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A bilinear form is alternating if $B(v,v)=0$ for every vector $v$. Expanding $B(v+w,v+w)$ implies $B(v,w)=-B(w,v)$. The converse requires characteristic different from two. In that characteristic a skew-symmetric matrix of odd integer size has zero determinant, since $\det A=\det (-A^T)=(-1{)}^n\det A$. Thus an alternating form on a finite odd-dimensional vector space is degenerate. A nondegenerate alternating form defines a symplectic vector space.

A scalar-valued map $b:V\times W\to F$ is bilinear when it is linear in either argument with the other held fixed. A bilinear form is the case $V=W$. 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 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 $|B(u,v)|\le 4(rs{)}^{-1}\Vert u\Vert \Vert v\Vert$.

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 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.

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

Represent one-dimensional homology classes on a closed oriented topological surface by transverse oriented curves and sum their signed crossing points. This gives a bilinear form that is skew-symmetric. A symplectic handle basis has matrix $\left(\begin{array}{cc}0& I_g\\ -I_g& 0\end{array}\right)$, which is unimodular. In particular, a class detected with intersection $1$ or $-1$ by another integral class is a primitive homology class: the detecting homomorphism splits its rank-one subgroup.

For a symmetric bilinear form on a Lie algebra, invariance is the displayed identity. It equivalently says that each Adjoint representation endomorphism is skew-adjoint with respect to $B$. The Killing form has this property by cyclic invariance of the trace.

A subspace $W$ of a symplectic vector space $(V,\Omega )$ is isotropic when $\Omega {|}_{W\times W}=0$. It has at most half the dimension of $V$; equality makes it a Lagrangian subspace. This use of isotropic concerns an alternating bilinear form, rather than an isotropic quadratic form.

A bilinear form on a Lie algebra representation is invariant when it satisfies the displayed identity. Equivalently, $T_B(v)=B(v,-)$ is an intertwining operator from the representation to its dual Lie algebra representation. For a finite-dimensional irreducible representation over an algebraically closed field, any nonzero such form is nondegenerate, and the Schur lemma makes all invariant forms proportional. In characteristic different from two, transposing twice then proves that the form is a symmetric bilinear form or an alternating bilinear form.

A bilinear form $B$ on a finite-dimensional vector space is nondegenerate whenequivalently when $w\mapsto B(-,w)$ is an isomorphism from $V$ to its dual space.