Bilinear Forms over a field F

Let V be a vector space. A bilinear form on V is a set map B : V × V ⟶ F which is linear in each slot. This means that for λ ∈ F and x, x′, y, y′ ∈ F we have

λB(x, y) = B(λx, y) = B(x, λy),
B(x + x′, y) = B(x, y) + B(x′, y)
B(x, y + y′) = B(x, y) + B(x, y′).

We call B symmetric if B(x, y) = B(y, x) and alternating if B(x, y) = −B(y, x).

The most famous symmetric bilinear form is the dot product B(x, y) = x · y on F^{n}. Another symmetric bilinear form is the “Lorentz metric” on ℝ^{4}: B(x, y) = x_{1}y_{1} + x_{2}y_{2} + x_{3}y_{3} − x_{4}y_{4}. The most famous alternating bilinear form is the cross product B(x, y) = x × y = x_{1}y_{2} − x_{2}y_{1} on F^{2}. Every n × n matrix A = (a_{ij}) gives rise to a bilinear form on the vector space F^{n} of column vectors by the formula B(x, y) = x^{t}Ay = x · Ay.

A variation of this definition is often used for vector spaces over ℂ. Let λ^{∗} denote the complex conjugate of λ ∈ ℂ. We call a set map B : V × V ⟶ F sesquilinear if it is linear in the second slot and anti-linear in the first; that is we replace the scalar condition for a bilinear form by the condition

λB(x, y) = B(λ^{∗}x, y) = B(x, λy)

We call B hermitian if B(x, y) = B(y, x)^{∗}. The most famous hermitian form on ℂ^{n} is given by x^{∗} · y = ∑ x_{i}^{∗}y_{i}.

Any square matrix A gives rise to a sesquilinear form: B(x, y) = x^{∗} · Ay. If A is a hermitian matrix (a matrix with A^{t} = A^{∗}) then this is a hermitian form. Our discussion will concentrate on bilinear forms, because the sesquilinar/hermitian cases are all proven the same way (with conjugation thrown in where needed).

Proposition. If dim(V) = n, there is a 1-1 correspondence between bilinear forms and n × n matrices. The symmetric and alternating bilinear forms correspond to symmetric and alternating matrices.

There is also a 1-1 correspondence between sesquilinear forms and n×n matrices, in which the hermitian forms correspond to hermitian matrices.

To make this correspondence, choose a basis e_{1}..., e_{n} for V. The matrix A = (a_{ij}) associated to a bilinear form B has a_{ij} = B(e_{i}, e_{j}). The formula B(x, y) = x^{t}Ay follows from bilinearity of B.

Change of basis. A change of basis for V is carried out by an invertible matrix P. Writing x = P x_{0}, y = P y_{0} we see that x^{t}Ay = x_{0}^{t}(P^{t}AP)y_{0}. Thus the change of basis replaces the matrix A by the matrix P^{t}AP.

Warning: the use of A to describe a linear transformation and a bilinear form result in two distinct equivalence relations on matrices: A is similar to P^{−1}AP as a linear transformation, and is equivalent to P^{t}AP as a bilinear form (or to P^{∗t}AP as a sesquilinear form.)

We call a form B non-degenerate if its corresponding matrix A has a nonzero determinant. Note that det(A) is only well-defined up to a square, since det(P^{t}AP) = det(A) det(P)^{2}; det(A) is called the discriminant of B. If B is nondegenerate, the discriminant is well-defined in F^{∗}/F^{∗2}.

Typeset by AMS-TEX

Lemma. The following are equivalent for a bilinear (or even sesquilinear) form B. (a) B is a non-degenerate form on V

(b) For x ≠ 0, the linear map f : V ⟶ F defined by f(y) = B(x, y) is nonzero.

(c) Every linear map f : V ⟶ F is f(y) = B(x, y) for some unique x ∈ V.

Quadratic Forms. A set map q : V ⟶ F is called a quadratic form if q(λx) = λ^{2}x for λ ∈ F, x ∈ V and if

B(x, y) = q(x, y) − q(x) − q(y)

is a symmetric bilinear form. B is called the associated bilinear form of q.

Every symmetric bilinear form determines a quadratic form, namely q(x) = B(x, x), whose associated bilinear form is 2B. Thus if char(F) ≠ 2 there is a 1-1 correspondence between quadratic forms and symmetric bilinear forms.

Associated to the dot product (a bilinear form) is the quadratic form q(x) = x · x = ∥x∥^{2}. This corresponds to the usual Riemannian metric on ℝ^{n}. Associated to the Lorentz metric on ℝ^{4} is the quadratic form q(x) = x_{1}^{2} + x_{2}^{2} + x_{3}^{2} − x_{4}^{2}.

Diagonalizability of symmetric and hermitian forms. Every real symmetric matrix, and every complex hermitian matrix, is diagonalizable as a linear transformation. That is, V has a basis of eigenvectors for A, and all the eigenvalues of A are real numbers.

To see this, note that if A were not diagonalizable then there would exist a Jordan block of length ≥ 2 for some eigenvalue λ. Thus there would exist a vector x which is not an eigenvector, but such that y = (A−λ)x is an eigenvector. But we must have (y^{∗} · Ax) − (x^{∗} · Ay) = y^{∗} · y equal to zero, contradiction. Finally, if x is an eigenvector for λ ≠ 0 with ∥x∥ = 1 then we have λ = x^{∗} · Ax = (x^{∗} · Ax)^{∗} = λ^{∗}. A non-diagonalizable symmetric matrix over ℂ is given below.

Classification of symmetric bilinear forms. If B is a symmetric bilinear form, and char(F) ≠ 2, then there is a basis of V, and an integer r (the rank of B) such that the matrix of B has the diagonal form

( \begin{matrix}a_{1} & & & & 0 \\ & . & . & . \\ 0 & & & & a_{r} \\ & & & & & 0 \end{matrix} ) B(x, y) = ∑_{i=1}^{r} a_{r}x_{i}y_{i},

To construct the diagonal matrix, we use the notion of orthogonality relative to B. Write x ⊥ y if B(x, y) = 0. If B = 0 we are done. Otherwise choose e_{1} such that a_{1} = B(e_{1}, e_{1}) is nonzero, and set U = e_{1}^{⊥} = {x ∈ V : x ⊥ e_{1}}. Then dim(U) = dim(V) − 1, so by induction on n = dim(V) we can pick e_{2}, ..., e_{n} to diagonalize B.

The diagonal elements a_{i} constructed by this process are not well-defined. For example, if we replace e_{i} by re_{i} in our basis, the effect is to divide x_{i}, y_{i} by r and change a_{i} to a_{i}/r^{2}. Thus over an algebraically closed field like ℂ every nondegenerate symmetric bilinear form is the dot product (up to change of basis).

Over ℝ, this only shows that we can replace each a_{i} by ±1. We will see that the number of +1′s and −1′s is an invariant, typically described using the signature of B, defined as

signature σ = (number of +1 entries) − (number of −1 entries).

Sylvester’s Law of Inertia. The rank r and signature σ of a symmetric bilinear form on V = ℝ^{n} are well-defined. If B and B_{1} are two symmetric forms with the same rank and signature, then they differ only by a change of basis matrix P: B_{1}(x, y) = B(P x, P y). If p = (r + σ)/2, there is a basis for V such that

B(x, y) = (x_{1}y_{1} + ... + x_{p}y_{p}) − (x_{p+1}y_{p+1} + ... + x_{r}y_{r}).

Indeed, if we follow the algorithm classifying symmetric forms, always choosing eigenvectors, then we see that p = (r + σ)/2 is the number of positive eigenvalues of A, and the number of negative eigenvalues of A is (r − σ)/2. As an example, the Lorentz metric on ℝ^{4} has rank 4 and signature σ = 2, so it is really different than the Riemann metric.

Classification of Hermitian forms. If B is a Hermitian form, there is a basis of V such that the matrix of B is diagonal, with the diagonal entries either 1 or 0. In particular, every nondegenerate hermitian form is (up to change of basis) the dot product x^{∗} · y.

Classification of alternating bilinear forms. If B is an alternating bilinear form, and char(F) ≠ 2, then there is a basis of V such that the matrix of B has the block form

( \begin{matrix}0 & 1 & 0 \\ −1 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} )

In particular, if B is a nondegenerate alternating form then dim(V) is even, and

B (( \begin{matrix}x_{1} \\ x_{2} \end{matrix} ) , ( \begin{matrix}y_{1} \\ y_{2} \end{matrix} )) = x_{1} · y_{2} − x_{2} · y_{1}.

Exercises. 1) A real symmetric form (or a complex hermitian form) is called positive definite when B(x, x) > 0 for all nonzero x. Show that r = σ, and that there is an invertible matrix P such that B(x, y) = (P x) · (P y).

2) Suppose that a symmetric n × n matrix A has n distinct eigenvalues λ_{i} over some field F. Show that there is a basis of eigenvectors e_{i} for F^{n} such that the associated bilinear form is described by a diagonal matrix, with a_{i} = λ_{i}e_{i} · e_{i}.

3) Let A be a real symmetric (or complex hermitian) matrix. Show that there is basis of orthonormal eigenvectors, i.e., eigenvectors e_{i} for A such that e_{i} · e_{j} = δ_{ij}.

4) Over ℂ, show that the symmetric matrix A = ( \begin{matrix}1 & i \\ i & −1 \end{matrix} ) has Jordan form

( \begin{matrix}0 & 1 \\ 0 & 0 \end{matrix} ). Then show that ℂ^{2} cannot have a basis of orthonormal eigenvectors.

Orthogonal and Unitary matrices. If B is a symmetric bilinear form over a field F, its orthogonal group is the group O_{B} of all invertible matrices P such that B(P x, P y) = B(x, y) for all x, y. When B is the dot product, we write O_{n}(F) for this group, and can also describe it as {P : P^{−1} = P^{t}}. The special orthogonal group is the group SO_{n}(F) of orthogonal matrices with determinant 1; O_{n} is the semidirect product of SO_{n} and the group {±1}.

For example, O_{1}(ℝ) = {±1}, and SO_{2}(ℝ) is the circle group of all rigid rotations in the plane. More generally, SO_{n}(ℝ) is the group of rigid rotations in the usual sense about the origin. The entire orthogonal group O_{n} includes reflections about the origin. The Lorentz group O_{31} is the orthogonal group O_{B} for the Lorentz metric B, and plays a fundamental role in physics.

The unitary group U_{n} = U_{n}(ℂ) is the corresponding group for the canonical hermitian form: a quick calculation shows that U_{n} = {P : P^{−1} = P^{∗t}}. For example, U_{1} = {z ∈ ℂ : ∥z∥ = 1}. The special unitary group is the subgroup of all unitary matrices with determinant 1; if we write S^{1} for the group of n × n matrices of the form {zI_{n} : ∥z∥ = 1}, then as a group U_{n} = S^{1} × SU_{n}.

Normal matrices. A matrix A over ℂ is called normal if AA^{∗t} = A^{∗t}A. The main theorem states that A is normal if and only if ℂ^{n} has a basis of orthonormal eigenvectors, i.e., eigenvectors e_{i} for A such that e_{i} · e_{j} = δ_{ij}.

Both real orthogonal and complex unitary matrices are trivially normal, the columns of these matrices forming an orthonormal basis. It is also clear that real symmetric and complex hermitian matrices are normal; the orthonormal bases were found in exercise 3 above. Real alternating matrices are also clearly normal, although they have purely imaginary eigenvalues and no real eigenvectors (why?).

If there is an orthonormal basis of eigenvectors u_{i} for the matrix A, the change-of-basis matrix U has the u_{i} as its columns. Since u_{i}^{∗} · u_{j} = δ_{ij}, U is a unitary matrix. Note that the matrix U^{−1}AU = U^{∗t}AU is the diagonal matrix D of eigenvectors of A. A quick calculation shows that A is normal.

Conversely, suppose that A is normal. If we write A = B + iC with B = \frac{1}{2}(A + A^{∗t}) and C = \frac{1}{2i}(A − A^{∗t}) both hermitian, then BC = CB and hence B preserves the eigenspaces of C: if Cv = λv then w = Bv satisfies

Cw = CBv = BCv = B(λv) = λw.

Hence ℂ^{n} is the direct sum of subspaces V_{ij} which are eigenspaces for both B and C. By exercise 3, each V_{ij} has an orthonormal basis diagonalizing both B and C, hence A. Concatenating these yields an orthonormal basis of ℂ^{n} (why?), and this basis diagonalizes A, B and C.