--- title: Bilinear Form area: Linear Algebra Foundations type: chapter --- Linear algebra begins with linear maps — functions from a vector space to a field that respect addition and scaling. But many of the most important structures in mathematics arise not from maps on one vector space but from maps on *pairs* of vectors. The dot product on $\mathbb{R}^n$, the determinant of a $2 \times 2$ matrix formed from two column vectors, the pairing between a vector space and its dual, the symplectic form of classical mechanics, the Killing form on a Lie algebra — all of these are examples of a single notion: a bilinear form. What makes a bilinear form more than just a function of two variables is the requirement that it be linear in each variable separately. This structural constraint is strong enough to reduce all questions about bilinear forms to questions about matrices, yet flexible enough to capture wildly different phenomena depending on whether the form is symmetric, skew-symmetric, nondegenerate, or positive definite. The classification of bilinear forms is one of the gems of linear algebra. Over the real numbers, the story of symmetric forms ends with Sylvester's Law of Inertia: up to a change of basis, every symmetric bilinear form is determined by two numbers — the number of positive and negative entries in its diagonal representation. Over the complex numbers, the story is even simpler. Over finite fields, it is different again. These classification results are structurally deep, and they will occupy much of this chapter. [example: The Dot Product as a Bilinear Form] The standard inner product on $\mathbb{R}^n$ is \begin{align*} B: \mathbb{R}^n \times \mathbb{R}^n &\to \mathbb{R} \\ (v, w) &\mapsto v \cdot w = \sum_{i=1}^n v_i w_i. \end{align*} Fix any vector $v \in \mathbb{R}^n$ and consider the map $w \mapsto B(v, w) = \sum_{i=1}^n v_i w_i$. This is a linear function of $w$: if $w = \alpha w_1 + \beta w_2$, then \begin{align*} B(v, \alpha w_1 + \beta w_2) &= \sum_{i=1}^n v_i(\alpha (w_1)_i + \beta (w_2)_i) = \alpha \sum_{i=1}^n v_i (w_1)_i + \beta \sum_{i=1}^n v_i (w_2)_i \\ &= \alpha B(v, w_1) + \beta B(v, w_2). \end{align*} The same calculation shows linearity in $v$ when $w$ is fixed. So the dot product is linear in each argument separately. Moreover $B(v, w) = B(w, v)$ for all $v, w$ — it is symmetric. And if $B(v, w) = 0$ for all $w \in \mathbb{R}^n$, then in particular $B(v, v) = \sum_{i=1}^n v_i^2 = 0$, so $v = 0$. This is nondegeneracy. Now consider what happens if we replace $\sum_{i=1}^n v_i w_i$ by $\sum_{i=1}^n (-1)^i v_i w_i$. The result is still bilinear and symmetric, but no longer positive definite: the vector $e_1 = (1, 0, \ldots, 0)$ satisfies $B(e_1, e_1) = -1 < 0$. The bilinear structure is the same; the geometry is completely different. This shows why the definition abstracts away from the specific formula to the algebraic properties alone. [/example] ## Definition Before giving the formal definition, it is worth asking: why should we study maps of two variables that are linear in each separately, rather than maps of two variables that are jointly linear? A map $T: V \times W \to \mathbb{R}$ is jointly linear if $T(v_1 + \lambda v_2, w_1 + \mu w_2) = T(v_1, w_1) + \lambda T(v_2, w_1) + \mu T(v_1, w_2) + \lambda\mu T(v_2, w_2)$... but that equation is not linear: the term $\lambda\mu T(v_2, w_2)$ is quadratic in the scalars. Joint linearity in that sense is not the right notion. The correct notion is *separate* linearity in each variable, which gives an algebraically tractable theory while capturing the examples we care about. [definition: Bilinear Form] Let $V$ be a vector space over a field $k$. A **bilinear form** on $V$ is a map \begin{align*} B: V \times V &\to k \end{align*} such that for all $u, v, w \in V$ and all $\lambda \in k$: \begin{align*} B(u + \lambda v, w) &= B(u, w) + \lambda B(v, w), \\ B(u, v + \lambda w) &= B(u, v) + \lambda B(u, w). \end{align*} That is, $B$ is linear in its first argument and linear in its second argument. [/definition] [remark: Bilinear Forms over Different Fields] The definition works over any field $k$, but the classification theory depends heavily on which field we are working over. Over $\mathbb{R}$, the theory of symmetric forms is governed by Sylvester's law of inertia. Over $\mathbb{C}$, every nondegenerate symmetric form is equivalent to the standard one. Over $\mathbb{F}_q$, the classification involves quadratic residues. In this chapter, we work primarily over $\mathbb{R}$ and $\mathbb{C}$, mentioning the general case where the results are field-independent. [/remark] The most natural way to study bilinear forms is through their matrix representations. Once we choose a basis, a bilinear form becomes a matrix, and a change of basis transforms the matrix by a congruence operation — not conjugation. This is a crucial distinction from the theory of linear maps. [definition: Matrix of a Bilinear Form] Let $V$ be a finite-dimensional vector space over $k$ with ordered basis $\mathcal{B} = (e_1, \ldots, e_n)$. The **matrix of $B$ with respect to $\mathcal{B}$** is the $n \times n$ matrix $A$ with entries \begin{align*} A_{ij} := B(e_i, e_j), \qquad 1 \le i, j \le n. \end{align*} Given vectors $v = \sum_{i=1}^n v_i e_i$ and $w = \sum_{j=1}^n w_j e_j$ with coordinate vectors $[v]_\mathcal{B} = (v_1, \ldots, v_n)^\top$ and $[w]_\mathcal{B} = (w_1, \ldots, w_n)^\top$, we have \begin{align*} B(v, w) = [v]_\mathcal{B}^\top A\, [w]_\mathcal{B} = \sum_{i,j=1}^n v_i A_{ij} w_j. \end{align*} [/definition] [explanation: Congruence vs. Conjugation] When a linear map $T: V \to V$ is represented by a matrix $A$ in basis $\mathcal{B}$, then in a new basis $\mathcal{B}' = P\mathcal{B}$ (where $P$ is the change-of-basis matrix), the new matrix is $P^{-1}AP$. This is called *conjugation* or *similarity*. For a bilinear form, the transformation law is different. If $\mathcal{B}'$ is a new basis with $[v]_{\mathcal{B}} = P[v]_{\mathcal{B}'}$, then \begin{align*} B(v, w) = [v]_\mathcal{B}^\top A [w]_\mathcal{B} = (P[v]_{\mathcal{B}'})^\top A (P[w]_{\mathcal{B}'}) = [v]_{\mathcal{B}'}^\top (P^\top A P) [w]_{\mathcal{B}'}. \end{align*} So the new matrix is $P^\top A P$, not $P^{-1}AP$. This operation — replacing $A$ by $P^\top A P$ — is called *congruence*. Two matrices $A$ and $A'$ are congruent if there exists an invertible $P$ with $A' = P^\top A P$. The distinction matters because congruence and similarity are genuinely different equivalence relations: two matrices can be similar without being congruent, and congruent without being similar. The theory of bilinear forms is the theory of matrix congruence. [/explanation] The space of all bilinear forms on $V$ is itself a vector space. If $B_1$ and $B_2$ are bilinear forms and $\lambda \in k$, then $B_1 + \lambda B_2$ defined by $(B_1 + \lambda B_2)(v, w) = B_1(v, w) + \lambda B_2(v, w)$ is again a bilinear form. The matrix of $B_1 + \lambda B_2$ is $A_1 + \lambda A_2$, so the map $B \mapsto A$ (choice of basis fixed) is a linear isomorphism between bilinear forms on $V$ and $n \times n$ matrices. The space of bilinear forms on an $n$-dimensional space is therefore $n^2$-dimensional. ## Symmetry and Skew-Symmetry Not all bilinear forms are created equal. The most important structural property a bilinear form can have is a symmetry condition relating $B(v, w)$ and $B(w, v)$. There are three fundamental cases, and they lead to very different theories. The dot product satisfies $B(v, w) = B(w, v)$ for all $v, w$. The determinant of the $2 \times 2$ matrix with columns $v$ and $w$ satisfies $B(v, w) = -B(w, v)$. Most bilinear forms satisfy neither. Before defining these formally, note that any bilinear form $B$ can be decomposed uniquely as $B = B_+ + B_-$ where \begin{align*} B_+(v, w) = \frac{B(v,w) + B(w,v)}{2}, \qquad B_-(v, w) = \frac{B(v,w) - B(w,v)}{2}. \end{align*} The form $B_+$ satisfies $B_+(v,w) = B_+(w,v)$ (it is symmetric) and $B_-$ satisfies $B_-(v,w) = -B_-(w,v)$ (it is skew-symmetric). This decomposition shows that every bilinear form is a sum of a symmetric part and a skew-symmetric part, and this decomposition is unique. Understanding symmetric and skew-symmetric forms separately therefore suffices to understand all forms. [definition: Symmetric Bilinear Form] A bilinear form $B: V \times V \to k$ is **symmetric** if \begin{align*} B(v, w) = B(w, v) \qquad \text{for all } v, w \in V. [/definition] \end{align*} In terms of the matrix $A$ of $B$ with respect to any basis, $B$ is symmetric if and only if $A = A^\top$. [definition: Skew-Symmetric Bilinear Form] A bilinear form $B: V \times V \to k$ is **skew-symmetric** (or **alternating**) if \begin{align*} B(v, w) = -B(w, v) \qquad \text{for all } v, w \in V. \end{align*} In terms of the matrix $A$, this is the condition $A = -A^\top$. [/definition] Note that setting $v = w$ gives $B(v, v) = -B(v, v)$, so $2B(v,v) = 0$. Over a field with $\operatorname{char}(k) \neq 2$ (in particular over $\mathbb{R}$ or $\mathbb{C}$), this forces $B(v, v) = 0$ for all $v$. [remark: Alternating vs. Skew-Symmetric] In characteristic $2$, a form can satisfy $B(v,w) = B(w,v)$ (symmetric) and $B(v,v) = 0$ (alternating) simultaneously, since $-1 = 1$. In characteristic not equal to $2$, the conditions $B(v,w) = -B(w,v)$ and $B(v,v) = 0$ are equivalent. On this page we always work over $\mathbb{R}$ or $\mathbb{C}$, so the two conditions coincide. [/remark] [example: The Standard Symplectic Form] On $\mathbb{R}^{2n}$ with standard basis $e_1, \ldots, e_n, f_1, \ldots, f_n$, define the **standard symplectic form** $\omega$ by \begin{align*} \omega(e_i, e_j) &= 0, & \omega(f_i, f_j) &= 0, & \omega(e_i, f_j) &= \delta_{ij}, & \omega(f_i, e_j) &= -\delta_{ij}. \end{align*} The matrix of $\omega$ in the ordered basis $(e_1, \ldots, e_n, f_1, \ldots, f_n)$ is the $2n \times 2n$ block matrix \begin{align*} \Omega = \begin{pmatrix} 0 & I_n \\ -I_n & 0 \end{pmatrix}, \end{align*} where $I_n$ is the $n \times n$ identity matrix. Since $\Omega^\top = -\Omega$, the form $\omega$ is skew-symmetric. To check that it is nondegenerate: if $\omega(v, w) = 0$ for all $w \in \mathbb{R}^{2n}$, write $v = \sum_{i=1}^n a_i e_i + b_i f_i$. Setting $w = f_j$ gives $\omega(v, f_j) = \sum_{i=1}^n a_i \omega(e_i, f_j) = a_j = 0$. Setting $w = e_j$ gives $\omega(v, e_j) = \sum_{i=1}^n b_i \omega(f_i, e_j) = -b_j = 0$. So $v = 0$, confirming nondegeneracy. This form is the foundation of symplectic geometry: Hamiltonian mechanics takes place on symplectic manifolds, and the standard symplectic form on $\mathbb{R}^{2n}$ is the local model for all of them. [/example] The classification of skew-symmetric forms is remarkably clean: every nondegenerate skew-symmetric form looks like the standard symplectic form above. In particular, nondegenerate skew-symmetric forms can only exist on even-dimensional spaces. [quotetheorem:3295] This theorem says there is essentially one nondegenerate skew-symmetric form in each even dimension — there is no further invariant like a signature. The theory of symmetric forms is far richer. ## Nondegeneracy and the Radical A bilinear form that is identically zero is bilinear, but it is also completely useless: it carries no information about the vectors. The key property that makes a bilinear form geometrically useful is *nondegeneracy* — the form separates distinct vectors. To understand what can go wrong, consider the form $B: \mathbb{R}^2 \times \mathbb{R}^2 \to \mathbb{R}$ with matrix \begin{align*} A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}. \end{align*} Then $B(v, w) = v_1 w_1$. The second coordinate is completely invisible to $B$: $B((0,1), w) = 0$ for every $w$. Any theorem that tries to invert $B$ or use it to define a dual pairing will fail on the vector $(0,1)$. [definition: Radical of a Bilinear Form] Let $B: V \times V \to k$ be a bilinear form. The **left radical** of $B$ is \begin{align*} \operatorname{rad}_L(B) = \{ v \in V : B(v, w) = 0 \text{ for all } w \in V \}, \end{align*} and the **right radical** is \begin{align*} \operatorname{rad}_R(B) = \{ w \in V : B(v, w) = 0 \text{ for all } v \in V \}. \end{align*} When $B$ is symmetric or skew-symmetric, the left and right radicals coincide, and we write $\operatorname{rad}(B)$ for their common value. [/definition] [definition: Nondegenerate Bilinear Form] A bilinear form $B: V \times V \to k$ is **nondegenerate** if $\operatorname{rad}_L(B) = \{0\}$ and $\operatorname{rad}_R(B) = \{0\}$. [/definition] Equivalently, $B$ is nondegenerate if the matrix $A$ of $B$ with respect to any (equivalently, every) basis satisfies $\det A \neq 0$. [explanation: Nondegeneracy and the Associated Linear Map] A bilinear form $B: V \times V \to k$ on a finite-dimensional space $V$ defines a linear map \begin{align*} \hat{B}: V &\to V^* \\ v &\mapsto B(v, \cdot), \end{align*} where $V^*$ is the dual space of $V$ and $B(v, \cdot): w \mapsto B(v, w)$ is a linear functional. The map $\hat{B}$ is linear because $B$ is linear in its first argument. Its kernel is exactly the left radical: $\ker \hat{B} = \operatorname{rad}_L(B)$. Since $\dim V = \dim V^*$, the map $\hat{B}$ is an isomorphism if and only if $\ker \hat{B} = \{0\}$, i.e., if and only if $B$ is nondegenerate on the left. When $B$ is symmetric, the same map $\hat{B}$ controls both the left and right radical, and nondegeneracy of $B$ is equivalent to $\hat{B}: V \xrightarrow{\sim} V^*$ being an isomorphism. This perspective is crucial for PDEs: the Lax–Milgram theorem finds solutions to elliptic PDEs by showing that a bilinear form $B[u, v]$ on a Hilbert space defines an invertible linear map from the space to its dual. [/explanation] [example: Degenerate Form and Its Radical] Let $V = \mathbb{R}^3$ and let $B$ have matrix \begin{align*} A = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 3 & 6 & 9 \end{pmatrix}. \end{align*} The rows of $A$ are $(1,2,3)$, $2(1,2,3)$, $3(1,2,3)$ — they are all multiples of the same vector. So $\det A = 0$ and $B$ is degenerate. To find $\operatorname{rad}(B)$: we need $Av = 0$. The row-reduced form of $A$ has rank $1$, and the solution space is \begin{align*} Av = 0 \iff v_1 + 2v_2 + 3v_3 = 0, \end{align*} which is a $2$-dimensional subspace of $\mathbb{R}^3$. For instance, $v = (-2, 1, 0)$ satisfies $A(-2,1,0)^\top = (-2+2, -4+4, -6+6)^\top = 0$. So the radical is $\{v \in \mathbb{R}^3 : v_1 + 2v_2 + 3v_3 = 0\}$, a plane through the origin. Any theorem about nondegenerate forms fails when applied to $B$. [/example] ## Symmetric Forms and Diagonalization Symmetric bilinear forms are the most important class, and their classification is the main theorem of this subject. The key result is that every symmetric bilinear form can be diagonalized: there exists a basis in which the matrix of $B$ is diagonal. The proof of this fact is constructive and provides an algorithm for finding the diagonalizing basis. Why should diagonalization hold for bilinear forms when it does not hold for all linear maps? The answer is that the matrix of a symmetric bilinear form transforms by congruence, not similarity. A matrix $A$ is diagonalizable by similarity only if it has $n$ linearly independent eigenvectors; this fails for, say, $A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$. But the same matrix, viewed as the matrix of a symmetric bilinear form, cannot arise because $A \neq A^\top$. The constraint of symmetry is precisely what allows diagonalization. [quotetheorem:427] [quotetheorem:3296] The proof of the diagonalization theorem now proceeds by induction on $\dim V$. The lemma supplies a vector $v$ with $B(v,v) \neq 0$; set $f_1 = v$ and let $W = (\operatorname{span}(f_1))^\perp$. One checks $V = \operatorname{span}(f_1) \oplus W$ (using $B(f_1, f_1) \neq 0$) and applies the inductive hypothesis to $B|_{W \times W}$. The diagonal entries $d_i = B(f_i, f_i)$ are the values of the quadratic form $Q(v) = B(v,v)$ on the basis vectors. Over $\mathbb{R}$, these entries can be positive, negative, or zero. Over $\mathbb{C}$, any nonzero diagonal entry can be rescaled to $1$ by replacing $f_i$ by $f_i / \sqrt{d_i}$. This makes the complex case simple — but over $\mathbb{R}$, we cannot take square roots of negative numbers, and the signs of the diagonal entries are genuine invariants. [example: Diagonalizing a Symmetric Form by Completing the Square] Let $B: \mathbb{R}^3 \times \mathbb{R}^3 \to \mathbb{R}$ have matrix \begin{align*} A = \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}. \end{align*} So $B(v, w) = v_1 w_2 + v_2 w_1 + v_3 w_3$. The associated quadratic form is $Q(v) = B(v,v) = 2v_1 v_2 + v_3^2$. Notice $Q(e_1) = 0$ and $Q(e_2) = 0$, so we cannot immediately use $e_1$ or $e_2$ as a basis vector. But $B(e_1, e_2) = 1 \neq 0$, so set $f_1 = e_1 + e_2$ and $f_2 = e_1 - e_2$. Then \begin{align*} Q(f_1) &= 2(f_1)_1 (f_1)_2 = 2 \cdot 1 \cdot 1 = 2, \\ Q(f_2) &= 2(f_2)_1 (f_2)_2 = 2 \cdot 1 \cdot (-1) = -2. \end{align*} We compute \begin{align*} B(f_1, f_2) &= B(e_1 + e_2, e_1 - e_2) = B(e_1, e_1) - B(e_1, e_2) + B(e_2, e_1) - B(e_2, e_2) = 0 - 1 + 1 - 0 = 0. \end{align*} Set $f_3 = e_3$. We check $B(f_1, f_3) = B(e_1 + e_2, e_3) = 0$ and $B(f_2, f_3) = 0$. So in the basis $(f_1, f_2, f_3)$, the matrix of $B$ is \begin{align*} \begin{pmatrix} 2 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & 1 \end{pmatrix}. \end{align*} This is diagonal, as promised. Rescaling: $g_1 = f_1/\sqrt{2}$, $g_2 = f_2/\sqrt{2}$, $g_3 = f_3$ gives diagonal entries $(1, -1, 1)$. The form $B$ has signature $(2, 1)$: two positive entries and one negative entry. [/example] ## Sylvester's Law of Inertia The diagonalization theorem shows that every symmetric bilinear form over $\mathbb{R}$ can be written, in some basis, as \begin{align*} B(v, w) = v_1 w_1 + \cdots + v_p w_p - v_{p+1} w_{p+1} - \cdots - v_{p+q} w_{p+q} \end{align*} for some non-negative integers $p, q$ with $p + q \le n = \dim V$. Different choices of diagonalizing basis will in general give different diagonal entries — but Sylvester's Law of Inertia says that the numbers $p$, $q$, and $n - p - q$ are invariants: they do not depend on the choice of diagonalizing basis. This is what makes the classification meaningful. The triple $(p, q, n - p - q)$ is the **signature** of $B$. Two symmetric bilinear forms over $\mathbb{R}$ are congruent (equivalent by change of basis) if and only if they have the same signature. The classification of symmetric bilinear forms over $\mathbb{R}$ is therefore complete and given entirely by the signature. [quotetheorem:429] The proof of the law of inertia uses the following key lemma: if $W^+ \subset V$ is a subspace on which $B$ is positive definite, and $W^- \subset V$ is a subspace on which $B$ is negative definite, then $W^+ \cap W^- = \{0\}$ (since $B(v,v) > 0$ and $B(v,v) < 0$ cannot hold simultaneously). Hence $\dim W^+ + \dim W^- \le \dim V$. The maximal dimension of a positive definite subspace is $p$ (independent of the diagonalization), and this dimension can be bounded above and below by considering the radical and subspace intersections. [example: Naive Diagonalization Fails Without Completing the Square] A naive approach to diagonalizing a symmetric bilinear form might be: pick any basis, read off the diagonal of the matrix, and declare those to be the diagonal entries. This fails because the off-diagonal entries spoil everything — the matrix is not diagonal just because we chose a convenient basis. Consider $B: \mathbb{R}^2 \times \mathbb{R}^2 \to \mathbb{R}$ with matrix $A = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ in the standard basis. If we blindly take the diagonal, we read $(0, 0)$: two zeros. But $B$ is nondegenerate ($\det A = -1 \neq 0$), so the radical is trivial — the zero entries are not genuine invariants. The diagonal entries $(0, 0)$ suggest $B$ is identically zero, which is false: $B(e_1, e_2) = 1$. The naive approach has given us the wrong answer. The correct diagonalizing change of basis is $f_1 = e_1 + e_2$, $f_2 = e_1 - e_2$, giving $B(f_1, f_1) = 2$ and $B(f_2, f_2) = -2$ with $B(f_1, f_2) = 0$. The signature is $(1, 1)$, not $(0, 0)$. One must actually perform the completing-the-square algorithm — reading diagonal entries of an arbitrary matrix representation is meaningless. [/example] [example: Two Diagonalizations of the Same Form] Consider $B: \mathbb{R}^2 \times \mathbb{R}^2 \to \mathbb{R}$ with matrix $A = \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix}$. **Diagonalization 1.** The standard basis gives diagonal entries $B(e_1, e_1) = 1$, but the form is not diagonal since $B(e_1, e_2) = 2 \neq 0$. Completing the square in $Q(v) = v_1^2 + 4v_1 v_2 + v_2^2$: set $u_1 = v_1 + 2v_2$ so $v_1 = u_1 - 2v_2$, giving \begin{align*} Q = (u_1 - 2v_2)^2 + 4(u_1 - 2v_2)v_2 + v_2^2 = u_1^2 - 4u_1 v_2 + 4v_2^2 + 4u_1 v_2 - 8v_2^2 + v_2^2 = u_1^2 - 3v_2^2. \end{align*} So in the basis $f_1 = e_1$, $f_2 = -2e_1 + e_2$ (where $u_1 = v_1 + 2v_2$ corresponds to the first coordinate), the diagonal entries are $(1, -3)$: one positive, one negative. **Diagonalization 2.** The eigenvectors of $A$ are $v_+ = (1,1)/\sqrt{2}$ with eigenvalue $3$ and $v_- = (1,-1)/\sqrt{2}$ with eigenvalue $-1$. In the eigenbasis, $B(v_+, v_+) = 3$, $B(v_-, v_-) = -1$, $B(v_+, v_-) = 0$. Diagonal entries: $(3, -1)$. Both diagonalizations give exactly one positive and one negative entry. The magnitudes $1, 3$ and $3, 1$ differ — but the signature $(1,1)$ is the same, as Sylvester's law guarantees. [/example] The example above shows that different diagonalizing bases produce different magnitudes — but the count of positive, negative, and zero entries is always the same. This is striking enough that it deserves its own definition: the pair $(p, q)$ is a genuine invariant of the form, not an artifact of the choice of basis. We give it a name so we can state the classification of real symmetric forms cleanly: two forms are equivalent if and only if they share this invariant. [definition: Signature of a Symmetric Bilinear Form] Let $B: V \times V \to \mathbb{R}$ be a symmetric bilinear form on a finite-dimensional real vector space $V$ of dimension $n$. The **signature** of $B$ is the pair $(p, q) \in \mathbb{Z}_{\ge 0}^2$ where $p$ is the number of positive diagonal entries and $q$ is the number of negative diagonal entries in any diagonal representation of $B$. By Sylvester's Law of Inertia, this is well-defined. The integer $p - q$ is called the **index** of $B$. [/definition] [definition: Positive Definite Symmetric Form] A symmetric bilinear form $B: V \times V \to \mathbb{R}$ is **positive definite** if $B(v,v) > 0$ for all $v \in V \setminus \{0\}$. [/definition] Equivalently, the signature of $B$ is $(n, 0)$ where $n = \dim V$: all diagonal entries in any diagonalization are strictly positive. A positive definite symmetric bilinear form is called an **inner product** on $V$. A real vector space equipped with an inner product is called a **real inner product space**. Positive definiteness is the strongest possible condition on a symmetric form: every nonzero vector has positive self-pairing. But many naturally occurring forms are not positive definite. The Minkowski metric of special relativity assigns negative self-pairing to timelike vectors; the signature of a conic determines whether its zero set is an ellipse or a hyperbola; the second derivative test in multivariable calculus classifies critical points by whether the Hessian is positive definite, negative definite, or *neither* — the saddle point case. This last case, where the form takes both positive and negative values, deserves its own name: indefiniteness is not a failure of the theory, but a genuine geometric phenomenon. [definition: Indefinite Symmetric Form] A symmetric bilinear form $B: V \times V \to \mathbb{R}$ is **indefinite** if it takes both positive and negative values: there exist $v, w \in V$ with $B(v,v) > 0$ and $B(w,w) < 0$. [/definition] Equivalently, $p \ge 1$ and $q \ge 1$ in the signature $(p,q)$. The Minkowski metric of special relativity is an indefinite symmetric bilinear form on $\mathbb{R}^4$ with signature $(3,1)$ (or $(1,3)$ depending on convention). The indefiniteness encodes the distinction between spacelike and timelike directions. ## The Associated Quadratic Form Every symmetric bilinear form $B$ gives rise to a *quadratic form* $Q: V \to k$ defined by $Q(v) = B(v,v)$. Conversely, over a field of characteristic not $2$, any quadratic form determines its associated symmetric bilinear form via the **polarization identity**. Why single out $Q(v) = B(v,v)$? Because the quadratic form is what appears in the applications. The kinetic energy $\frac{1}{2} \langle v, v \rangle$ in mechanics, the squared norm $\|v\|^2$ in geometry, the discriminant of a conic section — all of these are quadratic forms. Studying bilinear forms and quadratic forms together is essential because each determines the other (over suitable fields). [definition: Quadratic Form] Let $V$ be a vector space over a field $k$. A **quadratic form** on $V$ is a map $Q: V \to k$ such that: 1. $Q(\lambda v) = \lambda^2 Q(v)$ for all $\lambda \in k$ and $v \in V$, and 2. the map $B_Q: V \times V \to k$ defined by \begin{align*} B_Q(v, w) = \frac{1}{2}(Q(v + w) - Q(v) - Q(w)) \end{align*} is bilinear (this requires $\operatorname{char}(k) \neq 2$). The bilinear form $B_Q$ is called the **polarization** of $Q$, and satisfies $B_Q(v,v) = Q(v)$. [/definition] [explanation: The Polarization Identity] The polarization identity \begin{align*} B(v, w) = \frac{1}{2}(B(v+w, v+w) - B(v, v) - B(w, w)) = \frac{1}{2}(Q(v+w) - Q(v) - Q(w)) \end{align*} shows that $B$ is completely determined by $Q$. This is not obvious: $B$ is a function of two variables, while $Q$ is a function of one. The key is that expanding $B(v+w, v+w)$ via bilinearity and symmetry gives \begin{align*} B(v+w, v+w) = B(v,v) + 2B(v,w) + B(w,w) = Q(v) + 2B(v,w) + Q(w), \end{align*} so $B(v,w) = \frac{1}{2}(Q(v+w) - Q(v) - Q(w))$. This identity fails in characteristic $2$ because dividing by $2$ is not permitted. This is why the theory of quadratic forms in characteristic $2$ requires separate treatment: the bijection between symmetric bilinear forms and quadratic forms breaks down. [/explanation] [example: Conic Sections as Quadratic Forms] A conic section in $\mathbb{R}^2$ is the zero set of a degree-$2$ polynomial $ax^2 + bxy + cy^2 + dx + ey + f = 0$. The homogeneous degree-$2$ part $Q(x,y) = ax^2 + bxy + cy^2$ is a quadratic form on $\mathbb{R}^2$ with associated bilinear form \begin{align*} B\bigl((x_1, y_1), (x_2, y_2)\bigr) = ax_1 x_2 + \frac{b}{2}(x_1 y_2 + x_2 y_1) + cy_1 y_2. \end{align*} The matrix of $B$ is $\frac{1}{2}\begin{pmatrix} 2a & b \\ b & 2c \end{pmatrix} = \begin{pmatrix} a & b/2 \\ b/2 & c \end{pmatrix}$, whose discriminant is $ac - b^2/4$. The signature of $B$ classifies the type of conic: - Signature $(2, 0)$: $B$ is positive definite, $Q > 0$ off the origin. The zero set is an ellipse (or empty, or a point). - Signature $(1, 1)$: $B$ is indefinite. The zero set is a hyperbola (or two crossing lines). - Signature $(1, 0)$ (degenerate): $B$ has a radical. The zero set is a parabola (or parallel lines, etc.). For $Q(x,y) = x^2 - y^2$, we get $a = 1$, $b = 0$, $c = -1$, matrix $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$, signature $(1,1)$, and $Q = 0$ is the pair of lines $y = \pm x$ — a degenerate hyperbola. [/example] ## Orthogonality and Orthogonal Complements One of the central geometric notions arising from a bilinear form is orthogonality. Two vectors are orthogonal with respect to $B$ if $B(v,w) = 0$. This definition makes sense for any bilinear form, not just inner products — but without positive definiteness, orthogonality loses its familiar Euclidean meaning. [definition: Orthogonality with Respect to a Bilinear Form] Let $B: V \times V \to k$ be a bilinear form and let $S \subset V$ be a subset. The **orthogonal complement** of $S$ with respect to $B$ is \begin{align*} S^\perp = \{ v \in V : B(v, w) = 0 \text{ for all } w \in S \}. \end{align*} Two vectors $v, w \in V$ are **$B$-orthogonal** if $B(v, w) = 0$. A set $S \subset V$ is **$B$-orthogonal** if $B(v, w) = 0$ for all distinct $v, w \in S$. [/definition] Over a nondegenerate form, we have the usual dimension formula. [quotetheorem:3297] The proof uses the map $\hat{B}: V \to V^*$ defined by $\hat{B}(v)(w) = B(v,w)$, which is an isomorphism when $B$ is nondegenerate. Then $W^\perp = \hat{B}^{-1}(W^0)$ where $W^0 = \{f \in V^* : f(w) = 0 \text{ for all } w \in W\}$ is the annihilator of $W$, and $\dim W^0 = n - \dim W$ by the standard theory of dual spaces. A critical difference from the Euclidean case: when $B$ is indefinite, we can have $W \cap W^\perp \neq \{0\}$, and even $W \subset W^\perp$ (such subspaces are called **isotropic**). In the Euclidean case, $W \cap W^\perp = \{0\}$ always, so $V = W \oplus W^\perp$. Over an indefinite form, this decomposition can fail. [example: Isotropic Subspaces of the Minkowski Form] Let $B: \mathbb{R}^4 \times \mathbb{R}^4 \to \mathbb{R}$ be the Minkowski form $B(v,w) = v_1 w_1 + v_2 w_2 + v_3 w_3 - v_4 w_4$. Consider the vector $v = (1, 0, 0, 1)$. Then $B(v, v) = 1 + 0 + 0 - 1 = 0$: $v$ is isotropic, lying on the light cone. The orthogonal complement of $\operatorname{span}(v)$ is $\{w : B(v, w) = 0\} = \{w : w_1 - w_4 = 0\}$, a $3$-dimensional hyperplane. Since $B(v,v) = 0$, we have $v \in v^\perp$, so $\operatorname{span}(v) \subset (\operatorname{span}(v))^\perp$. This is impossible in a positive definite setting. This failure of $V = W \oplus W^\perp$ is not a pathology but a feature: the geometry of the light cone, with its isotropic directions, is the mathematical foundation of special relativity. [/example] [definition: Isotropic Vector and Isotropic Subspace] Let $B: V \times V \to k$ be a symmetric bilinear form. A vector $v \in V$ is **isotropic** (or **null**) with respect to $B$ if $B(v,v) = 0$. A subspace $W \subset V$ is **isotropic** if every vector in $W$ is isotropic [/definition] Equivalently if $B|_{W \times W} = 0$. ## References - Bilinear forms and their classification: Hoffman and Kunze, *Linear Algebra*, 2nd ed. (1971), Chapter 10. - Sylvester's Law of Inertia and quadratic forms: Artin, *Geometric Algebra* (1957), Chapter III. - Symplectic forms and skew-symmetric forms: Guillemin and Pollack, *Differential Topology* (1974), Appendix on linear algebra. - Minkowski forms and indefinite geometry: O'Neill, *Semi-Riemannian Geometry* (1983), Chapter 2. - Bilinear forms in the context of elliptic PDEs: Evans, *Partial Differential Equations*, 2nd ed. (2010), §6.2.