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For a doubly stochastic matrix, the Hall marriage theorem supplies a perfect matching in its positive-entry support. Subtract the smallest matched entry times the corresponding permutation matrix. The residual has equal row and column sums and smaller support unless it vanishes. Repeating gives a convex combination of permutation matrices. With $r$ initially positive entries, there are at most $r-n+1$ terms, since a positive-mass residual has at least $n$ positive entries. Elementary augmenting-path matching gives $O(n^5)$ arithmetic operations.

Every doubly stochastic matrix is a convex combination of permutation matrices. Equivalently, the extreme points of the doubly stochastic convex polytope are exactly the permutation matrices. The extreme-point assertion follows from an alternating-cycle perturbation of a doubly stochastic matrix; the convex-hull assertion follows because a bounded finite-dimensional polytope is the convex hull of its vertices. Permutation matrices are extreme since their zero entries force the same zeros in every convex decomposition.

A square real matrix is doubly stochastic if its entries are nonnegative and every row and column sums to $1$. Its positive-entry support satisfies the condition of the Hall marriage theorem: for a row subset $S$, $|S|\le |N(S)|$ by summing column capacities. Consequently its support contains a permutation matrix.

Let $S_M|y⟩=|-y\mathrm{mod}M⟩$. The modular-addition quantum oracle obeysThe three steps replace $y$ by $-y$, then $-y+h(x)$, then $y-h(x)$. Thus a query to the inverse costs one forward query and two known unitary operators. The negation operator is a permutation matrix with $S_M^2=I$.

Write the quantum circuit as $U=U_T\cdots U_1$, and let $V_0=I$, $V_t=U_t\cdots U_1$. Use a nonlocal quantum clock with orthonormal states $|0⟩,\dots ,|T⟩$. Let the work space include the quantum witness and the ancilla qubits. The input penalty of a history Hamiltonian is built fromIt annihilates precisely the correctly initialized ancilla qubits, leaving the quantum witness unrestricted. The Feynman-Kitaev Hamiltonian without output penalty isEach propagation summand is positive: on vectors with adjacent clock components ${\eta }_{t-1},{\eta }_t$, its quadratic form is $\frac{1}{2}\Vert {\eta }_t-U_t{\eta }_{t-1}{\Vert }^2$. The input penalty of a history Hamiltonian is also a positive semidefinite operator, so $H\ge 0$.

To verify that this is a stoquastic Hamiltonian, use the work computational basis and the clock basis. Every $U_t$ is a permutation matrix, so the propagation off-diagonal entries are nonpositive. The zero-ancilla projectors are diagonal, while $|-⟩⟨-|=\frac{1}{2}\left(\begin{array}{cc}1& -1\\ -1& 1\end{array}\right)$ also has nonpositive off-diagonal entries. No positive off-diagonal entry is introduced by summing these terms. Thus $H$ is positive semidefinite and stoquastic, with no output penalty.

The construction uses the abstract $T+1$ dimensional clock space. A binary implementation needs diagonal penalties for unused clock labels. The clock transitions are nonlocal; the construction alone does not establish fixed qubit locality of an ordinary local Hamiltonian problem.

For a normalized computational history state, the output projector sees only its final clock component. With $p_{+}(\zeta )$ the quantum circuit's acceptance probability,This proves the hint, including for a computational basis quantum witness. Moreover $H_{\mathrm{out}}=|-⟩⟨-{|}_1\otimes |T⟩⟨T|$ is a stoquastic Hamiltonian term, so adding it with positive coefficient preserves stoquasticity.

There are two substantive problems with the stated promise. First, for any basis quantum witness the initialized state has nonnegative amplitudes, and every permutation matrix preserves them. Write the output as $|0⟩|r_0⟩+|1⟩|r_1⟩$, with both vectors entrywise nonnegative. ThenThe stoquastic acceptance floor rules out the printed one-third soundness condition. There are no NO instances of that literal promise. Second, the ground space includes histories of arbitrary quantum witnesses. It is the maximum over those quantum witnesses, not the maximum over basis quantum witnesses, that determines the lowest output energy. Even the identity quantum circuit accepts each basis input with probability $1/2$, but accepts a $|+⟩$ quantum witness with probability one. Thus a basis-witness soundness bound would not control this Hamiltonian, even if its numerical threshold were repaired.

The requested nontrivial hardness statement therefore needs the standard quantum-witness StoqMA promise, with $1/2\le b<a\le 1$ and inverse-polynomial gap $\gamma =a-b$. The intended reduction can be completed precisely under that corrected promise. Define the quantum witness embedding $J|\zeta ⟩=|\zeta ⟩|a⟩$ and the witness acceptance operatorIt is a positive contraction, and $p_{\max }={\lambda }_{\max }(F)$. The minimum expectation of $H_{\mathrm{out}}$ on the history ground space is $\nu =(1-p_{\max })/(T+1)$. This is also consistent with a nonnegative optimal StoqMA quantum witness: $F$ is entrywise nonnegative, so replacing amplitudes by their absolute values cannot decrease its quadratic form.

A uniform ground-space perturbation bound is needed because the gap shrinks with quantum circuit length. Put $g=(T+1{)}^{-3}$, and choose a positive inverse-polynomial coefficientFor $P$ the history ground projector and any unit vector $\psi =p+q$ with $p=P\psi$, the operator norm bound $\Vert H_{\mathrm{out}}\Vert =1$ givesThe last inequality completes the square in $\Vert q\Vert$ and uses $0\le \nu \le 1$. Testing a ground space minimizing vector gives the upper bound ${\lambda }_{\min }(H^{\prime })\le \delta \nu$. Since $\delta <g/2$, we obtainYES instances of the corrected promise have energy at most $\delta (1-a)/(T+1)$; NO instances have energy at leastThe separation is at least $7\delta \gamma /[8(T+1)]$, an inverse polynomial. The quantum circuit and all these Hamiltonian terms have polynomial-size descriptions. This proves StoqMA hardness for the local Hamiltonian problem variant that permits the nonlocal clock, under the corrected quantum witness and acceptance promise.

The printed perturbation formula also has $H_{\mathrm{out}}$ where a general perturbation $V$ belongs. Its unspecified $O({\delta }^2)$ cannot be treated as uniform in a closing gap. Likewise a circuit-length-independent constant $\delta$ cannot in general satisfy the stated small-perturbation requirement for an arbitrary long quantum circuit. The explicit bound above avoids both issues; it does not claim the defective literal promise defines standard StoqMA.

True. The sign of a permutation is a group homomorphism $\mathrm{sgn}:S_3\to \{1,-1\}$, as follows, for example, from the multiplicative determinant of permutation matrices. Its kernel of a group homomorphism is the alternating groupA transposition maps to $-1$, so this homomorphism is onto.

Form a bipartite graph with one vertex for each row and one for each column, putting an edge $ij$ exactly where $a_{ij}>0$. For a set $S$ of row vertices, let $N(S)$ be its neighbouring column vertices. Since $A$ is a doubly stochastic matrix,The Hall marriage theorem therefore supplies a perfect matching. Its incidence entries define a permutation matrix $P$ supported on the positive entries of $A$, hence on the ones of $B$. Entrywise $P\le B$, so

The preceding perfect matching gives a permutation matrix $P$ supported on the positive entries of $A$. LetIf $\alpha =1$, every row's selected entry is one and all other entries vanish, so $A=P$. Otherwiseis a doubly stochastic matrix with strictly fewer positive entries. Induct on the number of positive entries: the base case has exactly $n$ positive entries and is a permutation matrix. By induction, write $A^{\prime }={\sum }_i{\beta }_i{P}_i$ as a convex combination. Thenis another convex combination, with nonnegative coefficients summing to one. ThusThis constructive proof of the Birkhoff-von Neumann theorem is Birkhoff decomposition by support matchings.