Bilinear form

A bilinear form on a vector space $V$ over a field $F$ is a map $B:V\times V\to F$ that is linear in each argument separately.

A symmetric real bilinear form is positive-definite when $B(v,v)>0$ for every nonzero vector. It defines an inner product and the associated norm $\sqrt{B(v,v)}$. The complex Killing form is not positive-definite on the entire complex Cartan subalgebra; its restriction to the Euclidean subspace of a Cartan subalgebra is a real positive-definite form.

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 form $B$ is symmetric when $B(v,w)=B(w,v)$ for every pair of vectors $v,w$.

A bilinear form $B:X\times X\to \mathbb{R}$ on a normed vector space is bounded when some $C<\infty$ satisfies $|B(u,v)|\le C\Vert u\Vert \Vert v\Vert$ for all $u,v\in X$.

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 $|B(u,v)|\le 4(rs{)}^{-1}\Vert u\Vert \Vert v\Vert$.

A bilinear form $B$ on a normed vector space is coercive when some $\alpha >0$ satisfies $B(v,v)\ge \alpha \Vert v{\Vert }^2$ for every $v$.

A bilinear form is degenerate when its radical contains a nonzero vector.

The radical of a symmetric bilinear form $B$ isThe form is nondegenerate exactly when its radical is zero.

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.

A symplectic vector space is a vector space equipped with a nondegenerate antisymmetric bilinear form. Every finite-dimensional symplectic vector space has even dimension: splitting off one symplectic two-plane leaves a smaller nondegenerate antisymmetric space, and induction completes the decomposition.

If $T$ preserves a nondegenerate bilinear form $\omega$ that is alternating and satisfies $T^2=I$ on a finite-dimensional vector space over a field of characteristic different from two, thenThe two eigenspaces are orthogonal for $\omega$, since $\omega (v_{+},v_{-})=\omega (T{v}_{+},T{v}_{-})=-\omega (v_{+},v_{-})$. Each restricted form is nondegenerate: a vector annihilating its own eigenspace also annihilates the other and hence all of $V$. Both dimensions are even, say $2a$ and $2b$. Consequently $\mathrm{tr}T=2a-2b=\dim V-4b$ as an integer obtained from the eigenspace dimensions. Over the real numbers this gives the congruence $\mathrm{tr}T\equiv \dim V(\mathrm{mod}4)$.

The symplectic orthogonal complement of a subspace $W$ is $W^{\omega }=\{v:\omega (v,w)=0\text{ for every }w\in W\}$. The nondegenerate bilinear form gives $\dim W^{\omega }=\dim V-\dim W$. An isotropic subspace of a symplectic vector space satisfies $W\subset W^{\omega }$; a Lagrangian subspace satisfies equality.

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 symplectic basis $(e_1,\dots ,e_n,f_1,\dots ,f_n)$ has pairings $\omega (e_i,f_j)={\delta }_{ij}$ and all pairings between two $e_i$ or two $f_i$ equal to zero.

If $B_1$ is a nondegenerate bilinear form on a finite-dimensional vector space and $B_2$ is any bilinear form, there is a unique linear map $\alpha$ such thatThe right radical of $B_2$ is $\ker \alpha$.

The inertia of a real symmetric bilinear form is the triple giving the numbers of positive, negative, and zero squares in a diagonalization. Sylvester's law of inertia makes it independent of the chosen basis.

For a real symmetric bilinear form, let $n_{+}$ and $n_{-}$ be the dimensions of its positive and negative diagonal subspaces. This pair gives the first two components of the inertia of a bilinear form and is independent of basis by the Sylvester's law of inertia. For a nondegenerate form their sum is the full dimension; otherwise the nullity is an additional invariant. Another convention calls the difference $n_{+}-n_{-}$ the signature of a quadratic form. The pair convention describes the positive and negative harmonic spaces of the real de Rham intersection form in dimension four.