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The right radical is $\{w\in V:B(v,w)=0\text{ for every }v\in V\}$. If $A$ is the matrix of a bilinear form, this condition is $Aw=0$. For $B(v,w)=v_1{w}_2$ from the left radical example, the right radical is $\mathrm{span}\{(1,0)\}$. The left radical and right radical differ here because the bilinear form is neither symmetric nor alternating. For a symmetric bilinear form or an alternating bilinear form, reversing a zero pairing preserves zero and the two linear subspaces coincide.

A linear subspace $W\subseteq {\mathbb{F}}_2^n$ is self-orthogonal if $u\cdot v=0$ for every $u,v\in W$. The standard bilinear form is nondegenerate, whence $\dim W+\dim W^{\perp }=n$ and $\dim W\le \lfloor n/2\rfloor$. The condition with $u=v$ means every vector has even Hamming weight. Such a binary linear code gives the linear algebra behind the Eventown theorem.

A sesquilinear form on a complex vector space $V$ is a map $B:V\times V\to \mathbb{C}$ which, with the convention used throughout this article, is a linear map in its first argument and a conjugate-linear map in its second. For vectors $u,v,w$ and complex numbers $\alpha ,\beta$, the defining identities areThe bars denote complex conjugation. No symmetry or positivity is required. Some authors choose the opposite convention; every formula here keeps the first argument linear.^[ref]

For a basis $e_1,\dots ,e_n$ of a finite-dimensional vector space, write $u={\sum }_i{x}_i{e}_i$, $v={\sum }_j{y}_j{e}_j$, and $A_{ij}=B(e_i,e_j)$. Expanding by the two identities givesEvery complex matrix $A$ gives a sesquilinear form this way. The superscript $T$ means transpose, not conjugate transpose. Equivalently, the scalar is $y^{\ast }{A}^{T}x$, where $y^{\ast }={\overline{y}}^T$. This is the convention used for the matrix of a Hermitian form when symmetry is imposed. The coordinate formula follows directly by expanding the basis coefficients.^[ref]

For a concrete example on ${\mathbb{C}}^2$, let $B(z,w)=z_1\overline{w_2}$. Its matrix is $\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$: $B(e_1,e_2)=1$ but $B(e_2,e_1)=0$, so it is not a Hermitian form. Also $B((1,i),(1,i))=-i$, so even real diagonal values cannot be assumed for an arbitrary sesquilinear form. By comparison, the bilinear form $F(z,w)=z_1{w}_1+z_2{w}_2$ satisfies $F(iz,w)=F(z,iw)=iF(z,w)$ and has $F((i,0),(i,0))=-1$; it cannot supply positive squared lengths on this complex vector space.

The effect of changing the complex argument of a multiplier is particularly revealing. For real numbers $\theta ,\varphi$, sesquilinear forms satisfyThus a common change of complex argument cancels from the diagonal, whereas the two arguments rotate the output in opposite directions. Conjugating one argument is what replaces the ${\alpha }^2$ scaling of a bilinear form by the nonnegative $|\alpha {|}^2$ scaling needed for a Hermitian inner product. This mechanism also works in function spaces: for complex continuous functions on $[0,1]$, $B(f,g)={\int }_0^{1}f(t)\overline{g(t)}dt$ obeys the same identities, letting integration measure overlap rather than merely multiply coefficients.

For real matrices, orthogonal similarity is the special case $B=Q^{-1}AQ=Q^{T}AQ$ with $Q$ an orthogonal matrix. Then a symmetric matrix remains symmetric, because $B^T=Q^T{A}^{T}Q=B$. The expression $QA{Q}^T$ describes the reverse conversion and has the same property. An orthonormal basis retains the ordinary dot product formula for its coordinate vectors.

An arbitrary change of basis need not preserve entrywise symmetry. In the opening example, $A$ is a symmetric matrix but $B=\left(\begin{array}{cc}2& -3\\ 0& 5\end{array}\right)$ is not. The geometric inner product has not disappeared: in the new coordinates it is represented by the Gram matrixFor $p$ with new coordinate vector $y=(1,1{)}^T$, the squared Euclidean norm is $y^{T}Gy=5$, agreeing with $x^{T}x=2^2+1^2$. The naive coordinate sum $y^{T}y=2$ would use the wrong inner product.

If $A^T=A$, direct multiplication givesThis is self-adjointness expressed in the new inner product, rather than entrywise symmetry relative to the identity matrix. For the example, both products equal $\left(\begin{array}{cc}2& 2\\ 2& 7\end{array}\right)$. Thus arbitrary matrix similarity preserves the linear operator, while its geometric interpretation requires transforming the Gram matrix as well. Matrix congruence supplies the corresponding rule for a bilinear form and should be distinguished from matrix similarity.

A skew-coupled elliptic system has zeroth-order couplings with opposite signs. For the displayed example with homogeneous Dirichlet boundary conditions, its weak formulation on $V=H_0^1(\Omega )\times H_0^1(\Omega )$ has the bilinear formThe coupling cancels for equal arguments, so $b((p,q),(p,q))=\Vert \nabla p{\Vert }_0^2+\Vert \nabla q{\Vert }_0^2$. On a domain satisfying the Poincaré inequality, this is a coercive bilinear form. With the product first-order Sobolev norm, the Cauchy-Schwarz inequality applied to $(\nabla p,\nabla q,-q,p)$ and $(\nabla v,\nabla z,v,z)$ gives $|b(P,T)|\le \Vert P{\Vert }_V\Vert T{\Vert }_V$. The Lax-Milgram theorem therefore applies even though the bilinear form is not symmetric. Eliminating $w=-\Delta u$ recovers ${\Delta }^{2}u+u=f$, involving the biharmonic operator.

For a strictly positive Dirichlet realization of an elliptic operator with smooth coefficients and smooth boundary, put $X_0=L^2(U)$. For each integer $k\ge 1$,These are the domains $D(A_D^{k/2})$, with norms equivalent to the indicated Sobolev space norms. The bilinear forms $((u,v){)}_{2l}=(L^{l}u,L^{l}v{)}_{L^2}$ and $((u,v){)}_{2l+1}=B[L^{l}u,L^{l}v]$ are inner products on $X_{2l}$ and $X_{2l+1}$. The higher boundary conditions are essential: for $L=-d^2/d{x}^2$ and $u=x(\pi -x)$, $u$ is smooth and zero at the boundary but $L^{2}u=0$, so $\Vert L^{2}u{\Vert }_2$ is not a norm on all of $H^4\cap H_0^1$. Repeated elliptic regularity proves $A_D:X_{k+2}\to X_k$ is an isomorphism, which gives the norm equivalence inductively from $X_0$ and $X_1=H_0^1$.

The displayed inner product on the first-order Sobolev space induces its usual Sobolev norm. A continuous reaction-diffusion weak formulation with unit diffusion and unit reaction has this exact bilinear form, so its conforming Galerkin method gives a best approximation with constant one in the first-order Sobolev norm.

For the B2 root system with long simple root first, the four-dimensional spinor representation has highest weight $(0,1)$. Its symmetric square has highest weight $(0,2)$ and dimension ten, so it is the Adjoint representation. The exterior square has the four nonzero defining weights and two zero weights; it splits into the five-dimensional vector representation of highest weight $(1,0)$ and a trivial representation. Thus the tensor square decomposes as displayed. The invariant alternating bilinear form reflects the pseudoreal representation structure. The spinor representation is a group representation of the Spin group $\mathrm{Spin}(5)$, while its tensor square descends to $SO(5)$.

The stiffness matrix represents a variational bilinear form $a$ in a finite element basis. A symmetric coercive bilinear form gives a positive-definite matrix. Local support of the piecewise-linear hat functions produces a sparse matrix, allowing efficient numerical solution.

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.

A bilinear form $B$ is symmetric when $B(v,w)=B(w,v)$ for every pair of vectors $v,w$. Equivalently, its matrix of a bilinear form in any basis is a symmetric matrix, since $A_{ij}=B(e_i,e_j)=B(e_j,e_i)=A_{ji}$. Symmetry makes the relation $B(v,w)=0$ reciprocal, giving the bilinear orthogonal complement a geometric interpretation.

Over the real numbers, set $q(v)=B(v,v)$. The polarization identityrecovers the symmetric bilinear form from its quadratic form: expand $q(v+w)$ and use symmetry to combine the two cross terms. For $q(x,y)=x^2-y^2$, take $v=(2,1)$ and $w=(1,3)$. Then $q(v+w)=-7$, $q(v)=3$, and $q(w)=-8$, so $B(v,w)=(-7-3+8)/2=-1=2\cdot 1-1\cdot 3$. The same formula works whenever the characteristic is not two.^[ref]

Three real diagonal matrices show how much variety symmetry allows. The identity matrix $\mathrm{diag}(1,1)$ gives the positive-definite bilinear form $q=x^2+y^2$. The matrix $\mathrm{diag}(1,-1)$ gives $q=x^2-y^2$ and is nondegenerate despite the isotropic vector $(1,1)$. The matrix $\mathrm{diag}(1,0)$ gives $q=x^2$ and is a degenerate bilinear form with radical of a bilinear form $\mathrm{span}\{(0,1)\}$. Their level sets make the distinction visible: a circle, two branches of a hyperbola, and two parallel lines for $q=1$.

The inertia of a bilinear form records the numbers of positive, negative, and zero directions after matrix congruence diagonalization. By Sylvester's law of inertia, these counts survive change of basis; the three examples have inertia of a bilinear form $(2,0,0)$, $(1,1,0)$, and $(1,0,1)$. Historically, Sylvester stated this invariance in an 1852 paper on reducing real quadratic forms to sums of positive and negative squares. Ghys and Ranicki reproduce his statement and distinguish it from his subsequent 1853 proof.^[ref]

Construct a symplectic basis two vectors at a time. Choose $e\ne 0$; the nondegenerate bilinear form supplies $g$ with $\omega (e,g)\ne 0$, and set $f=g/\omega (e,g)$. For any vector $x$, remove its paired components byThen $\omega (x^{\prime },e)=\omega (x^{\prime },f)=0$. Thus, for $U=\mathrm{span}\{e,f\}$, the symplectic orthogonal complement gives a direct sum $V=U\oplus U^{\omega }$; its restricted bilinear form remains nondegenerate. Repeat there until no vectors remain, then put all $e_i$ before all $f_i$. This proves even dimension and existence together. Only the nonzero pairing is divided out, so the construction also works in characteristic two. The pair-splitting argument is described in Bilinear Forms by Keith Conrad (Keith Conrad, Bilinear Forms, §5, proof of Theorem 5.1 and Theorem 5.4).

The tabloid bilinear form makes the tabloid basis orthonormal and restricts to an invariant bilinear form on each Specht module. In positive characteristic of a field, its radical controls the corresponding simple quotient.

The real matrix space defined by $C+C^{\ast }=\mathrm{tr}(C)I$ consists exactly of $C=aI+i\mathbf{v}\cdot \mathbf{\sigma }$, with $a$ and the three components of $\mathbf{v}$ real. The Pauli matrices have $\mathrm{tr}({\sigma }_i{\sigma }_j)=2{\delta }_{ij}$, giving the displayed real symmetric bilinear form of signature $(3,1)$. The real isomorphism $C\mapsto aI-\mathbf{v}\cdot \mathbf{\sigma }$ to Hermitian matrices preserves this form when the latter have the polarized negative-determinant form $[\mathrm{tr}(AB)-\mathrm{tr}A\mathrm{tr}B]/2$.

For a finite field extension $L/K$, the trace pairing is the bilinear formIt is nondegenerate exactly when $L/K$ is separable.

For continuous real coefficients with $p>0$ and $q,r\ge 0$ on a bounded interval, use the clamped second-order Sobolev space and the displayed bilinear form. A forcing in $L^2$ defines $\ell (v)=\int fv$. The weak solution satisfies $a(u,v)=\ell (v)$ for every clamped test vector. Compactly supported test functions give $(p{u}^{\prime \prime }{)}^{\prime \prime }-(q{u}^{\prime }{)}^{\prime }+ru=f$ as a distribution; density extends that equation to the energy space. Continuous coefficients alone do not justify differentiating them classically. The clamped interval derivative norm bound gives coercivity, so Lax-Milgram theorem supplies a unique solution and the quadratic energy has that unique minimum.

For $Lu=-{\partial }_j(a^{ij}{\partial }_{i}u)+b^i{\partial }_{i}u+cu$ with smooth coefficients on a bounded smooth domain, the weak formulation of $(L+\lambda )u=f$ and conormal Robin boundary condition $a^{ij}{\partial }_{i}u{\nu }_j+\beta u=0$ isThe bilinear form is bounded by the Sobolev trace theorem. If $a^{ij}$ satisfy the positive uniformly elliptic operator inequality, the multiplicative trace inequality on a bounded smooth domain and the Young inequality show that the form is coercive when $\lambda$ is sufficiently large, even if $\beta$ and $c$ have negative parts. The Lax-Milgram theorem then gives a unique weak solution for every $f\in L^2(U)$.