Multilinear algebra

Multilinear algebra studies maps linear in each of several arguments, including determinants, tensors, and exterior products.

A comonoid in the monoidal category of modules over a commutative ring, with tensor ${\otimes }_k$ and unit $k$.

A right comodule is an object $M$ with a coaction $\rho :M\to M\otimes C$ satisfying the coassociativity and counit laws. This definition extends from coalgebras to comonoids in a monoidal category.

A natural underlying transformation of tensor bifunctors on right comodules over a field has components ${\phi }_{M,N}(u\otimes v)=\sum u_{(0)}\otimes v_{(0)}\gamma (u_{(1)},v_{(1)})$. Recover $\gamma =(\epsilon \otimes \epsilon ){\phi }_{H,H}$; colinearity and monoidal axioms impose further equations.

For a transformation from the tensor built using $n$ to the tensor built using $m$, colinearity is $m\ast \gamma =\gamma \ast n$: $\sum m(a_{(1)},b_{(1)})\gamma (a_{(2)},b_{(2)})=\sum \gamma (a_{(1)},b_{(1)})n(a_{(2)},b_{(2)})$. Scalar maps enter this convolution product for coalgebra maps via the common algebra unit.

A comonoid morphism $f:H\to K$ induces a functor $f_{\ast }$ on right comodules, replacing $\rho$ by $(1\otimes f)\rho$ and preserving underlying objects and arrows. For bimonoids, it is strict monoidal exactly when $f$ is also a monoid morphism.

A natural transformation between corestriction functors over a field has components ${\beta }_M(u)=\sum u_{(0)}\alpha (u_{(1)})$, recovered by $\alpha =\epsilon {\beta }_H$. Monoidality forces $\alpha$ to be a unital algebra homomorphism over a field; for a Hopf algebra, $\alpha S$ supplies the inverse.

The structure map of a comodule; a right coaction satisfies $(\rho \otimes 1)\rho =(1\otimes \Delta )\rho$ with coherent parentheses, and $(1\otimes \epsilon )\rho$ is the inverse right unitor.

For a coalgebra $C$ and a unital associative algebra over a commutative ring $A$, maps $f,g:C\to A$ have convolution $f\ast g=m_A(f\otimes g){\Delta }_C$, with unit $j_A{\epsilon }_C$. The two-sided convolution inverse of $1_H$ is the antipode.

The notation $\Delta h=\sum h_{(1)}\otimes h_{(2)}$ suppresses the summation index of a comultiplication; iterated indices use coassociativity. For a comodule, write $\rho (u)=\sum u_{(0)}\otimes u_{(1)}$.

A multilinear map $F:V_1\times \cdots \times V_r\to W$ is a map between vector spaces that is linear in each argument when all other arguments are fixed. For example the evaluation $(\omega ,v)\mapsto \omega (v)$ is bilinear, and the determinant is multilinear in its columns. The universal property of a tensor product replaces this separate linearity by a single linear map from $V_1\otimes \cdots \otimes V_r$.

A bilinear map is a map $B:V\times W\to U$ between vector spaces that is linear in each argument when the other is fixed. The nonlinear temperature advection can be written as the diagonal of a bilinear map, $\mathcal{N}(\varphi ,\chi )=-\mathbf{u}[\varphi ]\cdot \nabla \chi$, when velocity depends linearly on temperature. Unlike a bilinear form, its output need not be a scalar.

For nonzero finite-dimensional complex vector spaces $U,V$, suppose a bilinear map never vanishes on a pair of nonzero vectors. Equivalently, each map obtained by fixing one nonzero argument is injective. If its associated map on $U\otimes V$ has image dimension $r$, projectivization gives a map to ${\mathbb{CP}}^{r-1}$ whose positive degree-two class pulls back to $x+y$. The nonzero top power $(x+y{)}^{\dim U+\dim V-2}$ in the product cohomology ring of complex projective space proves the bound. Multiplication of polynomials of bounded degree attains equality. Nonzero hypotheses matter: with a zero factor, slice conditions can be vacuous.

An alternating multilinear map vanishes whenever two arguments are equal. Swapping two arguments reverses its sign.

The tensor product $V\otimes W$ of vector spaces is generated by symbols $v\otimes w$ subject to bilinearity. It is characterized by the property that every bilinear map from $V\times W$ factors uniquely through a linear map from $V\otimes W$.

For vector spaces $V_1,\dots ,V_r$, every multilinear map $F:V_1\times \cdots \times V_r\to W$ factors uniquely through a linear map $\tilde{F}:V_1\otimes \cdots \otimes V_r\to W$ sending a decomposable tensor to $F$ of its factors. Thus multilinear formulas define linear maps on tensor products without choosing a basis. Tensor contraction is an example, using the evaluation pairing between a vector and a covector.

If $\{e_i\}$ and $\{f_j\}$ are bases of finite-dimensional vector spaces $V,W$, then $\{e_i\otimes f_j\}$ is a basis of $V\otimes W$. Bilinearity gives spanning, and applying products of dual coordinate functionals gives independence. Thus $\dim (V\otimes W)=\dim V\dim W$.

The tensor algebra of a vector space $V$ is the graded algebrawith multiplication given by concatenation of tensors.

The nth tensor power is the iterated tensor product of modules $V^{\otimes n}=V\otimes \cdots \otimes V$.

The tensor square is the second tensor power $V^{\otimes 2}=V\otimes V$. Over a field of characteristic different from two it decomposes as the direct sum of the symmetric square and exterior square.

The symmetric algebra is the quotient of $T(V)$ by the ideal generated by $v\otimes w-w\otimes v$. It is graded by symmetric powers:

The nth symmetric power $S^{n}V$ is the quotient of $V^{\otimes n}$ that identifies tensors differing by a permutation of their factors. Equivalently, over characteristic zero it is the subspace of symmetric tensors.

For eigenvalues ${\lambda }_1,\dots ,{\lambda }_d$ of a finite-order linear action, its symmetric power characters satisfy ${\sum }_{n\ge 0}{\chi }_{S^{n}V}{t}^n={\prod }_j(1-{\lambda }_{j}t{)}^{-1}$. The denominator is the alternating generating polynomial of its exterior power characters.

The symmetric square is the quotient of $V\otimes V$ by $v\otimes w-w\otimes v$, and records symmetric tensors of rank two.

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.

For the defining real representation of $SO(d)$, the symmetric square splits as a scalar trace plus symmetric trace-free tensors: ${\mathrm{Sym}}^2({\mathbb{R}}^d)=\mathbb{R}\oplus {\mathrm{Sym}}_0^2({\mathbb{R}}^d)$. The trace projection is $S\mapsto S-(\mathrm{tr}S)I/d$, so the second summand has dimension $d(d+1)/2-1$.

For $d\ge 3$ the trace-free summand is an irreducible real group representation. To see this, diagonalize a nonzero symmetric trace-free matrix in a nonzero invariant subspace. Two diagonal entries differ. Infinitesimal rotation in their plane yields a nonzero symmetric off-diagonal matrix; rotations carry it to every coordinate pair, and a rotation through $\pi /4$ produces a difference of two diagonal entries. These off-diagonal matrices and diagonal differences span all symmetric trace-free matrices, so the invariant subspace is the whole summand. For $d=24$, its dimension is 299, the third-level nontrivial multiplet of the lowest levels of a fully transverse ND bosonic string.

The exterior algebra is the graded algebrawith multiplication given by the antisymmetric wedge product.

The exterior product is the multiplication ${\bigwedge }^{p}V\otimes {\bigwedge }^{q}V\to {\bigwedge }^{p+q}V$ of the exterior algebra. For homogeneous elements of a graded algebra, $u\wedge v=(-1{)}^{pq}v\wedge u$.

For a polynomial ring $A$, a degree-$p$ polynomial polyvector field is an element of ${\bigwedge }_A^p{\mathrm{Der}}_k(A)$. Its product is the exterior product. Together with the Schouten-Nijenhuis bracket these spaces form a Gerstenhaber algebra.

The left Schouten-Nijenhuis bracket on polynomial polyvector fields is the degree-minus-one Gerstenhaber bracket extending the commutator of derivations and $[D,f]=D(f)$ by graded antisymmetry and the left graded Leibniz rule. For $\Pi ={\partial }_X\wedge {\partial }_Y$, this convention gives $[h\Pi ,f]=h(f_Y{\partial }_X-f_X{\partial }_Y)$ and $[F{\partial }_X+G{\partial }_Y,h\Pi ]=(D(h)-h(F_X+G_Y))\Pi$. A right insertion convention reverses the degree-two-with-function formula.

A Grassmann algebra is an exterior algebra regarded as an algebra of anticommuting generators. Its odd elements satisfy ${\theta }_i{\theta }_j=-{\theta }_j{\theta }_i$.

The parity grading records whether a homogeneous element has even or odd Grassmann degree. Homogeneous elements obey $XY=(-1{)}^{|X||Y|}YX$. An odd differential uses the graded Leibniz rule; this is the sign bookkeeping behind left-acting BRST differential calculations.

A Grassmann derivative is an odd differentiation operation on a Grassmann algebra. A left Grassmann derivative obeys ${\partial }_{\theta }^L(ab)=({\partial }_{\theta }^{L}a)b+(-1{)}^{|a|}a({\partial }_{\theta }^{L}b)$ for homogeneous $a$. Right derivatives use the corresponding right-sided product rule. Their placement must be fixed when differentiating fermionic sources in a generating functional, since commuting ordinary derivatives would lose fermionic signs.

Left differentiation with respect to an odd Grassmann variable is the odd linear operation ${\partial }_{\theta }^L\theta =1$, with zero derivative on every other generator. For a homogeneous element $a$ of parity $|a|$, it obeys the graded product ruleFor independent odd generators, ${\partial }_{\bar{\theta }}^L(\theta \bar{\theta })=-\theta$. This sign comes from moving the derivative through the first odd factor in the Grassmann algebra. It fixes chirality in the chiral-superfield component expansion; an ordinary commuting-variable product rule would give the wrong sign.

A Grassmann variable is an odd generator $\theta$ of a Grassmann algebra. In characteristic zero it obeys ${\theta }^2=0$, so functions of finitely many Grassmann variables have finite Taylor expansions.

The $k$th exterior power ${\Lambda }^{k}V$ is the quotient of $V^{\otimes k}$ that makes every tensor with two equal factors zero. It represents alternating $k$-linear maps and has dimension $\left(\genfrac{}{}{0ex}{}{\dim V}{k}\right)$ in finite dimensions.

The exterior square is the quotient of $V\otimes V$ by tensors $v\otimes v$, and records antisymmetric tensors of rank two.

The exterior square of a direct sum splits into wedges of two vectors from $V$, two from $W$, or one from each. The mixed map sends $v\otimes w$ to $(v,0)\wedge (0,w)$. A combined basis proves it is an isomorphism, and the exterior-power Lie algebra representation proves equivariance for modules. No division by $2$ is used.

For a symplectic vector space $(V,\omega )$, contraction is the map $c:{\Lambda }^{2}V\to \mathbb{C}$ defined by $c(v\wedge w)=\omega (v,w)$. It is equivariant for the symplectic Lie algebra. In dimension four its five-dimensional kernel is irreducible of highest weight ${\omega }_2$, while the invariant inverse-form bivector spans a complementary trivial Lie algebra representation.

The primitive exterior square of a symplectic vector space is the kernel of its symplectic contraction of an exterior square. In dimension four, the invariant inverse-form bivector $\Omega$ has nonzero wedge square. Choose $\mathrm{vol}=\Omega \wedge \Omega /2$; the symmetric bilinear form $\xi \wedge \eta =B(\xi ,\eta )\mathrm{vol}$ is nondegenerate on ${\Lambda }^{2}V$. The primitive subspace is ${\Omega }^{\perp }$, and therefore inherits a nondegenerate form of dimension five. The symplectic Lie algebra acts on it by infinitesimal orthogonal transformations.

A rank-$n$ tensor on a vector space $V$ is a multilinear map of $n$ vector or covector arguments, equivalently an element of an $n$-fold tensor product. Its components transform by one copy of the change-of-basis matrix for each index.

A Cartesian second-rank tensor invariant under every proper rotation preserving an unoriented axis $n$ has this form. Rotations about the axis give a planar block $aI+bJ$ and eliminate mixed axial-planar entries. A half-turn about a transverse axis reverses $J$, forcing $b=0$. If only rotations about the oriented axis are imposed, this planar antisymmetric term may survive.

The six-element dihedral group generated by the displayed rotations already forces every invariant second-rank tensor to have the same form as an axially invariant second-rank tensor. The axial threefold rotation eliminates mixed entries and leaves only a planar scalar plus a planar antisymmetric part; the transverse half-turn eliminates the latter. No smaller finite subgroup suffices: cyclic rotation groups retain an axial antisymmetric tensor, while a four-element noncyclic rotation group retains arbitrary diagonal tensors along its three mutually perpendicular half-turn axes.

A Cartesian second-rank tensor has component matrix $T$ in each orthonormal basis, with transformation $T\prime =RT{R}^T$ under an orthogonal component change $R$. Both indices transform; this is why assigning arbitrary matrices independently in different bases does not define a tensor. The determinant, trace, and Frobenius inner product with another such tensor are unchanged by these basis transformations.

A Cartesian second-rank tensor invariant under half-turns about all three coordinate axes is diagonal. Each half-turn has diagonal signs, and each off-diagonal entry changes sign under at least one of them. Invariance therefore kills all off-diagonal entries. It does not make the diagonal entries equal: that requires additional rotational invariance, as for an isotropic second-rank tensor.

A type-$(1,1)$ tensor identifies with an endomorphism of a finite-dimensional vector space. Composing two such maps produces another type-$(1,1)$ tensor, because $(X,\eta )\mapsto \eta (B(A(X)))$ is a multilinear map. In output-first components the result is $C^a{}_b=B^a{}_c{A}^c{}_b$; the contracted index implements composition, and a basis change conjugates the corresponding matrices.

Under an orthogonal coordinate transformation $R$, a Cartesian pseudotensor has the usual tensor transformation law multiplied by $\det R$. Thus proper rotation matrices act as on ordinary tensors, while an orientation-reversing change contributes an extra minus sign. Interpreted as a rank-three pseudotensor, the Levi-Civita symbol has the same numerical components in every orthonormal frame, because its ordinary three-index transformation already contributes a factor $\det R$.

A symmetric tensor is unchanged by every permutation of its slots. A symmetric second-rank tensor is the rank-two case; symmetry for higher rank requires invariance under every such slot exchange.

Over a field of characteristic zero, degree-$n$ symmetric tensors are spanned by pure powers $a^{\otimes n}$. Inclusion-exclusion expands ${\sum }_{J\subseteq [n]}(-1{)}^{n-|J|}({\sum }_{j\in J}{a}_j{)}^{\otimes n}$ into the sum of all permutation orderings of $a_1\otimes \cdots \otimes a_n$. Every surviving term uses each of the $n$ labels once. This converts an invariant-tensor spanning problem into pure tensor powers.

A second-rank tensor is symmetric when $T_{ij}=T_{ji}$. Every second-rank tensor has the unique decompositioninto symmetric and antisymmetric parts.

Suppose an array $T_{ij}$ is specified in each right-handed orthonormal basis. If $w_i=T_{ij}{v}_j$ transforms as a vector for every vector $v$, then $T$ is a second-order Cartesian tensor under those basis changes. Indeed $v^{\prime }=Rv$, $w^{\prime }=Rw$ imply $T^{\prime }Rv=RTv$ for every $v$, hence $T^{\prime }=RT{R}^T$. The condition on every test vector is essential; one contraction is insufficient.

If contraction of an array with every symmetric second-rank tensor is a vector in every orthonormal frame, its part symmetric in the contracted slots obeys the rank-three tensor transformation law. The difference between its proposed transformation and actual components is symmetric and contracts to zero against every symmetric matrix, so it vanishes. The antisymmetric part is invisible, by the blindness of symmetric contraction tests to antisymmetric arrays, and need not be tensorial.

If an array specified in each orthonormal basis has invariant Frobenius inner product with every Cartesian second-rank tensor, it transforms as such a tensor. For a component change $R$, invariance says $(R^{T}A\prime R-A):B=0$ for every matrix $B$. Nondegeneracy of the Frobenius inner product forces the difference to vanish. Testing every tensor, rather than a single selected tensor, is essential.

If an array is antisymmetric in every orthonormal basis and its contractions with every antisymmetric second-rank tensor are invariant scalars, it obeys the Cartesian second-rank tensor transformation law. The difference $D=R^{T}A\prime R-A$ is antisymmetric and orthogonal to every antisymmetric test matrix. Choosing the test matrix equal to $D$ gives $D:D=0$, hence $D=0$. The Frobenius inner product is nondegenerate on the antisymmetric subspace.

The Frobenius inner product of a symmetric matrix and an antisymmetric matrix is zero. Thus invariant contractions of an array with all symmetric second-rank tensors test only its symmetric part; an arbitrary basis-dependent antisymmetric part is invisible. A nonzero antisymmetric matrix in one basis and the zero matrix in another passes all such tests with scalar zero but is not a Cartesian second-rank tensor.

A second-rank Cartesian tensor is traceless when the contraction $T_{ii}$ vanishes. For a symmetric second-rank tensor, this is equivalent to its eigenvalues summing to zero. The orientational nematic order parameter is both symmetric and traceless, excluding an isotropic scalar contribution.

For finite sets $X,Y,Z$ and a field $F$, a three-variable tensor $T:X\times Y\times Z\to F$ has slice rank at most $r$ when it is a sum of $r$ terms, each of which separates one variable from the other two. Such terms have one of the formsThe slice rank is the least possible $r$.

A three-variable tensor is diagonal when it vanishes unless its three arguments are equal.

Ifon a finite set $X$, and every $c_x$ is nonzero, then the slice rank of $T$ is $|X|$. The upper bound uses one $x$-slice for each diagonal entry. For the lower bound, restrict any shorter slice decomposition to a common kernel of the coefficient functions in two slice directions; the remaining diagonal matrix has rank larger than the number of slices available in the third direction, a contradiction.

Covariance and contravariance describe how vector and covector components transform when a basis changes.

A covariant tensor is multilinear in vector arguments and transforms with one covariant index for each argument.

The Kronecker delta is ${\delta }_{ij}=1$ when $i=j$ and zero otherwise. It is the component matrix of the identity map in a basis.

A second-rank Cartesian tensor is antisymmetric when its components form a skew-symmetric matrix, $T_{ij}=-T_{ji}$. Orthogonal coordinate changes preserve this property because $T^{\prime }=RT{R}^T$ implies $(T^{\prime }{)}^T=R{T}^T{R}^T=-T^{\prime }$.

A tensor changes sign whenever two indices are interchanged. In three dimensions, a rank-two antisymmetric tensor is $A_{ij}=a_k{ϵ}_{kij}$, a rank-three one is $A_{ijk}=a{ϵ}_{ijk}$, and every totally antisymmetric tensor of rank greater than three vanishes.

An isotropic tensor is invariant under every proper orthogonal change of basis. A rank-four isotropic tensor has the formRank-five isotropic pseudotensors are formed by multiplying one Kronecker delta by one Levi-Civita symbol and permuting the five indices.

A second-rank Cartesian tensor invariant under every proper rotation matrix is a scalar multiple of the Kronecker delta. Half-turns about the coordinate axes kill its off-diagonal entries, and quarter-turns make its diagonal entries equal. A tensor of the form $v_i{w}_j$ is an outer product with matrix rank at most one, so it is isotropic in three dimensions only when it vanishes, equivalently when $v=0$ or $w=0$.

Einstein notation sums automatically over an index repeated once up and once down, or twice in Euclidean coordinates.

The metric trace contracts two indices of a covariant rank-two tensor: $T=g^{\mu \nu }{T}_{\mu \nu }$. For the metric itself in $D$ dimensions, $g^{\mu \nu }{g}_{\mu \nu }=D$.

For a basis $e_1,e_2,e_3$ with scalar triple product $\Delta$, the reciprocal vectors are cyclic cross products divided by $\Delta$ and satisfy $f_i\cdot e_j={\delta }_{ij}$. This construction is used for the basis of a reciprocal lattice.

The scalar triple product $a\cdot (b\times c)$ is the signed volume of the parallelepiped spanned by three vectors.

An orientation of a finite-dimensional real vector space distinguishes its ordered bases into two classes according to the sign of their change-of-basis determinant.

An oriented volume retains the sign of a determinant; reversing the orientation changes that sign.

The determinant is the alternating multilinear volume scale of a square matrix; it is nonzero exactly for invertible matrices.

The block matrix $M=\left(\begin{array}{cc}A& -B\\ B& A\end{array}\right)$ is similar over $\mathbb{C}$ to $\mathrm{diag}(A+iB,A-iB)$ via $T=\left(\begin{array}{cc}I& I\\ -iI& iI\end{array}\right)$. Therefore its determinant is $\det (A+iB)\det (A-iB)$ for arbitrary square matrices $A,B$. The argument uses a constant change-of-basis matrix and multiplicativity of the determinant.

For a differentiable invertible matrix $A(t)$, $\frac{d}{dt}\log \det A=\mathrm{tr}(A^{-1}\dot{A})$ wherever a consistent logarithm is defined. In particular a positive spatial metric tensor satisfies $h^{ij}{\dot{h}}_{ij}=\dot{h}/h$. This separates its local volume change from shape change.

For an invertible matrix $A$ and any matrix $H$,Indeed $\det (A+tH)=\det (A)\det (I+t{A}^{-1}H)$, and multilinearity shows that the linear coefficient of $\det (I+tB)$ is $\mathrm{tr}B$. At a singular matrix the derivative is instead written $\mathrm{tr}(\mathrm{adj}(A)H)$; this polynomial formula holds for all $A$.

On positive-definite matrices, the log-determinant has differential $d\log \det A=\mathrm{Tr}(A^{-1}dA)$ and is a strictly concave function.

The determinant is linear in each row and in each column when all the others are held fixed.

The determinant is multiplicative: for square matrices of the same size, $\det (AB)=\det (A)\det (B)$.

For an $n$ by $n$ matrix,

For diagonal entries $a_i$, superdiagonal entries $b_i$, and subdiagonal entries $c_i$, the leading principal determinants satisfy