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For an explicit algorithm, keep an unnormalized residual $R$, initially $A$, whose row and column sums all equal a common mass $\tau$, initially one. While $\tau >0$, find a perfect matching in its positive support, form its permutation matrix $P$, and subtract $\alpha P$, where $\alpha$ is the smallest selected entry. Record $(\alpha ,P)$ and replace $\tau$ by $\tau -\alpha$.

If the residual mass is positive, the same proof of the Hall marriage theorem condition applies after division by $\tau$. Every subtraction removes at least one positive entry and creates no new positive entries. Before the last subtraction, any positive-mass residual has at least $n$ positive entries; the last step removes all its entries. If $r$ is the initial number of positive entries, thenSince the residual ultimately vanishes, summing the recorded subtractions gives $A=\sum {\alpha }_i{P}_i$, and the row sums give $\sum {\alpha }_i=1$.

A perfect matching can be found by at most $n$ searches for an augmenting path in a matching, each costing $O(n^2)$ in a graph with at most $n^2$ edges. Thus each iteration costs $O(n^3)$, including forming the support and updating the residual, and there are $O(n^2)$ iterations. HenceThis bound uses exact real arithmetic. For rational input, a common denominator lets all residual subtractions be performed on integers of polynomial bit length, giving a polynomial bit-time implementation as well.

Every permutation is a product of transpositions, since a cycle can be writtenwith rightmost factors acting first. Define the sign of a permutation to be $(-1{)}^r$ for a product of $r$ transpositions. To prove that it is well defined without assuming parity, let $P_{\sigma }$ be the permutation matrix specified by $P_{\sigma }{e}_i=e_{\sigma (i)}$. Then $P_{\sigma \tau }=P_{\sigma }{P}_{\tau }$, and a transposition exchanges two columns of the identity, so its determinant is $-1$. Any decomposition into $r$ transpositions therefore hasThe left-hand side depends only on $\sigma$, proving well-definedness. Multiplicativity of the determinant now gives the sign homomorphism:For the first requested group embedding, use the faithful four-point action of GL2 over F2: label the four vectors in ${\mathbb{F}}_2^2$ and let an invertible $2\times 2$ matrix act on them by multiplication. Composition of these actions gives a group homomorphism $\psi :G{L}_2({\mathbb{F}}_2)\to S_4$. If a matrix fixes every vector, it fixes both standard basis vectors and is the identity, so the action is faithful. The matrixfixes $(0,0)$ and $(1,1)$ and exchanges $(1,0)$ with $(0,1)$. Its image is a single transposition, hence $\mathrm{sgn}(\psi (B))=-1$. The composite sign map is nontrivial.

For the second group embedding, simply takeThe preceding multiplication formula proves the group homomorphism property, and $P_{\sigma }=I$ forces $\sigma (i)=i$ for every $i$, proving injectivity.

An extreme point of a convex set cannot be written as a nontrivial convex combination of two distinct members. If a permutation matrix $P=\theta B+(1-\theta )C$ with $0<\theta <1$ and $B,C$ doubly stochastic, each zero entry of $P$ forces the corresponding entries of $B,C$ to be zero by nonnegativity. The unique possible nonzero entry in each row must then be one by the row sum. Thus $B=C=P$, proving that every permutation matrix is extreme.

Conversely, let $A$ be a doubly stochastic matrix that is not a permutation matrix. At least one entry lies strictly between zero and one. Form a bipartite graph whose two vertex classes are rows and columns, with edges precisely at these fractional entries. Every incident row has at least two fractional entries: a single fractional entry, with all other entries zero or one, could not give row sum one. The same holds for columns.

A finite nonempty graph whose incident vertices all have degree at least two contains a cycle. In a bipartite graph that cycle has even length. Assign alternating signs $+1,-1$ to its entries, giving a nonzero matrix $D$ with every row and column sum zero. Choose $\epsilon >0$ smaller than the distance of each cycle entry from both zero and one. Then $A+\epsilon D$ and $A-\epsilon D$ remain doubly stochastic matrices, are distinct, and satisfySo $A$ is not extreme. ThereforeThis alternating-cycle perturbation of a doubly stochastic matrix proves the extreme-point part of the Birkhoff-von Neumann theorem directly.

With rightmost-first composition, a permutation cycle satisfiesTracing each $a_j$ verifies the equality, and every other letter is fixed. Thus every cycle is a product of transpositions; decomposing a permutation into disjoint permutation cycles proves that every element of $S_n$ is such a product. The identity is the empty product.

To establish that parity is well defined, use the permutation matrix $P_{\sigma }$ defined by $P_{\sigma }{e}_i=e_{\sigma (i)}$. Its determinant is $\pm 1$, and $P_{\sigma \tau }=P_{\sigma }{P}_{\tau }$. A transposition swaps two columns of the identity matrix and has determinant $-1$. Therefore if $\sigma$ is expressed as a product of $r$ transpositions,The left side depends only on $\sigma$, so any two decompositions have the same parity. DefineMultiplicativity of the determinant makes this the sign homomorphism. Its kernel of a group homomorphism is the alternating groupIt is a normal subgroup consisting of the even permutations. For $n\ge 2$ the sign map is surjective because a transposition has sign $-1$, so $[S_n:A_n]=2$. For $n=1$ both groups are trivial.

For $\sigma \in S_n$, let $P_{\sigma }$ be its permutation matrix and defineSince a transposition exchanges two columns, it has determinant $-1$. Thus any expression of $\sigma$ as $r$ transpositions has sign $(-1{)}^r$, proving that the parity is independent of the expression. Moreover, $P_{\sigma \tau }=P_{\sigma }{P}_{\tau }$, so the determinant provesHence the sign homomorphism is a surjective homomorphism for $n\ge 2$.

For any homomorphism $\phi :S_n\to \{\pm 1\}$, all transpositions have the same image because they are conjugate. Since transpositions generate $S_n$, image $+1$ would make $\phi$ trivial. Surjectivity therefore forces every transposition to map to $-1$, and hence $\phi =\mathrm{sgn}$.

The map $x\mapsto -x(\mathrm{mod}K)$ is a bijection and an involution, so its permutation matrix satisfiesIn particular $S^{†}S=I$, which proves that $S$ is a unitary operator. Equivalently, it permutes the computational basis and therefore preserves every inner product.

Here the only nonzero coefficients of $a$ areTherefore, for each $(m,n)\in {\Lambda }_N$, the truncated system iswhere a coefficient is omitted when its index is outside ${\Lambda }_N$ or equals $(0,0)$.

For wavevectors $\mathbf{k}=(m,n)$ and $\mathbf{\ell }=(p,q)$, the matrix entry isBecause ${\hat{a}}_{\mathbf{r}}={\hat{a}}_{-\mathbf{r}}$, $A_{\mathbf{k},\mathbf{\ell }}=A_{\mathbf{\ell },\mathbf{k}}$, so the matrix is symmetric. Its diagonal entry is $5{\pi }^2|\mathbf{k}{|}^2$. The sum of absolute off-diagonal entries is at mostbecause $|\mathbf{k}{|}^2\pm k_j\ge 0$ for nonzero integer $\mathbf{k}$; truncation can only reduce the sum. Hence $A$ is strictly diagonally dominant with positive diagonal. The Gershgorin circle theorem places every eigenvalue in the positive intervalfor at least one row. Since the symmetric matrix has only real eigenvalues, all are positive, and $A$ is a positive-definite matrix. Any one-dimensional ordering merely conjugates $A$ by a permutation matrix, preserving symmetry and positive definiteness.

For subgroups $H,P\le G$, define $x\sim y$ when $y\in HxP$. The identity elements show reflexivity, inverses show symmetry, and multiplication shows transitivity. Thus the double cosets $HxP$ partition $G$.

Let $H$ act by left multiplication on the set of left cosets $G/P$. The orbit of $xP$ consists of the cosets contained in $HxP$, so its size is $|HxP|/|P|$. Its stabilizer isThe orbit-stabilizer theorem therefore gives

Suppose $P$ is a Sylow subgroup of $G$, with $|P|=p^a$, and write the largest power of $p$ dividing $|H|$ as $p^b$. If every $H\cap xP{x}^{-1}$ had order at most $p^{b-1}$, the displayed formula would make every double-coset size divisible by $p^{a+1}$. Their sum $|G|$ would then also be divisible by $p^{a+1}$, contradicting the choice of $P$. Hence some $H\cap xP{x}^{-1}$ has order $p^b$ and is a Sylow $p$-subgroup of $H$. This is the Sylow subgroup of a subgroup from double cosets argument.

To count the general linear group over a finite field $G{L}_n({\mathbb{F}}_p)$, choose its columns successively. There are $p^n-1$ choices for the first, $p^n-p$ for the second, and $p^n-p^k$ for column $k+1$. ConsequentlyNone of the factors $p^j-1$ is divisible by $p$, so the upper unitriangular group is a Sylow $p$-subgroup: it has one arbitrary field entry in each of the $n(n-1)/2$ positions above the diagonal and hence order $p^{n(n-1)/2}$. The permutation matrices form a subgroup isomorphic to the symmetric group $S_n$.

By Cayley theorem, every finite group $G$ embeds in $S_{|G|}$, and permutation matrices embed this symmetric group in $G{L}_{|G|}({\mathbb{F}}_p)$. The latter has the explicit Sylow $p$-subgroup just described, so the result proved in the first part, applied to the embedded copy of $G$, proves that every finite group has a Sylow $p$-subgroup.

The counting part of the Sylow theorems says that if $p^a$ is the largest power of $p$ dividing $|G|$, then the number $n_p$ of Sylow $p$-subgroups satisfies

Finally, let $|G|=pq$ with prime numbers $p>q$. The Sylow counts giveso $n_p=1$ and the Sylow $p$-subgroup is a normal subgroup. Also $n_q\mid p$ and $n_q\equiv 1(\mathrm{mod}q)$. If $n_q=1$, both Sylow subgroups are normal; their elements commute, so $G$ is their direct product and is abelian. In the nonabelian case one must therefore have $n_q=p$, whenceor equivalently $q\mid p-1$, as recorded by nonabelian group of order pq.

Because $A{x}^{\ast }=b$, the error $e^{(k)}=x^{\ast }-x^{(k)}$ obeysThus the heavy-ball error propagation matrix is

If $A=\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_n)$, permute the coordinates fromtoFor the corresponding permutation matrix $P$,

A signed permutation matrix has exactly one nonzero entry in each row and column, and every nonzero entry is $1$ or $-1$. It is a product of a diagonal sign matrix and a permutation matrix.

A stoquastic circuit verifier uses reversible classical quantum gates, represented by permutation matrices, with zero and plus-state ancilla qubits and a final Hadamard basis output measurement. A basis input remains an entrywise nonnegative state; a general quantum witness need not be a basis input. The resulting acceptance structure is used in StoqMA.

A stoquastic Hamiltonian is a real Hermitian Hamiltonian operator with nonpositive off-diagonal entries in a specified basis, usually the computational basis. The property depends on that basis. Permutation matrices in a quantum circuit's propagation terms and $|-⟩⟨-|$ input penalties give a useful family of such operators.