Sylvester's Law of Inertia

Sylvester's Law of Inertia states that the signature of a real quadratic form is invariant under change of basis. This means that while the specific diagonal entries may change when we diagonalize a quadratic form, the number of positive, negative, and zero entries remains the same. This gives a complete classification of real quadratic forms up to congruence.

Statement

Theorem8.2Sylvester's Law of Inertia

Let $Q$ be a real quadratic form on $\mathbb{R}^n$ with symmetric matrix $M$ . If $Q$ can be reduced to diagonal form in two different ways:

$Q = a_1 y_1^2 + \cdots + a_n y_n^2 = b_1 z_1^2 + \cdots + b_n z_n^2,$

then the number of positive $a_i$ , the number of negative $a_i$ , and the number of zero $a_i$ are the same as for the $b_j$ .

Equivalently: if $P^T M P = D$ and $Q^T M Q = D'$ are two diagonalizations (where $D, D'$ are diagonal), then $D$ and $D'$ have the same numbers of positive, negative, and zero diagonal entries.

Definition8.11Signature

The signature of $Q$ is the triple $(p, q, z)$ where $p$ is the number of positive diagonal entries, $q$ the number of negative entries, and $z$ the number of zero entries after diagonalization. Since $p + q + z = n$ and $r = p + q$ is the rank, we often write just $(p, q)$ .

Two real quadratic forms are congruent (equivalent) if and only if they have the same signature.

Proof sketch

ProofProof of Sylvester's Law of Inertia

Suppose $Q = x_1^2 + \cdots + x_p^2 - x_{p+1}^2 - \cdots - x_{p+q}^2$ in one basis and $Q = y_1^2 + \cdots + y_{p'}^2 - y_{p'+1}^2 - \cdots - y_{p'+q'}^2$ in another. We prove $p = p'$ .

Assume for contradiction $p' > p$ . Let $V^+ = \operatorname{span}\{e_1, \ldots, e_{p'}\}$ (the " $y$ " coordinates where $Q$ is positive) and $V^- = \operatorname{span}\{f_{p+1}, \ldots, f_n\}$ (the " $x$ " coordinates where $Q$ is $\leq 0$ ).

$\dim V^+ = p'$ and $\dim V^- = n - p$ . Since $p' > p$ , we have $\dim V^+ + \dim V^- = p' + n - p > n$ , so $V^+ \cap V^- \neq \{0\}$ .

Pick $v \in V^+ \cap V^-$ , $v \neq 0$ . Since $v \in V^+$ : $Q(v) > 0$ (the form is a sum of positive squares on $V^+$ ). Since $v \in V^-$ : $Q(v) \leq 0$ (the form is non-positive on $V^-$ ). Contradiction.

Therefore $p' \leq p$ . By symmetry, $p \leq p'$ , so $p = p'$ . Similarly $q = q'$ . $\blacksquare$

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Examples

ExampleSignature of 2x2 forms

$M = \begin{pmatrix} 1 & 2 \\ 2 & 3 \end{pmatrix}$ : eigenvalues $2 \pm \sqrt{5}$ , one positive and one negative. Signature $(1, 1)$ .

$M = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$ : eigenvalues $2, 0$ . Signature $(1, 0)$ , rank $1$ .

$M = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$ : eigenvalues $3, 1$ . Signature $(2, 0)$ (positive definite).

$M = \begin{pmatrix} -1 & 0 \\ 0 & -3 \end{pmatrix}$ : eigenvalues $-1, -3$ . Signature $(0, 2)$ (negative definite).

ExampleSignature of 3x3 forms

$M = \operatorname{diag}(1, -1, 0)$ : signature $(1, 1)$ , rank $2$ . The form is $Q = x^2 - y^2$ .

$M = \operatorname{diag}(2, 3, -5)$ : signature $(2, 1)$ , rank $3$ . Equivalent to $\operatorname{diag}(1, 1, -1)$ .

$M = \operatorname{diag}(1, 1, 1)$ : signature $(3, 0)$ , positive definite. The form $Q = x^2 + y^2 + z^2$ .

ExampleTwo diagonalizations, same signature

$Q(x, y) = 2x^2 + 4xy + 3y^2$ .

Diagonalization 1 (completing the square): $Q = 2(x + y)^2 + y^2$ . Diagonal form: $2u^2 + v^2$ . Positive coefficients: $2$ .

Diagonalization 2 (eigenvalues of $M = \begin{pmatrix} 2 & 2 \\ 2 & 3 \end{pmatrix}$ ): eigenvalues $\frac{5 \pm \sqrt{5}}{2} \approx 3.62, 1.38$ . Both positive.

Signature $(2, 0)$ in both cases ✓. Sylvester's Law guarantees this consistency.

Complete classification

Theorem8.3Classification of real quadratic forms

Two real quadratic forms on $\mathbb{R}^n$ are congruent if and only if they have the same signature $(p, q)$ (equivalently, the same rank $r = p + q$ and the same index of positivity $p$ ).

There are exactly $\frac{(n+1)(n+2)}{2}$ equivalence classes of quadratic forms on $\mathbb{R}^n$ : one for each pair $(p, q)$ with $p + q \leq n$ .

ExampleAll quadratic forms on R^2

The possible signatures on $\mathbb{R}^2$ are:

$(2, 0)$ : $x^2 + y^2$ (positive definite), represented by $I_2$ .
$(1, 1)$ : $x^2 - y^2$ (indefinite), represented by $\operatorname{diag}(1, -1)$ .
$(0, 2)$ : $-x^2 - y^2$ (negative definite), represented by $-I_2$ .
$(1, 0)$ : $x^2$ (degenerate, rank $1$ , positive).
$(0, 1)$ : $-x^2$ (degenerate, rank $1$ , negative).
$(0, 0)$ : $Q = 0$ (the zero form).

Total: $6 = \frac{3 \cdot 4}{2}$ equivalence classes.

ExampleCounting forms on R^3

On $\mathbb{R}^3$ : pairs $(p, q)$ with $p + q \leq 3$ :

$(3,0), (2,1), (1,2), (0,3), (2,0), (1,1), (0,2), (1,0), (0,1), (0,0)$ .

Total: $10 = \frac{4 \cdot 5}{2}$ . For example, the Lorentz form $-t^2 + x^2 + y^2$ has signature $(2, 1)$ , the same class as $x^2 + y^2 - z^2$ .

Sylvester's criterion for definiteness

Theorem8.4Sylvester's criterion

A symmetric matrix $M$ is positive definite if and only if all leading principal minors are positive:

$\Delta_1 = m_{11} > 0, \quad \Delta_2 = \det\begin{pmatrix} m_{11} & m_{12} \\ m_{21} & m_{22} \end{pmatrix} > 0, \quad \ldots, \quad \Delta_n = \det M > 0.$

$M$ is negative definite iff $(-1)^k \Delta_k > 0$ for all $k$ (the minors alternate in sign: $\Delta_1 < 0, \Delta_2 > 0, \Delta_3 < 0, \ldots$ ).

ExampleApplying Sylvester's criterion

$M = \begin{pmatrix} 4 & 2 & 0 \\ 2 & 5 & 3 \\ 0 & 3 & 6 \end{pmatrix}$ .

$\Delta_1 = 4 > 0$ ✓, $\Delta_2 = 20 - 4 = 16 > 0$ ✓, $\Delta_3 = 4(30 - 9) - 2(12) = 84 - 24 = 60 > 0$ ✓.

All positive, so $M$ is positive definite.

ExampleTesting negative definiteness

$M = \begin{pmatrix} -2 & 1 \\ 1 & -3 \end{pmatrix}$ .

$\Delta_1 = -2 < 0$ ✓ (should be $< 0$ for negative definite). $\Delta_2 = 6 - 1 = 5 > 0$ ✓ (should be $> 0$ ).

$(-1)^1 \Delta_1 = 2 > 0$ and $(-1)^2 \Delta_2 = 5 > 0$ . Negative definite ✓.

ExampleWhen Sylvester's criterion detects indefiniteness

$M = \begin{pmatrix} 1 & 3 \\ 3 & 1 \end{pmatrix}$ .

$\Delta_1 = 1 > 0$ , $\Delta_2 = 1 - 9 = -8 < 0$ .

$\Delta_2 < 0$ means $M$ is not positive definite. Since $\Delta_1 > 0$ (so it is not negative definite either), $M$ is indefinite. Eigenvalues: $4, -2$ , confirming signature $(1, 1)$ .

Congruence vs. similarity

ExampleCongruence is not similarity

$A = \begin{pmatrix} 1 & 0 \\ 0 & 4 \end{pmatrix}$ and $B = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}$ .

$A$ and $B$ are congruent (both positive definite with signature $(2, 0)$ , so $B = P^TAP$ for some $P$ ). Indeed, $P = \operatorname{diag}(\sqrt{2}, 1/\sqrt{2})$ gives $P^TAP = \operatorname{diag}(2, 2) = B$ ✓.

But $A$ and $B$ are not similar (eigenvalues $\{1, 4\}$ vs $\{2, 2\}$ ).

Congruence preserves the signature; similarity preserves the characteristic polynomial. These are different equivalence relations.

ExampleCongruence over Q vs over R

Over $\mathbb{R}$ : $Q_1 = x^2 + y^2$ and $Q_2 = x^2 + 2y^2$ are congruent (both signature $(2, 0)$ ).

Over $\mathbb{Q}$ : they are not congruent. To see this, $Q_1 = 1$ has the rational solution $(1, 0)$ , and $Q_1 = 3$ has no rational solutions (since $x^2 + y^2 = 3$ has no solutions in $\mathbb{Q}$ ), but $Q_2 = 3$ has the solution $(1, 1)$ . The forms represent different sets of rationals.

Over $\mathbb{R}$ , Sylvester's Law classifies forms by signature alone. Over $\mathbb{Q}$ (and other fields), the classification is much richer.

Summary

RemarkSylvester's Law as complete invariant

Sylvester's Law of Inertia provides a complete classification of real symmetric bilinear forms:

The signature $(p, q)$ is the unique congruence invariant.
Every real quadratic form is congruent to $x_1^2 + \cdots + x_p^2 - x_{p+1}^2 - \cdots - x_{p+q}^2$ .
Sylvester's criterion provides a practical test for definiteness via leading principal minors.
Over other fields (e.g., $\mathbb{Q}$ , $\mathbb{F}_p$ ), the classification is more complex (Hasse--Minkowski theorem, Witt cancellation).
The law connects to the Hodge Index Theorem in algebraic geometry, where the intersection form on a surface has signature $(1, \rho - 1)$ .