S$. Suppose at some step the endpoint lies in $S$, the area is block diagonal on $S\\oplus H$, and its block on $H$ remains invertible. Any legal input then belongs to $S$. The wedge of two vectors in $S$ acts only on $S$, so the update preserves this block structure and leaves the $H$ block fixed. Induction proves the trajectory assertions. The same induction compares any other invertible skew block on $H$, since an invertible block contributes no kernel vectors.\n\nFinally, any invariant subspace containing $x$ and $K$ contains $u=x-P_Kx$, and then every $A^ju$. It therefore contains $S$, proving the stated minimality.", "references_in_statement_or_proof": [], "source_file": "manuscript/Avenyra_Area_Feedback_v1.0.0.tex", "proof_audit_file": "audits/PROOF_AUDIT.txt", "priority_status": "No per-claim priority certification; see audits/NOVELTY_AUDIT.txt.", "verification_limit": "Manuscript proof status is not formal certification; finite code checks are separate."}
{"claim_id": "prop:subdivision", "title": "Reversal and subdivision", "environment": "proposition", "component": "area_feedback", "status": "manuscript_proved", "statement_latex": "\\label{prop:subdivision}\nLet $a$ be legal at $(x,A)$. The algebraic inverse displacement $-a$ is legal immediately after that move if and only if $x\\wedge a=0$. If $x\\wedge a\\neq0$ and $m\\geq2$ is an integer, replacing $a$ by $m$ successive copies of $a/m$ fails legality at the second copy.", "proof_latex": "Since $Aa=0$,\n\\[\n (A+x\\wedge a)a=\\norm{a}^2x-\\inner{x}{a}a.\n\\]\nIf $a\\neq0$, this is zero exactly when $x$ is parallel to $a$, equivalent to $x\\wedge a=0$. If $a=0$, both assertions hold trivially. When the reverse move is legal, substitution shows that it restores both $x$ and $A$.\n\nAfter the first fractional move the area is $A+x\\wedge(a/m)$. Applying it to the second proposed $a/m$ gives\n\\[\n \\frac1{m^2}\\bigl(\\norm{a}^2x-\\inner{x}{a}a\\bigr),\n\\]\nwhich is nonzero when $x\\wedge a\\neq0$.", "references_in_statement_or_proof": [], "source_file": "manuscript/Avenyra_Area_Feedback_v1.0.0.tex", "proof_audit_file": "audits/PROOF_AUDIT.txt", "priority_status": "No per-claim priority certification; see audits/NOVELTY_AUDIT.txt.", "verification_limit": "Manuscript proof status is not formal certification; finite code checks are separate."}
{"claim_id": "thm:continuous", "title": "Continuous collapse and a quantitative refinement bound", "environment": "theorem", "component": "area_feedback", "status": "manuscript_proved", "statement_latex": "\\label{thm:continuous}\nSuppose $x,A$ are absolutely continuous on a compact interval, $A(t)$ is skew-adjoint, and\n\\[\n \\dot x=v,\\qquad \\dot A=x\\wedge v,\\qquad Av=0\n\\]\nhold almost everywhere. If $A(0)=0$, then $A(t)=0$ everywhere and $x\\wedge\\dot x=0$ almost everywhere.\n\nFor a finite legal discrete trajectory with $A_0=0$, set\n\\[\n M=\\max_{0\\leq i\\leq n}\\norm{x_i},\\qquad\n L=\\sum_{i=1}^n\\norm{a_i},\\qquad\n \\delta=\\max_{1\\leq i\\leq n}\\norm{a_i},\n\\]\nwith $\\delta=0$ for the empty word. Then\n\\begin{equation}\n E_1(A_n)\\leq M^2L\\delta.\n \\label{eq:refinement}\n\\end{equation}\nThus a family with bounded $M,L$ and $\\delta\\to0$ has area tending to zero at every prefix, uniformly within each trajectory.", "proof_latex": "Absolute continuity permits the chain rule for the quadratic function $\\norm A_F^2/2$. Its derivative is $\\inner{A}{x\\wedge v}_F=2\\inner{x}{Av}=0$ almost everywhere. Hence it remains zero, and positive definiteness forces $A=0$. The area differential equation then gives $x\\wedge\\dot x=0$.\n\nIn the discrete case, sum the exact $E_1$ increments from~\\eqref{eq:scalar}. Each is at most $\\norm{x_{i-1}}^2\\norm{a_i}^2$. Bound the first factor by $M^2$ and $\\norm{a_i}^2$ by $\\delta\\norm{a_i}$, obtaining~\\eqref{eq:refinement}. The same upper bound applies to every prefix because every summand is nonnegative.", "references_in_statement_or_proof": ["eq:refinement", "eq:scalar"], "source_file": "manuscript/Avenyra_Area_Feedback_v1.0.0.tex", "proof_audit_file": "audits/PROOF_AUDIT.txt", "priority_status": "No per-claim priority certification; see audits/NOVELTY_AUDIT.txt.", "verification_limit": "Manuscript proof status is not formal certification; finite code checks are separate."}
{"claim_id": "prop:closure", "title": "A nonclosed endpoint set", "environment": "proposition", "component": "area_feedback", "status": "manuscript_proved", "statement_latex": "\\label{prop:closure}\nFor the state $x_0=(1,1,0)^T$, $A_0=e_1\\wedge e_2$ in $\\R^3$, the closure of $\\Reach(x_0,A_0)$ contains points other than $x_0$ of norm $\\sqrt2$. None of those other points is reachable by a finite legal word. Consequently this reachable endpoint set is not closed.", "proof_latex": "Consider the polynomial ordinary differential equation\n\\[\n \\dot x=k,\\qquad \\dot k=x\\times k,\n \\qquad x(0)=x_0,\\ k(0)=e_3.\n\\]\nIt has a smooth solution on a sufficiently small interval. Direct differentiation yields\n\\[\n \\norm{k(t)}^2=1,\\qquad x(t)\\cdot k(t)=t,\n \\qquad \\norm{x(t)}^2=2+t^2.\n\\]\nDefine $u(t)=x(t)-t k(t)$. Then $\\norm{u(t)}^2=2$, while\n\\[\n u'(t)=-t\\dot k(t),\\qquad\n u''(0)=-x_0\\times e_3=(-1,1,0)^T.\n\\]\nThus $u(T)\\neq x_0$ for every sufficiently small positive $T$.\n\nFor each positive integer $n$, take $n$ steps of~\\eqref{eq:axial} with scalar $T/n$. These are finite legal histories and their axial vectors stay nonzero. Their endpoints $(x^{(n)},k^{(n)})$ converge to $(x(T),k(T))$. Here is the relevant elementary convergence argument. On a compact tube around the short smooth solution, the polynomial vector field has a Lipschitz constant $C_1$ and bounded derivative along the solution. A one-step Taylor estimate gives local error at most $C_2h^2$, with $h=T/n$. The error recurrence\n\\[\n e_{j+1}\\leq(1+C_1h)e_j+C_2h^2,\\qquad e_0=0,\n\\]\ngives $e_n\\leq C_2h(e^{C_1T}-1)/C_1$ when $C_1>0$, or $e_n\\leq C_2Th$ when $C_1=0$. For large $n$ this estimate also keeps the approximations inside the chosen tube, closing the induction.\n\nAppend to each discrete history the exact projection move along its current kernel. The final endpoint becomes\n\\[\n y_n=x^{(n)}-\n \\frac{x^{(n)}\\cdot k^{(n)}}{\\norm{k^{(n)}}^2}k^{(n)}.\n\\]\nContinuity and $\\norm{k(T)}=1$ imply $y_n\\to u(T)$. Hence $u(T)$ lies in the closure. At the initial state, $\\rho=\\sqrt2$ and the unique finite-word minimizer is $x_0$, by Theorem~\\ref{thm:minimum}. Since $u(T)\\neq x_0$ has the same norm, it is not finitely reachable.", "references_in_statement_or_proof": ["eq:axial", "thm:minimum"], "source_file": "manuscript/Avenyra_Area_Feedback_v1.0.0.tex", "proof_audit_file": "audits/PROOF_AUDIT.txt", "priority_status": "No per-claim priority certification; see audits/NOVELTY_AUDIT.txt.", "verification_limit": "Manuscript proof status is not formal certification; finite code checks are separate."}
{"claim_id": "conj:interior", "title": "Interior for cyclic states in odd dimension", "environment": "conjecture", "component": "area_feedback", "status": "conjecture", "statement_latex": "\\label{conj:interior}\nLet $d\\geq5$ be odd, let $\\rank A=d-1$, and suppose\n\\[\n \\ker A\\oplus\\Span\\{A^jP_{\\im A}x:j\\geq0\\}=V.\n\\]\nThen the finite reachable endpoint set has nonempty interior in $V$.", "proof_latex": "", "references_in_statement_or_proof": [], "source_file": "manuscript/Avenyra_Area_Feedback_v1.0.0.tex", "proof_audit_file": "audits/PROOF_AUDIT.txt", "priority_status": "No per-claim priority certification; see audits/NOVELTY_AUDIT.txt.", "verification_limit": "Manuscript proof status is not formal certification; finite code checks are separate."}
{"claim_id": "thm:alt-interval", "title": "Interval update and terminal formula", "environment": "theorem", "component": "algebraic_alternative", "status": "manuscript_proved", "statement_latex": "\\label{thm:alt-interval}\nFor $1\\leq m\\leq N-1$,\n\\begin{equation}\n D(U_m)=U_{\\min(m,N-m-1)},\\qquad\n T(U_m)=U_{\\max\\{m,\\,2\\min(m,N-m-1)\\}}.\n \\label{eq:alt-interval-step}\n\\end{equation}\nPut $n=N-1$, let $r$ be the smallest nonnegative integer for which\n$2^rm\\geq n/2$, and set $a=2^rm$. Then\n\\begin{equation}\n F(U_m)=U_{\\max\\{a,\\,2(n-a)\\}}.\n \\label{eq:alt-interval-final}\n\\end{equation}\nThe nonzero intervals with $m\\leq N-1$ are terminal exactly when\n$3m\\geq2(N-1)$. The boundary states $U_0$ and $U_N$ are terminal.", "proof_latex": "Each $t^i\\in U_m$ has only one possible nonzero pairing with a basis element,\nnamely $t^{N-i}$. That partner is absent from $U_m$ exactly when\n$i\\leq N-m-1$. Distinct exponents have distinct partners, so there is no\nadditional cancellation between basis columns. This proves the radical formula.\nIf $d=\\min(m,N-m-1)$, products in $U_d^2$ span exactly the exponents\n$2,\\ldots,2d$ when $d>0$, and give zero when $d=0$. Since $2d\\leq N-1$,\nno product in this range is truncated. Adding $U_m$ proves the update formula.\n\nWhile $m<n/2$, the endpoint of the interval doubles. At its first value\n$a\\geq n/2$, the update becomes $a\\mapsto\\max\\{a,2(n-a)\\}$.\nIf $a\\geq2n/3$, it is already fixed. Otherwise the next value $2(n-a)$ is\nstrictly greater than $2n/3$ and is fixed. This proves both the terminal formula\nand the fixed-interval criterion. The zero space is fixed. Finally $U_N$ is\na subalgebra, and is therefore fixed; equivalently, its radical is $kt^N$,\nwhose square is zero.", "references_in_statement_or_proof": [], "source_file": "manuscript/Avenyra_Area_Feedback_v1.0.0.tex", "proof_audit_file": "audits/PROOF_AUDIT.txt", "priority_status": "No per-claim priority certification; see audits/NOVELTY_AUDIT.txt.", "verification_limit": "Manuscript proof status is not formal certification; finite code checks are separate."}
{"claim_id": "thm:alt-tensor", "title": "An exact tensor activation law", "environment": "theorem", "component": "algebraic_alternative", "status": "manuscript_proved", "statement_latex": "\\label{thm:alt-tensor}\nLet $B=k[t]/(t^5)$, $\\tau=[t^4]$, and $U=\\operatorname{span}\\{t,t^2\\}$.\nThe state $U$ is terminal. For an integer $r\\geq1$, put\n\\[\n W_r=U^{\\otimes r}\\subset\n B^{\\otimes r}=k[x_1,\\ldots,x_r]/(x_1^5,\\ldots,x_r^5),\n\\]\nwith the tensor trace extracting $x_1^4\\cdots x_r^4$. Then\n\\begin{equation}\n F(W_r)=T(W_r),\\qquad\n \\dim F(W_r)=2^r+3^r-r-2.\n \\label{eq:alt-tensor}\n\\end{equation}\nFor $r\\geq2$, its terminal radical has dimension\n$3^r-2^r-r-1$ and has square zero.", "proof_latex": "On $U$, the only nonzero basis pairing is $\\beta(t^2,t^2)=1$.\nThus $D(U)=kt$ and $D(U)^2=kt^2\\subset U$, proving terminality.\n\nThe initial monomial exponent set of $W_r$ is $S_r=\\{1,2\\}^r$.\nAn initial monomial can pair with another initial monomial only when both\nhave every exponent equal to two. Hence $D(W_r)$ contains every initial\nbasis monomial except that all-two monomial.\n\nA sum of two radical exponent vectors lies in $\\{2,3,4\\}^r$. If a coordinate\nof the sum is two, both factors have a one there and are radical. If there is\nno two, coordinates four force the split $2+2$, while coordinates three permit\n$1+2$ or $2+1$. Both factors can be radical exactly when there are at least\ntwo coordinates three. Therefore the product exponent set is\n\\[\n P_r=\\{c\\in\\{2,3,4\\}^r:\n       c\\text{ has a two or at least two threes}\\}.\n\\]\nIts complement consists of the all-four vector and the $r$ vectors having\none three and otherwise fours. Thus $|P_r|=3^r-r-1$.\nThe intersection $S_r\\cap P_r$ consists only of the all-two vector.\nDistinct surviving monomials are linearly independent, so\n$\\dim T(W_r)=2^r+(3^r-r-1)-1$, as claimed.\n\nIt remains to prove terminality. The case $r=1$ was checked above.\nFor $r\\geq2$, every element of $S_r$ has its complementary exponent vector\n$4-c$ in $P_r$: the complement has a two, or it is all threes. Every vector\nin $P_r$ without a four has its complement in $S_r$. A vector in $P_r$ with\na four has a complementary zero coordinate, which occurs in neither set.\nThe new radical therefore consists exactly of product monomials having a\nfour. Every coordinate of every product monomial is at least two; multiplying\ntwo new radical monomials exceeds exponent four in a coordinate and gives zero.\nThus the new radical has square zero, and the first update is terminal.\nThere are $2^r$ vectors in $P_r$ without a four when $r\\geq2$, giving the\nstated radical dimension.", "references_in_statement_or_proof": [], "source_file": "manuscript/Avenyra_Area_Feedback_v1.0.0.tex", "proof_audit_file": "audits/PROOF_AUDIT.txt", "priority_status": "No per-claim priority certification; see audits/NOVELTY_AUDIT.txt.", "verification_limit": "Manuscript proof status is not formal certification; finite code checks are separate."}
{"claim_id": "thm:spectral-rank-one", "title": "Positive rank-one spectral representation", "environment": "theorem", "component": "spectral_appendix", "status": "manuscript_proved", "statement_latex": "\\label{thm:spectral-rank-one}\nFor $A'=A+x\\wedge a$, the positive squared skew frequencies of $A'$,\ncounting multiplicities, are as follows.\n\\begin{enumerate}\n\\item If $D_0=0$, they are the eigenvalues of\n\\[\n H^+=H+zz^T.\n \\tag{S2}\n\\]\nThe rank of $A'$ is $2m$, and its kernel has dimension $k$.\n\\item If $D_0>0$, define\n\\[\n \\widetilde H=\n \\begin{pmatrix}H&0\\\\0&0\\end{pmatrix},\n \\qquad\n \\widetilde z=\n \\begin{pmatrix}z\\\\\\sqrt{D_0}\\end{pmatrix}.\n\\]\nThe positive squared skew frequencies are the eigenvalues of\n\\[\n H^+=\\widetilde H+\\widetilde z\\widetilde z^T.\n \\tag{S3}\n\\]\nThe rank of $A'$ is $2m+2$, and its kernel has dimension $k-2$.\n\\end{enumerate}\nIn both cases the displayed matrix $H^+$ is positive definite on its\nstated dimension, with the zero-dimensional case interpreted as above.", "proof_latex": "For $t>0$, the matrix $M=tI-A$ is invertible, because the eigenvalues of\na real skew-adjoint operator are zero or purely imaginary. Write\n\\[\n \\beta=\\norm a^2,\\qquad \\gamma=\\inner x a=\\inner w a,\\qquad\n f(t)=x^TM^{-1}x.\n\\]\nThe admissibility condition $Aa=0$, together with skew-adjointness,\ngives both\n\\[\n M^{-1}a=\\frac at,\\qquad a^TM^{-1}=\\frac{a^T}{t}.\n \\tag{S4}\n\\]\nThe update can be written as a rank-two perturbation:\n\\[\n tI-A'=M+[-x,\\ a]\\,[a,\\ x]^T.\n\\]\nThe matrix determinant lemma\n$\\det(M+UV^T)=\\det M\\det(I+V^TM^{-1}U)$ now gives\n\\[\n\\begin{split}\n \\frac{\\det(tI-A')}{\\det(tI-A)}\n &=\\det\n \\begin{pmatrix}\n 1-\\gamma/t&\\beta/t\\\\\n -f(t)&1+\\gamma/t\n \\end{pmatrix}\\\\\n &=1-\\frac{\\gamma^2}{t^2}+\\frac{\\beta f(t)}{t}.\n\\end{split}\n\\tag{S5}\n\\]\nThere is no symmetry assumption on $M^{-1}$ in this calculation; the\ntwo identities in~(S4) supply the two off-diagonal contractions.\n\nOn $K$, $M^{-1}$ is $I/t$. On the plane $E_i$,\n\\[\n (tI-A|_{E_i})^{-1}\n =\\frac{tI+A|_{E_i}}{t^2+\\sigma_i^2}.\n\\]\nIts skew part makes no contribution to a quadratic form. Orthogonality\nof the invariant decomposition therefore yields\n\\[\n f(t)=\\frac{\\norm w^2}{t}\n      +\\sum_{i=1}^m\n        \\frac{t\\norm{u_i}^2}{t^2+\\sigma_i^2}.\n \\tag{S6}\n\\]\nSubstituting into~(S5), using $z_i^2=\\beta\\norm{u_i}^2$, gives the exact\nrational identity\n\\[\n \\boxed{\\quad\n \\frac{\\det(tI-A')}{\\det(tI-A)}\n =1+\\frac{D_0}{t^2}\n  +\\sum_{i=1}^m\\frac{z_i^2}{t^2+\\sigma_i^2}.\n \\quad}\n \\tag{S7}\n\\]\nAll quantities have been defined for $t>0$, where no denominator\nvanishes.\n\nThe original characteristic polynomial is\n\\[\n \\det(tI-A)=t^k\\det(t^2I_m+H).\n \\tag{S8}\n\\]\nIf $D_0=0$, applying the rank-one determinant lemma to\n$t^2I_m+H+zz^T$ transforms~(S7)--(S8) into\n\\[\n \\det(tI-A')=t^k\\det(t^2I_m+H+zz^T).\n \\tag{S9}\n\\]\nFor $m>0$, $H$ is positive definite, and adding $zz^T$ preserves positive\ndefiniteness. Thus the determinant on the right has exactly $m$ positive\nsquared frequencies and no additional zero root. If $m=0$, its determinant\nis one. In either case the zero root has multiplicity exactly $k$.\nThis proves the first case.\n\nIf $D_0>0$, apply the same lemma in dimension $m+1$:\n\\[\n\\begin{split}\n&\\det(t^2I_{m+1}+\\widetilde H+\\widetilde z\\widetilde z^T)\\\\\n&\\qquad=\nt^2\\det(t^2I_m+H)\n\\left(1+\\frac{D_0}{t^2}\n      +\\sum_{i=1}^m\\frac{z_i^2}{t^2+\\sigma_i^2}\\right).\n\\end{split}\n\\]\nSince $k\\geq2$, equations~(S7)--(S8) imply\n\\[\n \\det(tI-A')\n =t^{k-2}\n  \\det(t^2I_{m+1}+\\widetilde H+\\widetilde z\\widetilde z^T).\n \\tag{S10}\n\\]\nFor a vector $(v,s)\\in\\mathbb R^m\\oplus\\mathbb R$,\n\\[\n (v,s)^T H^+(v,s)\n =v^THv+\\bigl(z^Tv+\\sqrt{D_0}\\,s\\bigr)^2.\n\\]\nIf this is zero, positive definiteness of $H$ forces $v=0$, and\n$D_0>0$ then forces $s=0$. The same argument covers $m=0$ directly.\nThus $H^+$ is positive definite, and~(S10) gives exactly $m+1$\npositive squared frequencies and a zero root of multiplicity $k-2$.\n\nThe equalities have been proved for every positive real $t$. Both sides\nof~(S9) and~(S10) are polynomials, so they are identical polynomials.\nThey consequently determine the complete characteristic polynomial,\nnot merely its values on the positive real axis. Real skew-adjoint\nnormal form then identifies the positive squared frequencies with the\neigenvalues of $H^+$ as claimed.", "references_in_statement_or_proof": [], "source_file": "manuscript/Avenyra_Area_Feedback_v1.0.0.tex", "proof_audit_file": "audits/PROOF_AUDIT.txt", "priority_status": "No per-claim priority certification; see audits/NOVELTY_AUDIT.txt.", "verification_limit": "Manuscript proof status is not formal certification; finite code checks are separate."}