Inner products are symmetric bilinear forms with an extra positivity condition. We now drop positivity and classify symmetric bilinear forms themselves. Throughout this section, is a finite-dimensional vector space over a field with (characteristic appears only in a note), and is a symmetric bilinear form on .
Definition 9.6.1 (Quadratic form). A quadratic form on is a function of the form for some bilinear form on . If is a basis and is the matrix of , then a homogeneous polynomial of degree in the coordinates.
Proposition 9.6.2 (Polarization). Let . If is a symmetric bilinear form and , then Consequently, is a bijection from symmetric bilinear forms on to quadratic forms on .
Proof. Expanding, , and we may divide by . So determines , and the map is injective. For surjectivity, let for some bilinear form . Then is a symmetric bilinear form with . ∎
Example 9.6.3 (A quadratic form and its matrix). On , the quadratic form corresponds to the symmetric matrix , i.e., to the symmetric bilinear form The cross term is split equally between the two off-diagonal entries.
Note. In characteristic , polarization fails, and so does the classification below. Over , the form on is symmetric and nonzero, but for every , so it has the same quadratic form as the zero form. It also has no orthogonal basis in the sense of Theorem 9.6 below: in such a basis its matrix would be diagonal with entries , i.e., zero.
9.6.2 Orthogonal Bases
To classify symmetric bilinear forms, we look for a vector with and then pass to , as in the inductive construction of orthonormal bases.
Lemma 9.6.4 (Anisotropic vectors exist). If and is symmetric, then there is with .
Proof. Choose with . By polarization, , so at least one of , , is nonzero. ∎
Lemma 9.6.5 (Splitting off a vector). If , then , where .
Proof. The linear functional is nonzero (its value at is ), and since is symmetric its kernel is , which therefore has dimension . If , then , so ; hence , and the dimensions add up to . Explicitly, . ∎
Theorem 9.6.6 (Orthogonal bases for symmetric forms). Let and let be a symmetric bilinear form on a finite-dimensional space . Then has an orthogonal basis for : a basis with for . Equivalently, for every symmetric matrix there is an invertible such that is diagonal.
Proof. Induction on . If or , any basis works. Otherwise, Lemma 9.6.4 gives with , and Lemma 9.6.5 gives with of dimension . The restriction of to is symmetric and bilinear, so by induction has a basis with for . Also for , since . So is an orthogonal basis.
For the matrix statement, apply this to the form on and let be the matrix whose columns are the ; by Proposition 9.1.6, is the (diagonal) matrix of in the basis . ∎
Note. In an orthogonal basis the matrix of is with . So the rank of is the number of nonzero , and is nondegenerate iff every . The themselves are not determined by : replacing by () replaces by . The classification problem is to decide how far the can be normalized, and this depends on which elements of are squares.
9.6.3 Classification over and over
To state the classification, we say that a bilinear form on and a bilinear form on are equivalent if there is a linear isomorphism with for all . By Proposition 9.1.6, two forms on the same space are equivalent iff their matrices in a fixed basis are congruent; and every form on an -dimensional space is equivalent, via a coordinate isomorphism, to the form on , where is its matrix in some basis.
Over , every number is a square, so everything can be normalized.
Theorem 9.6.7 (Symmetric forms over ). Let be a nondegenerate symmetric bilinear form on a finite-dimensional complex vector space . Then has a basis with i.e., the matrix of is the identity. In particular, any two nondegenerate symmetric bilinear forms on complex vector spaces of the same finite dimension are equivalent.
Proof. Let be an orthogonal basis (Theorem 9.6.6) and , which is nonzero because is nondegenerate. Choose with , and set . Then , and for . ∎
Note. Over , symmetric bilinear forms have little to do with geometry: the form on (whose matrix is ) has for . The forms that measure lengths over are the Hermitian ones. The same proof shows that a possibly degenerate symmetric form on a complex space has, in a suitable basis, the matrix , where is its rank.
Over , only nonnegative numbers are squares, so the best we can do is normalize each to .
Theorem 9.6.8 (Symmetric forms over ). Let be a nondegenerate symmetric bilinear form on a real vector space of dimension . Then there is an integer with and a basis with for , for , and for . That is,
Proof. Let be an orthogonal basis with , and set . Then . Reorder the basis so that the ’s come first. ∎
The number does not depend on the basis. This is Sylvester’s law of inertia.
Theorem 9.6.9 (Sylvester’s law of inertia). In Theorem 9.6.8, is the largest dimension of a subspace on which is positive-definite ( for all nonzero ), and is the largest dimension of a subspace on which is negative-definite. In particular, is the same for every basis as in Theorem 9.6.8.
Proof. Let For , unless , so is positive-definite on , a subspace of dimension .
Now let be any subspace on which is positive-definite. For , . A nonzero vector of would satisfy both and ; so . Hence and . So is the largest dimension of a positive-definite subspace, which is defined without reference to any basis. Applying this to the form , whose basis as in Theorem 9.6.8 is , shows that is the largest dimension of a negative-definite subspace. ∎
Definition 9.6.10 (Signature). The signature of a nondegenerate symmetric bilinear form on a real vector space of dimension is the pair with , where and are as in Sylvester’s law of inertia.
Corollary 9.6.11 (Classification over ). Let and be nondegenerate symmetric bilinear forms on real vector spaces and of the same finite dimension. Then and are equivalent if and only if they have the same signature.
Proof. If both have signature , take bases of for and of for as in Theorem 9.6.8, and let be the linear isomorphism with ; then for all , hence for all by bilinearity. Conversely, if is an isomorphism with , it maps subspaces on which is positive-definite to subspaces of the same dimension on which is positive-definite, and does the reverse; so the largest such dimensions agree. ∎
Example 9.6.12 (Signatures).
- An inner product on has signature ; conversely, a form of signature is positive-definite. So Theorem 9.6.8 with recovers the existence of orthonormal bases.
- The Minkowski form on has signature .
- The quadratic form on comes from . In the basis , , we have , , and . So the signature is ; in coordinates, .
Note. Two practical remarks. First, by the real spectral theorem, a real symmetric matrix can be diagonalized by a matrix with , so is simultaneously the matrix of the operator and of the form in the new basis. Hence the signature of is (number of positive eigenvalues of , number of negative eigenvalues of ). Second, since , the sign of is . For degenerate forms the same argument as in Sylvester’s law shows that the numbers and of ’s and ’s in a diagonal form are well defined, with .
Over , much less can be normalized, because few rational numbers are squares, and number theory enters. The following example shows that the values a form takes can distinguish two forms that look alike.
Example 9.6.13 (Two forms over ). Let and be the forms on with matrices so and . Both are nondegenerate, and over both have signature .
takes the value : .
does not take the value : suppose with . Clearing denominators, with integers , , and we may assume they have no common factor. Squares of integers are or , so , while . Hence both sides are . Then is even (an odd would give ), and forces and to be even. This contradicts the assumption that they have no common factor.
So and are not equivalent over : if were a linear isomorphism with , then , which is impossible.