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A bilinear form is alternating if $B(v,v)=0$ for every vector $v$. Expanding $B(v+w,v+w)=0$ gives $B(v,w)+B(w,v)=0$, hence $B(v,w)=-B(w,v)$. If the characteristic is not two, the converse follows from $B(v,v)=-B(v,v)$ by dividing by two. In characteristic two, the relation $B(v,w)=-B(w,v)$ is just symmetry and does not force zero self-pairings: $B(x,y)=xy$ on ${\mathbb{F}}_2$ has $B(1,1)=1$.^[ref]

The matrixgives an alternating bilinear form over every field. Over the real numbers, its value is the signed area of the parallelogram with sides $v,w$: $B((2,1),(1,3))=6-1=5$, and swapping the sides gives $-5$. Every self-pairing is zero, yet $\det J=1$, so this bilinear form is nondegenerate. Directly, pairing $(x,y)$ with $(1,0)$ gives $-y$, and pairing it with $(0,1)$ gives $x$; both vanish only when $(x,y)=0$.

In characteristic different from two, a skew-symmetric matrix of odd integer size has zero determinant, since $\det A=\det (-A^T)=(-1{)}^n\det A$. Thus an alternating bilinear form on a finite odd-dimensional vector space is a degenerate bilinear form. The stronger statement that the rank of an alternating bilinear form is even holds in all characteristics: splitting off a two-dimensional block with matrix $J$ repeatedly proves it. A nondegenerate alternating bilinear form defines a symplectic vector space, the linear setting for signed area rather than squared length.

A nondegenerate invariant symmetric bilinear form on a Lie algebra extends by translation to a metric preserved by both left and right translation. It may have indefinite signature. Its Levi-Civita connection is ${\nabla }_{X}Y=\frac{1}{2}[X,Y]$ for left-invariant vector fields, and its Ricci tensor is $-K/4$. Choosing the nondegenerate Killing form of a semisimple Lie algebra yields a canonical indefinite Einstein manifold; compact groups instead admit the positive metric $-K$.

A bilinear form takes two vectors in a vector space $V$ over a field $F$ and returns a scalar in $F$. Its defining property is separate linearity: holding either input fixed gives a linear functional of the other. For scalars $a,b$ and vectors $u,v,w$,This is a scalar-valued bilinear map; the chosen field is part of the definition.^[ref]

Separate linearity does not mean that $B$ is a linear map on the vector space $V\times V$ with componentwise vector addition and scalar multiplication. Over the real numbers, $B(2v,2w)=4B(v,w)$, whereas a linear map on the pair would give $2B(v,w)$. The simplest example is $B(x,y)=xy$ on the one-dimensional vector space $\mathbb{R}$: fixing $y=3$ gives $x\mapsto 3x$, but doubling both $x$ and $y$ quadruples their product.

A bilinear form measures how two inputs interact. The dot product measures alignment; a symmetric bilinear form can describe a quadratic form; an alternating bilinear form can measure signed area. A nondegenerate bilinear form turns each vector into a unique linear functional and so connects geometry with the dual space. The same short definition therefore supports lengths, bilinear orthogonal complements, and symplectic vector spaces, depending on the additional properties imposed.

Over the complex numbers, a bilinear form is linear in both inputs without conjugating either scalar. A Hermitian form instead is a sesquilinear form and conjugates when its inputs are exchanged. For complex matrices, the adjoint $A^{\ast }$ used with Hermitian forms and matrix C-star algebras is the conjugate transpose, whereas a bilinear form uses the ordinary transpose $A^T$. For example, $B(z,w)=zw$ has $B(i,i)=-1$, while the Hermitian form $H(z,w)=z\overline{w}$ has $H(i,i)=1$.^[ref]

A map $B:V\times W\to Z$ between vector spaces is bilinear when it is linear in either argument with the other held fixed. A scalar-valued bilinear form is the case $V=W$ and $Z$ is the scalar field. Checking bilinearity is one of the algebraic steps in verifying a real inner product; being a symmetric bilinear form and having positive squared norm impose further conditions.

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.

A bilinear form $B:X\times X\to \mathbb{R}$ on a real normed vector space is bounded when one constant $0\le C<\infty$ satisfiesThis is an upper bound on the magnitude of every pairing. It says nothing about the sign of $B(v,v)$ and does not require a symmetric bilinear form. The chosen norm is part of the hypothesis. Boundedness is equivalent to joint continuity, as shown by the boundedness criterion for a bilinear form.^[ref]

For example, on ${\mathbb{R}}^2$ with the Euclidean norm, the bilinear formsboth satisfy the upper bound with $C=1$, by the Cauchy-Schwarz inequality. Yet their quadratic forms are $B_{-}(v,v)=x^2-y^2$ and $B_0(v,v)=x^2$: the former takes negative values and the latter vanishes at the nonzero vector $(0,1)$. Thus neither is a coercive bilinear form. The latter is also a degenerate bilinear form, whereas $B_{-}$ is nondegenerate. Boundedness controls size, not positivity or loss of directions.

Every bilinear form on a real or complex finite-dimensional vector space equipped with any norm is bounded. In coordinates, write $B(u,v)=u^{T}Av$ using its matrix of a bilinear form. The Cauchy-Schwarz inequality gives $|u^{T}Av|\le \Vert A{\Vert }_2\Vert u{\Vert }_2\Vert v{\Vert }_2$, with $\Vert A{\Vert }_2$ the spectral norm; finite-dimensional equivalence of norms transfers this to the chosen norm. The conclusion concerns each fixed norm and fixed bilinear form: it does not supply a single constant for an arbitrary family of matrices or changing dimensions.

On an infinite-dimensional vector space, the upper bound is an additional condition. Equip the finitely supported sequences $c_{00}$ with the norm inherited from the l2 sequence space and setEach pairing is a finite sum, so this is a well-defined bilinear form. For the basis vector $e_n$, however, $\Vert e_n{\Vert }_2=1$ and $B(e_n,e_n)=n$; no finite $C$ works. It is even a coercive bilinear form with $\alpha =1$, since $\sum n{v}_n^2\ge \sum v_n^2$. Consequently coercivity alone does not imply boundedness on a normed vector space. This example uses $c_{00}$, which is not a Hilbert space; it does not define this pairing on all of the l2 sequence space.

A bounded bilinear form defines a bounded linear operator $T:X\to X^{\prime }$ by $(Tu)(v)=B(u,v)$, where $X^{\prime }$ is the continuous dual space. Its operator norm is the smallest admissible $C$:This converts an upper bound for pairings into an upper bound for a linear map taking an unknown vector to a linear functional.

A bilinear form on a normed vector space is continuous exactly when it is bounded. For the nontrivial direction, continuity at $(0,0)$ gives $r,s>0$ such that $|B(x,y)|\le 1$ whenever $\Vert x\Vert <r$ and $\Vert y\Vert <s$; scaling nonzero $u,v$ into these balls gives $|B(u,v)|\le 4(rs{)}^{-1}\Vert u\Vert \Vert v\Vert$.

An adjoint-invariant symmetric bilinear form $B$ on a Lie algebra makes the displayed tensor alternating. If $B$ is invariant under the full group Adjoint representation, its left-translated three-form is also right-invariant. On left-invariant fields, the exterior derivative is $d\eta (W,X,Y,Z)=-2B(W,[X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]])=0$ by the Jacobi identity. The Killing form supplies such a $B$ on every semisimple Lie group. An overall sign change in the lowered structure-constant convention negates the form but not its invariance or closedness.

A real bilinear form $B$ on a normed vector space $X$ is coercive when there is one constant $\alpha >0$ such thatThe lower bound uniformly controls every direction in the chosen norm. Unlike the upper bound defining a bounded bilinear form, it forces positive diagonal values at every nonzero vector. No symmetry is assumed. For complex Hilbert spaces, the usual analogue uses a sesquilinear form and $\mathrm{Re}B(v,v)\ge \alpha \Vert v{\Vert }^2$; it is not the same assertion for a complex bilinear form.

A concrete bounded and coercive example on ${\mathbb{R}}^2$, with the Euclidean norm, isIts quadratic form isThe best lower constant is $\alpha =2$, attained at $(1,0)$. Nevertheless, $B(e_1,e_2)=1$ and $B(e_2,e_1)=-1$, so coercivity does not imply a symmetric bilinear form. The symmetric part of a matrix $S=(A+A^T)/2=\mathrm{diag}(2,3)$ determines $q$, while the skew-symmetric matrix part changes mixed pairings without changing $q$. This is the symmetric and alternating parts of a bilinear form decomposition.^[ref]

For the upper constant, direct multiplication gives $A^{T}A=\left(\begin{array}{cc}5& -1\\ -1& 10\end{array}\right)$. Its eigenvalues are $(15\pm \sqrt{29})/2$, so the optimal upper constant, the spectral norm, is $C=\sqrt{(15+\sqrt{29})/2}$. The two constants answer different questions: how large a mixed pairing can be, and how small a self-pairing can be relative to squared norm.

The plotted quadratic forms show the mechanism: $x^2-y^2$ descends in one direction, $x^2$ leaves one direction uncontrolled, and $2x^2+3y^2$ controls both. These are self-pairings; their level sets cannot reveal the skew-symmetric matrix part. Calling $q$ a quadratic energy here refers to this diagnostic quantity, not to a claim that a nonsymmetric equation comes from minimizing an energy functional.

On a real finite-dimensional vector space with any norm, strict positivity $B(v,v)>0$ for every nonzero $v$ does imply coercivity. Indeed the quadratic form is continuous, and its positive minimum on the compact unit sphere supplies $\alpha$. If $B$ is also a symmetric bilinear form, this is positive definiteness. For general $B$, it is the symmetric and alternating parts of a bilinear form decomposition that isolates the positive part. Switching between equivalent norms preserves coercivity, but usually changes its constant.

The infinite-dimensional vector space distinction is explicit on the real l2 sequence space:The Cauchy-Schwarz inequality gives boundedness with $C=1$, and $B(v,v)>0$ for every nonzero vector. But $B(e_n,e_n)=1/n$ with $\Vert e_n{\Vert }_2=1$, so no positive $\alpha$ works. On the span of $e_1,\dots ,e_N$, the best lower constant is $1/N$: each finite restriction is coercive, but the control disappears as $N$ grows. Strict positivity is therefore weaker than a uniform lower bound.

The practical payoff is a stability estimate for a weak formulation. Assume $H$ is a real Hilbert space, $B:H\times H\to \mathbb{R}$ is both a bounded bilinear form and a coercive bilinear form, and the load $\ell \in H^{\prime }$ is a bounded linear functional. The Lax-Milgram theorem gives a unique $u\in H$ satisfying $B(u,v)=\ell (v)$ for all $v\in H$, including nonsymmetric $B$.^[ref] The dual norm $\Vert \ell {\Vert }_{H^{\prime }}={\sup }_{v\ne 0}|\ell (v)|/\Vert v\Vert$ measures load size. Testing with $v=u$ gives the short energy estimateFor $u=0$ the bound holds directly; otherwise divide by $\Vert u\Vert$. Subtracting the equations for two loads and using the same argument gives $\Vert u_1-u_2\Vert \le \Vert {\ell }_1-{\ell }_2{\Vert }_{H^{\prime }}/\alpha$. Thus a positive lower constant limits the response to load errors. The upper bound and completeness of the Hilbert space enter the existence theorem; the estimate itself uses the lower bound and the load's dual norm.

For the displayed matrix $A$, take $\ell (v)=f^{T}v$. Then $B(u,v)=\ell (v)$ for every $v$ means $A^{T}u=f$, not $Au=f$, with this ordering of the bilinear form. Since $\det A=7$,For $f=t(1,0{)}^T$, the exact response is $u=t(3,-1{)}^T/7$ and $\Vert u{\Vert }_2=\sqrt{10}|t|/7<|t|/2$. For $f=t(0,1{)}^T$, it is $u=t(1,2{)}^T/7$, with norm $\sqrt{5}|t|/7$. The actual maximal amplification is the operator norm $\Vert (A^T{)}^{-1}{\Vert }_2=\sqrt{2/(15-\sqrt{29})}<1/2$: the coercivity estimate is valid but need not be sharp for a nonsymmetric bilinear form.

A coercive zero-Dirichlet weak form gives a bounded solution map from $L^2$ into $H_0^1$. Composing with the compact Sobolev embedding into $L^2$ gives the compact inverse $K$. Transposing the bilinear form produces its Hilbert-space adjoint inverse $K^{\ast }$. The Fredholm compatibility condition for a shifted equation is orthogonality of $f$ to the homogeneous adjoint solutions. Indeed $(Kf,w)=(f,w)/\lambda$ when $K^{\ast }w=w/\lambda$, and the zero spectral parameter is excluded at resonance by coercivity.

For sufficiently regular coefficients and a function $u$, this boundary quantity is the outward flux appearing when a divergence-form principal operator is integrated by parts. Its zero condition is the natural Neumann boundary condition for the corresponding bilinear form. With only bounded coefficients and a Sobolev solution, the weak formulation can encode that natural condition without asserting that a pointwise boundary derivative exists.

Let $U\subset {\mathbb{R}}^d$ be bounded, $A\in {\mathbb{R}}^d$ satisfy $\Vert A{\Vert }_2<2$, and $f\in L^2(U)$. The bilinear formon $H_0^1(U)$ is bounded. The Cauchy-Schwarz inequality and $2ab\le a^2+b^2$ giveso it is a coercive bilinear form. The Lax-Milgram theorem therefore gives a unique weak solution of $-\Delta v+v+A\cdot \nabla v=f$ with zero Dirichlet boundary condition.

A functional on a real vector space is convex when it is a convex function: $J(\theta v+(1-\theta )w)\le \theta J(v)+(1-\theta )J(w)$ for $0\le \theta \le 1$. For a symmetric nonnegative bilinear form $a$ and a linear functional $\ell$, the quadratic functional $J(v)=\frac{1}{2}a(v,v)-\ell (v)$ is convex, since its convexity difference is $\frac{1}{2}\theta (1-\theta )a(v-w,v-w)\ge 0$.

For a Lie subalgebra $\mathfrak{g}\subseteq \mathfrak{gl}(V)$, its defining representation is its given action on the vector space $V$. The Lie bracket identity $[X,Y]v=X(Yv)-Y(Xv)$ verifies the representation property. For the special linear Lie algebra, diagonal trace-zero matrices have coordinate weights of a representation $L_i$ on the standard basis; for a Special orthogonal Lie algebra, the same action preserves the defining symmetric bilinear form.

A bilinear form is degenerate if some nonzero vector pairs to zero with every vector on at least one side. For a symmetric bilinear form or an alternating bilinear form, this says its radical of a bilinear form contains a nonzero vector. On a finite-dimensional vector space, it is equivalent to a singular matrix of a bilinear form, or $\det A=0$.

For $A=\mathrm{diag}(1,0)$, $B((x,y),(s,t))=xs$. The nonzero vector $(0,1)$ is invisible to every pairing, and changing $y$ never changes $B((x,y),w)$. The quadratic form $q=x^2$ therefore has two parallel lines $x=\pm 1$ as its unit level set. This is a lost direction, a stronger condition than being an isotropic vector: with $A=\mathrm{diag}(1,-1)$, $B((1,1),(1,1))=0$ but $B((1,1),(1,0))=1$.

A Dirichlet form on a real $L^2$ space is a densely defined closed nonnegative symmetric bilinear form with the Markov contraction property. In the real case, truncating $u$ to $(0\vee u)\wedge 1$ keeps it in the form domain and cannot increase $\mathcal{E}(u,u)$. The Dirichlet form of a Markov chain is a discrete energy example; Laplacian energy on a region supplies a continuous example.

A smooth uniformly elliptic differential expression with symmetric Dirichlet bilinear form has an unbounded self-adjoint operator realization $A_D$ on $L^2(U)$, whose domain is $H^2(U)\cap H_0^1(U)$ on a smooth bounded domain. If the form is strictly positive, the Lax-Milgram theorem and elliptic regularity make $A_D^{-1}$ a bounded map from $L^2$ to $H^2\cap H_0^1$. Its action on $L^2$ is compact by the Rellich-Kondrachov compactness theorem and self-adjoint by symmetry of the form. The spectral theorem for compact Hermitian operators therefore gives an orthonormal basis of smooth Dirichlet eigenfunctions with positive eigenvalues tending to infinity.

A bilinear form $b:U\times V\to K$ with zero left and right radicals induces injective linear maps $U\to V^{\ast }$ and $V\to U^{\ast }$. Finite dimensions force $\dim U=\dim V$, so both maps are isomorphisms. For any other bilinear form $c$, composing its maps to the dual spaces with these inverse isomorphisms gives unique endomorphisms $S,T$ with $c(x,y)=b(Sx,y)=b(x,Ty)$. In bases with invertible pairing matrix $B$ and coefficient matrix $C$, $S=B^{-T}{C}^T$ and $T=B^{-1}C$.

Let $V_h\subset V$ be a conforming finite element space. Suppose the exact solution satisfies $a(u,v)=\ell (v)$ for every $v\in V$ and its Galerkin method approximation satisfies $a(u_h,v_h)=\ell (v_h)$ for every $v_h\in V_h$, using the same bilinear form and linear functional. Subtracting the two identities gives Galerkin orthogonality:This identity itself does not require symmetry. Its interpretation as an orthogonal projection in an energy norm does require a bounded symmetric coercive bilinear form: then $a$ is an inner product, $e=u-u_h$ is perpendicular to every available discrete correction, andConsequently $u_h$ is the unique closest point in $V_h$ in the energy norm. This Pythagorean theorem in a function space explains the beauty of the Galerkin method: integrated local equations compute a global best approximation. The Céa lemma supplies the corresponding estimate in a chosen reference norm.

For the finite element solution of the one-dimensional Poisson equation, $a(w,v)={\int }_0^1{w}^{\prime }{v}^{\prime }dx$ and $\Vert w{\Vert }_a=\Vert w^{\prime }{\Vert }_{L^2(0,1)}$. Testing the error against the $i$th piecewise-linear hat function on a uniform mesh givesIn this model $e$ vanishes at every node, so each of these integrals is individually zero, although $e$ is positive inside every element. Galerkin orthogonality controls the derivative-weighted balance; it does not say that the pointwise error vanishes.

The exact identity assumes conformity and exact evaluation of the continuous forms, including any numerical integration that is used. If assembly instead uses $a_h$ and ${\ell }_h$, subtraction gives the consistency defectThus inexact numerical integration need not preserve the projection. In the Petrov-Galerkin method the error is tested against a different test space, and for a nonsymmetric bilinear form the displayed identity need not describe perpendicularity in any energy norm.

For a finite-rank Coxeter system, let $V$ have basis $(e_i{)}_{i\in I}$ and symmetric bilinear formIts geometric representation sends the generator $x_i$ to the reflection