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For linear projections $E,F$, expanding the square proves that their sum is a linear projection exactly when $EF=-FE$. Multiplying this equality on the left and right by $E$ gives $EF=-EFE$ and $EFE=-FE=EF$, hence $2EF=0$. If the field has characteristic different from two, this forces $EF=FE=0$, and those conditions also suffice. In characteristic two, the condition instead reduces to commutation; $E=F=I$ has sum zero but nonzero product. For three commuting linear projections, the square-minus-sum is $2(EF+EG+FG)$, so vanishing of the triple product alone is not sufficient.

The sum-product phenomenon says that a finite subset of a field cannot usually have both a small sumset and a small product set unless it has additional subfield-like structure.

Over a field whose characteristic is not two, a bilinear form has a unique decomposition into a symmetric bilinear form $S$ and an alternating bilinear form $K$:Adding and subtracting the equations $B(v,w)=S(v,w)+K(v,w)$ and $B(w,v)=S(v,w)-K(v,w)$ both proves uniqueness and gives these formulas. In matrix notation, $S$ has matrix $(A+A^T)/2$ and $K$ has skew-symmetric matrix $(A-A^T)/2$.^[ref]

The diagonal values $q(v)=B(v,v)$ alone do not determine a general bilinear form, because $K(v,v)=0$ for every vector. For the matrix of a bilinear form in the example above,Thus two different bilinear forms, represented by $A$ and its symmetric matrix part, have exactly the same quadratic form. To recover a bilinear form from its diagonal, one needs extra information such as the hypothesis that it is a symmetric bilinear form.

Over every field, including characteristic two, there is a natural isomorphismThe middle summand maps $v\otimes w$ to the mixed product $vw$. Monomial bases prove bijectivity, and the Leibniz rule makes the map equivariant for Lie algebra representations.

Over a field of characteristic two, the linear map $T:M_n(F)\to M_n(F)$, $A\mapsto A+A^T$, has kernel of a linear map the symmetric matrices, of dimension $n(n+1)/2$. Its image of a linear map is the linear subspace of symmetric matrices with zero diagonal, of dimension $n(n-1)/2$. Indeed, $T(A)=0$ means $A=A^T$, and every symmetric zero-diagonal matrix is the image of the matrix obtained by retaining just its entries above the diagonal. Thus the two dimensions still add to $n^2$, as required by the rank-nullity theorem.

For a finite-dimensional symplectic vector space, the symplectic adjoint of a linear map $T$ is the unique linear map $T^{\ast }$ satisfying the displayed identity. Existence and uniqueness follow because the nondegenerate bilinear form identifies the vector space with its dual space. If $\omega$ has matrix $J$ in a basis, then $T^{\ast }=J^{-1}{T}^{\mathsf{T}}J$. Thus $(ST{)}^{\ast }=T^{\ast }{S}^{\ast }$ and $(T^{\ast }{)}^{\ast }=T$, and membership in the symplectic Lie algebra is equivalent to $T^{\ast }=-T$. This operation is linear over the ground field; it does not involve complex conjugation. It preserves diagonalizability and the property of being a nilpotent linear map, since it is conjugate to matrix transpose.

A symplectic basis organizes a nondegenerate alternating bilinear form into independent pairs. Let $(V,\omega )$ be a finite-dimensional vector space over a field $K$, with $\omega$ a nondegenerate bilinear form satisfying $\omega (v,v)=0$ for every vector $v$. Its dimension is even, say $2n$, and a symplectic basis is an ordered basis $\mathcal{B}=(e_1,\dots ,e_n,f_1,\dots ,f_n)$ such thatHere ${\delta }_{ij}$ is the Kronecker delta. Each $(e_i,f_i)$ is a hyperbolic pair; different pairs have zero mutual pairing. These conventions apply in every characteristic. In characteristic two, $-1=1$, but the condition $\omega (v,v)=0$ is still essential: a symmetric bilinear form need not be an alternating bilinear form. The existence theorem and the characteristic caveat are given in Bilinear Forms by Keith Conrad (Keith Conrad, Bilinear Forms, §1, Theorem 1.6; §5, Theorems 5.1 and 5.4 and Definition 5.3).

The benefit is a common coordinate language for every symplectic vector space of the same finite dimension over the same field. The span of all $e_i$, and separately of all $f_i$, is a Lagrangian subspace; each paired span is a symplectic subspace. In Hamiltonian mechanics, the corresponding pairing of position and momentum coordinates describes phase space and canonical transformations. On a symplectic manifold, a symplectic basis describes one tangent space; Darboux coordinates additionally describe a whole local neighborhood.^[ref]

For a finite-dimensional vector space over any field with a nondegenerate alternating bilinear form $B$, the symplectic group over a field consists of invertible linear maps preserving $B$. In a symplectic basis its matrices satisfy $g^{\mathsf{T}}Jg=J$. Matching two symplectic bases gives a unique such linear map, and a symplectic basis exists by successively splitting off paired vectors. This includes arbitrary-characteristic and infinite-field versions, beyond the real Lie group and the symplectic group over a finite field.

A tensor product turns a bilinear map into a linear map. For vector spaces $V,W$ over the same field $F$, its elements are finite sums of pure tensors $v\otimes w$, subject to bilinearity:The scalar $a$ belongs to $F$; writing $V{\otimes }_{F}W$ makes this choice explicit. These rules preserve separate linearity while allowing arbitrary linear combinations of products.^[ref]

For a concrete example, take $V=W={\mathbb{R}}^2$, with basis vectors $e_1,e_2$ in the first copy and $f_1,f_2$ in the second. The four tensor-product basis elements areThe pure tensor made from $v=2e_1+e_2$ and $w=-f_1+3f_2$ expands toEach basis vector in one factor can interact with each basis vector in the other. This is why dimensions multiply; the tensor-product coefficient matrix displays these four interactions as a two-by-two grid.

The tensor product is different from the Cartesian product $V\times W$. With componentwise vector addition and scalar multiplication, the latter contains pairs and has dimension $\dim V+\dim W$. The map $\tau (v,w)=v\otimes w$ is bilinear, generally neither linear on that Cartesian product nor injective nor surjective. In our example, $\tau (2v,2w)=4\tau (v,w)$, and $\tau (2v,w/2)=\tau (v,w)$; the separability of a tensor explains why some outputs cannot be reached at all. Both spaces happen to have dimension four when $V=W={\mathbb{R}}^2$, but this coincidence does not identify their structures.

For two unital associative algebras over a field $k$, their tensor product carries the multiplication in the title, extended by bilinearity, with identity $1\otimes 1$. Choosing a basis $(b_j)$ of the second factor gives the basis of a module $(1\otimes b_j)$ for the resulting left $A$-module; dual basis functionals recover the coefficients uniquely.

The theory of formally real fields consists of the field axioms and all sentences asserting that $-1$ is not a sum of $m$ squares, one for each positive integer $m$. The real closure theorem and model completeness make the Theory of real closed fields its model companion.

Over a field of characteristic different from two, three linear projections summing to the identity annihilate one another in both orders. Each pair sum is the complementary projection to the third. The sum of two linear projections criterion therefore proves zero products. Every vector is the sum of its three images; applying each projection to a zero sum proves that these images form a direct sum. Choosing a basis in each image gives a simultaneous eigenbasis, with triples of eigenvalues $(1,0,0)$, $(0,1,0)$ or $(0,0,1)$. This is algebraic orthogonality, and does not assume that the projections are orthogonal projections for a specified inner product.

Two canonically normalized complex scalar fields of vacuum modulus $v$ and charge rows $R_1=(1,-q,0)$, $R_2=(0,1,-q)$ couple to three canonically normalized Abelian vector fields. The scalar kinetic terms give $g^2{v}^2|RB{|}^2$, so the physical squared-mass matrix includes the factor two shown. Its nonzero eigenvalues are $2g^2{v}^2(1+q^2\pm q)$ and its normalized null vector is $(q^2,q,1)/\sqrt{q^4+q^2+1}$. The two scalar phases are absorbed into the massive vector bosons, leaving one unbroken local Abelian generator. A compact global gauge-group interpretation additionally requires a compatible charge lattice.

A toral subalgebra of a Lie algebra $\mathfrak{g}$ is a Lie subalgebra $\mathfrak{a}$ such that $\mathrm{ad}a$ is diagonalizable for every $a\in \mathfrak{a}$. Here diagonalizability is over the stated ground field, and refers to the Adjoint representation. In arguments that use a simultaneous weight decomposition, commutativity of $\mathfrak{a}$ is stated as well. For a semisimple Lie algebra over an algebraically closed field of characteristic zero, Cartan subalgebras are examples. Semisimplicity on the defining representation of a symplectic Lie algebra follows from adjoint semisimplicity in the symplectic Lie algebra.

A module $M$ over an integral domain $R$ is torsion-free when $rm=0$ with $0\ne r\in R$ implies $m=0$, equivalently when its torsion submodule is zero. Every free module over an integral domain is torsion-free, as one can check in each coefficient. Every vector space is torsion-free over its scalar field.

The converse requires hypotheses. A finitely generated module over a principal ideal domain is free if it is torsion-free, but the module $\mathbb{Q}$ over $\mathbb{Z}$ is torsion-free and not free. Any two nonzero rational numbers have a nontrivial integer relation, so a possible basis of a module could have at most one element; no single $q$ generates all rational numbers, since $q/2\notin \mathbb{Z}q$ for $q\ne 0$. The finite-generation hypothesis cannot be dropped. See Keith Conrad's Modules over a PID, Corollary 2.6.

Consider the real matrixThe Cayley-Hamilton theorem gives $A^2=3A-I$. ConsequentlyThusAll four entries come from just two scalar coefficients. If $A^n=a_{n}A+b_{n}I$, multiplying once by $A$ and replacing $A^2$ gives the linear recurrence relationEquivalently $a_{n+2}=3a_{n+1}-a_n$, with $a_0=0,a_1=1$. The appearance of Fibonacci numbers is explained by $A=Q^2$, where $Q=\left(\begin{array}{cc}1& 1\\ 1& 0\end{array}\right)$ advances the Fibonacci recurrence. No eigenvectors need to be computed.

The same identity computes an inverse. Rearrange it as $A(3I-A)=I$, givingThis works because the constant term of this characteristic polynomial is nonzero and the coefficients lie in a field. Generally, when $c_0\ne 0$ over a field, rearranging the Cayley-Hamilton theorem expresses $A^{-1}$ as a polynomial evaluated at a matrix. Over a general commutative ring, a nonzero constant need not be a unit; invertibility requires that extra condition.

The two-factor tensor rank is the smallest number of pure tensors whose sum equals $t$, with zero allowed as the empty sum. For two finite-dimensional vector spaces over the same field, it equals the matrix rank of the tensor-product coefficient matrix. A sum of $r$ outer products has column space contained in the span of their $r$ first factors, so its matrix rank is at most $r$. Conversely, choose a basis $u_1,\dots ,u_r$ for the column space, express each column in that basis, and collect the coefficients into row vectors $w_k^T$. This yields $C={\sum }_{k=1}^r{u}_k{w}_k^T$, hence a sum of $r$ pure tensors.

The pure tensor $(2,1)\otimes (-1,3)$ therefore has two-factor tensor rank one, whereas $e_1\otimes f_1+e_2\otimes f_2$ has two-factor tensor rank two. Both have two factors: the number of factors and the minimum number of product summands are different counts. Over the real numbers, the singular value decomposition supplies a useful decomposition $C={\sum }_k{\sigma }_k{u}_k{v}_k^T$, with one nonzero term per matrix rank. The two-factor rank equality does not extend to three or more factors by merely choosing one matrix layout.

Over a field of characteristic zero with $n\ge 4$, let $V_2$ be the permutation representation on unordered two-element subsets and $V_1$ the representation on singletons. The equivariant incidence map $D:V_2\to V_1$ sends $e_{ij}$ to $e_i+e_j$. It is surjective: if coefficients $c_i$ annihilate its image, then $c_i+c_j=0$ for all distinct $i,j$, forcing all coefficients to vanish. Thus its kernel has dimension $\left(\genfrac{}{}{0ex}{}{n}{2}\right)-n$.

A polytabloid of shape $(n-2,2)$, identified by its bottom-row pair, is an alternating four-edge vector $e_{cd}-e_{ad}-e_{cb}+e_{ab}$. Its incidence is zero. The hook-length formula gives the same dimension $\left(\genfrac{}{}{0ex}{}{n}{2}\right)-n$ for this Specht module, proving the identity. Maschke's theorem yields $V_2\cong S^{(n-2,2)}\oplus V_1$.

For a Young subgroup with block sizes ${\mu }_1,\dots ,{\mu }_l$, there are $l$ singleton orbits and $\left(\genfrac{}{}{0ex}{}{l}{2}\right)+\#\{i:{\mu }_i\ge 2\}$ pair orbits. Taking fixed vectors and using Frobenius reciprocity therefore gives the multiplicity of $S^{(n-2,2)}$ in $M^{\mu }$ as $\left(\genfrac{}{}{0ex}{}{l}{2}\right)+\#\{i:{\mu }_i\ge 2\}-l$. This derives the multiplicity without Young's rule; block sizes $(3,2,1)$ give exactly two.

For infinite spaces over a field of size $q$, two-variable types are determined by the kernel of the linear map sending $(a,b)$ to $ax+by$. Its ranks give the zero pair, $q+1$ one-dimensional directions and the independent-pair type. All $q+3$ types are isolated by finite formulas. Quantifier elimination makes the Lindenbaum algebra their full power-set Boolean algebra. The one-variable algebra has just the zero and nonzero atoms, hence four Elements.

Every nonzero nonunit in a unique factorization domain is a finite product of irreducible elements. Such a factorization is unique up to permutation and multiplication of the factors by units. Irreducibles are prime elements, so divisibility can be checked by comparing their multiplicities. This applies to a polynomial ring over a field and is useful for intersections of localizations inside its fraction field.