Bilinear forms, congruence and canonical geometry

All spaces are finite-dimensional over the stated field. A bilinear form is linear in both arguments; it is not a Hermitian inner product over $\mathbb C$.

1. A form changes by congruence, not similarity

A bilinear form on $V$ over $F$ is a map $B:V\times V\to F$ linear separately in both variables. For a basis $e_1,\ldots,e_n$, set $A_{ij}=B(e_i,e_j)$. Then in column coordinates

$$B(v,w)=[v]^{\mathsf T}A[w].$$

If old and new coordinates satisfy $[v]=P[v]'$, the new matrix is

$$A'=P^{\mathsf T}AP.$$

This is congruence. For an operator, the basis-change law is $P^{-1}AP$; the problems are different. Eigenvalues are invariants of similarity but generally not congruence. For example the real one-dimensional forms $[1]$ and $[4]$ are congruent by $P=[2]$ but do not have equal eigenvalues.

The map $B^\flat:V\to V^*$ defined by $B^\flat(v)(u)=B(u,v)$ is linear and has matrix $A$ in the chosen basis and its dual. The form is nondegenerate if this map is an isomorphism, equivalently $A$ is invertible. For a nonsymmetric form there are distinct left and right kernels, though their dimensions agree by rank equality. For a symmetric or alternating form, their underlying radical subspace agrees; write it as

$$\operatorname{rad}B=\{v:B(v,w)=0\text{ for all }w\}.$$

Rank and radical dimension are basis independent. On the quotient by the radical, such a form induces a well-defined nondegenerate form: changing a representative by a radical vector changes neither value.

2. Symmetric bilinear forms and quadratic forms

A form is symmetric if $B(u,v)=B(v,u)$, equivalently $A^{\mathsf T}=A$. It defines a quadratic form $q(v)=B(v,v)$. If $\operatorname{char}F\ne2$, polarization recovers the form:

$$B(u,v)=\frac{q(u+v)-q(u)-q(v)}2.$$

Thus in this characteristic the symmetric form and its quadratic form carry the same information. They do not in characteristic two.

Diagonalization by congruence. Over any field of characteristic different from two, a symmetric bilinear form has a basis in which its matrix is diagonal, possibly with zeros.

Proof. Induct on dimension. If $B=0$, every basis works. Otherwise some $v$ has $B(v,v)\ne0$: if every diagonal value were zero, polarization would make every $B(u,v)$ zero. For this $v$, every $w$ decomposes uniquely as

$$w=\frac{B(w,v)}{B(v,v)}v+\left(w-\frac{B(w,v)}{B(v,v)}v\right),$$

where the second term is perpendicular to $v$ for $B$. The sum is direct since $B(v,v)\ne0$. Restrict $B$ to this perpendicular subspace and apply induction. This argument allows the restriction to be degenerate and therefore proves the general statement. ∎

Worked example: completing squares. For $q(x,y)=2x^2+4xy-y^2$ over $\mathbb R$,

$$q=2(x+y)^2-3y^2.$$

The new coordinates $s=x+y,t=y$ show one positive and one negative square. This is a congruence change, not an eigenvalue computation. The associated symmetric matrix is $\begin{pmatrix}2&2\\2&-1\end{pmatrix}$ because the mixed coefficient is twice its off-diagonal entry.

Over $\mathbb C$, every nonzero diagonal coefficient has a square root, so scaling basis vectors reduces a symmetric form to $\operatorname{diag}(I_r,0)$. Its rank $r$ completely determines its congruence class. Over $\mathbb R$, signs cannot all be removed by real scaling; over $\mathbb Q$, even positive coefficients need not be rational squares. The coefficient field is part of the classification.

3. Sylvester's law of inertia over the reals

Rescaling the diagonal entries of a real symmetric form gives

$$q= x_1^2+\cdots+x_{p_+}^2-y_1^2-\cdots-y_{p_-}^2+0\cdot z_1^2+\cdots+0\cdot z_{p_0}^2.$$

The triple $(p_+,p_-,p_0)$ is its inertia and is independent of all basis choices.

Proof of uniqueness. In this representation, the positive coordinate subspace has dimension $p_+$ and the quadratic form $q$ is positive definite on it. No subspace of dimension greater than $p_+$ can carry a positive definite restriction: it must intersect the coordinate subspace where all positive coordinates vanish, whose dimension is $p_-+p_0$, because their dimensions sum to more than $n$. A nonzero vector in that intersection has quadratic value at most zero, a contradiction. Thus $p_+$ is intrinsically the largest possible dimension of a positive definite subspace. Similarly $p_-$ is the largest dimension for a negative definite restriction. Then $p_0=n-p_+-p_-$, equivalently the radical dimension. ∎

A form is positive definite exactly when $(p_+,p_-,p_0)=(n,0,0)$, negative definite when $(0,n,0)$, and indefinite when both $p_+,p_->0$. Positive semidefinite allows zeros but no negative squares. Nondegeneracy means $p_0=0$, not positivity.

For a real symmetric matrix, orthogonal spectral diagonalization is a particular congruence since $Q^{-1}=Q^{\mathsf T}$. It follows that inertia counts positive, negative and zero eigenvalues with multiplicity. General congruence preserves those counts, not the numerical eigenvalues themselves.

An isotropic vector is a nonzero $v$ with $q(v)=0$. It need not belong to the radical. In $q(x,y)=x^2-y^2$, the vector $(1,1)$ is isotropic, but $B((1,1),(1,0))=1$. Confusing isotropy with degeneracy loses essential geometry.

4. Alternating forms and symplectic pairs

A bilinear form is alternating if $B(v,v)=0$ for every $v$. Expanding $B(u+v,u+v)$ shows $B(u,v)=-B(v,u)$ in every characteristic. Conversely skew symmetry implies alternation only when the characteristic is not two; in characteristic two, diagonal entries must still be checked.

Alternating canonical form. There is a basis in which an alternating form is a direct sum of blocks

$$H=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$$

and a zero block. Consequently its rank is even, and a nondegenerate alternating space has even dimension.

Proof. If the form is zero, stop. Otherwise choose $u,v$ with $B(u,v)\ne0$ and rescale $v$ so $B(u,v)=1$. On $E=\operatorname{span}(u,v)$ the form has matrix $H$ and is nondegenerate. For every $w$, put

$$h=B(w,v)u-B(w,u)v.$$

Direct evaluation gives $B(w-h,u)=B(w-h,v)=0$. Thus $V=E\oplus E^\perp$, with a direct sum because the form on $E$ is nondegenerate. Restrict to the complement and induct. The remaining zero block is precisely the radical. This calculation uses no division by two and remains valid in characteristic two, where $-1=1$. ∎

A basis giving only the $H$ blocks in the nondegenerate case is a symplectic basis. It pairs directions rather than making vectors unit length; no positive norm is associated with an alternating form. The symmetric and alternating classifications should therefore not be described as variants of Gram–Schmidt without their different invariants and hypotheses.

5. Characteristic two is a genuine boundary, not a technical footnote

Over a field of characteristic two, the symmetric matrix $\begin{pmatrix}0&1\\1&0\end{pmatrix}$ is nonzero and nondegenerate, yet $q(x,y)=2xy=0$ for all vectors. Thus $B(v,v)$ does not determine a symmetric bilinear form, polarization by $1/2$ is undefined, and the diagonalization proof in §2 fails at its first step. Alternating canonical form still works. Whenever a theorem uses completing squares, first check that the scalar two is invertible.

6. Exercises

P01 — Foundation. Write the symmetric matrix of $q(x,y,z)=x^2+4xy+3y^2-2z^2$ over $\mathbb R$.

P02 — Core. Find the inertia and radical of the form in P01 by completing squares.

P03 — Core. Are $\operatorname{diag}(1,-1)$ and $\operatorname{diag}(2,-3)$ congruent over $\mathbb R$? Are they similar?

P04 — Core. For $B$ on $\mathbb R^3$ with matrix $\begin{pmatrix}0&1&0\\-1&0&0\\0&0&0\end{pmatrix}$, find its radical and induced nondegenerate quotient form.

P05 — Diagnosis. Give a nondegenerate symmetric form with an isotropic nonzero vector. Explain why the vector is not radical.

P06 — Proof. Prove that a nondegenerate alternating form cannot exist on an odd-dimensional space, including characteristic two.

P07 — Core. Over $\mathbb C$, show that the symmetric forms with matrices $\operatorname{diag}(1,-1)$ and $I_2$ are congruent.

P08 — Proof. For a symmetric form with radical $R$, verify that $\bar B(v+R,w+R)=B(v,w)$ is well-defined and nondegenerate on $V/R$.

P09 — Extension. Over $\mathbb Q$, prove that the one-dimensional forms $[1]$ and $[2]$ are not congruent, although they are congruent over $\mathbb R$.

7. Full solutions

P01. The matrix is $\begin{pmatrix}1&2&0\\2&3&0\\0&0&-2\end{pmatrix}$. The off-diagonal two produces the coefficient $4xy$ after symmetric expansion.

P02. Completing squares gives $q=(x+2y)^2-y^2-2z^2$. Thus inertia is $(1,2,0)$ and the radical is zero. In particular the form is indefinite and nondegenerate.

P03. They are congruent: $P=\operatorname{diag}(\sqrt2,\sqrt3)$ gives $P^{\mathsf T}\operatorname{diag}(1,-1)P=\operatorname{diag}(2,-3)$. They are not similar because their eigenvalue multisets differ (and their traces differ).

P04. The radical is $\operatorname{span}(e_3)$. In the quotient, $e_1+R,e_2+R$ form a basis with matrix $H$, a nondegenerate alternating form of rank two.

P05. For $B((x,y),(s,t))=xs-yt$, the matrix $\operatorname{diag}(1,-1)$ is invertible. The vector $(1,1)$ has quadratic value zero but pairs to one with $(1,0)$, so it is not in the radical.

P06. Alternating canonical form has rank twice the number of $H$ blocks. Nondegeneracy requires rank equal to dimension, which must therefore be even. Unlike a proof using $\det A=(-1)^n\det A$, this proof does not divide by two and remains valid in characteristic two.

P07. Let $P=\operatorname{diag}(1,i)$. Then $P^{\mathsf T}\operatorname{diag}(1,-1)P=I_2$, since $-i^2=1$. The transpose is not conjugate transpose; with the latter this would be a different form problem.

P08. Replacing $v,w$ by $v+r,w+s$ adds terms $B(r,w)+B(v,s)+B(r,s)$, all zero because symmetry makes radical vectors vanish in either argument. If $v+R$ pairs to zero with every quotient class, then $B(v,w)=0$ for every $w$, so $v\in R$ and its quotient class is zero. This is nondegeneracy.

P09. A one-dimensional rational basis change is multiplication by a nonzero $p\in\mathbb Q$, so congruence would require $p^2=2$. Writing $p=a/b$ in lowest terms yields $a^2=2b^2$, forcing both integers even, a contradiction. Over $\mathbb R$, multiplication by $\sqrt2$ supplies the congruence.

8. Summary

Bilinear forms correspond to linear maps into the algebraic dual, with congruence as their basis-change law. Symmetric forms diagonalize when two is invertible; real signs give inertia, while complex squares leave rank. Alternating forms decompose into two-dimensional pairs in every characteristic. The field, radical and distinction between transpose and conjugate transpose are indispensable parts of every classification.