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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 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.

For a finite-dimensional Lie algebra representation $\rho$ on $V$, the Trace form of a Lie algebra representation isWrite again $A=\rho (a)$, $B=\rho (b)$, and $C=\rho (c)=[A,B]$. The operator $C$ commutes with both $A$ and $B$. Direct use of the cyclic property of the trace givesIn the last line, cyclicity and $AC=CA$ turn $\mathrm{tr}(ABC)$ into $\mathrm{tr}(BAC)$. Thus the nonzero vector $c\in \mathfrak{g}$ is orthogonal to the basis $a,b,c$, and hence to all of $\mathfrak{g}$. The bilinear form is therefore degenerate.

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]