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Sylvester's law of inertia: every real symmetric form is congruent to $diag (I_{p}, - I_{q}, 0_{r})$ , and $(p, q, r)$ is unique

Statement

Every symmetric bilinear form on a finite-dimensional real vector space is congruent to exactly one normal form

$diag (I_{p}, - I_{q}, 0_{r}), p + q + r = \dim V .$

Equivalently, the numbers of positive, negative, and zero diagonal entries are independent of the diagonalizing basis.

Facts & Assumptions

Given: A symmetric bilinear form $B$ on a finite-dimensional real vector space $V$ .

[L1]

Every real symmetric matrix is congruent to a diagonal matrix (Over a field of characteristic not $2$ , every symmetric matrix is congruent to a diagonal matrix).

[L2]

Positive and negative definiteness and the inertia data have the meanings stated for real symmetric forms (Positive and negative definiteness, the inertia $(p, q, r)$ , rank $p + q$ , and signature $p - q$ of a real symmetric bilinear or quadratic form).

[L3]

The constructed real field has the least-upper-bound property and hence is complete ordered (The Cauchy-sequence reals have the least-upper-bound property), so every positive real has a nonzero positive square root (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ ).

[L4]

For finite-dimensional subspaces $U, W$ , $\dim (U + W) + \dim (U \cap W) = \dim U + \dim W$ (The dimension formula: for finite-dimensional linear subspaces $U$ and $W$ of $V$ , the subspaces $U + W$ and $U \cap W$ are finite-dimensional and $\dim_{F} (U + W) + \dim_{F} (U \cap W) = \dim_{F} U + \dim_{F} W$ ).

[L5]

A subspace of a finite-dimensional $n$ -dimensional space has dimension at most $n$ (If $\dim_{F} V = n$ and $U$ is a linear subspace of $V$ , then $U$ is finite-dimensional, $\dim_{F} U \leq n$ , and $\dim_{F} U = n$ if and only if $U = V$ , clause 1).

[L6]

Rank-nullity gives the dimension of a kernel as ambient dimension minus rank (Rank-nullity: $\dim_{F} V = nullity T + rank T$ ).

Proof

technique · diagonal normalization and intrinsic dimension bounds

1.1L1L3choosealgebra

By [L1], choose a basis in which the matrix is diagonal, say with positive entries $d_{1}, \dots, d_{p}$ , negative entries $d_{p + 1}, \dots, d_{p + q}$ , and $r$ zero entries. For each nonzero $d_{i}$ , [L3] supplies $s_{i} = \sqrt{∣ d_{i} ∣} > 0$ ; replacing the corresponding basis vector by $s_{i}^{- 1}$ times it changes $d_{i}$ to $1$ or $- 1$ . This proves existence of the displayed normal form, including $p = q = 0$ and the empty basis when $V = 0$ .

2.1step 1.1L2L4L5algebra

In this normal form, the positive coordinate subspace $P$ has dimension $p$ and the form is positive definite on it. Let $N_{0}$ be the span of the negative and zero coordinates, of dimension $q + r$ . If a positive-definite subspace $U$ had $\dim U > p$ , [L4] and [L5] applied to $U + N_{0} \leq V$ would give $\dim (U \cap N_{0}) \geq \dim U + q + r - \dim V > 0$ . A nonzero vector there has form value at most $0$ , a contradiction. Thus $p$ is the intrinsic maximum dimension of a positive-definite subspace.

3.1step 1.1step 2.1L2L6algebra

Applying step 2.1 to $- B$ shows that $q$ is the intrinsic maximum dimension of a negative-definite subspace. The radical is the kernel of the associated map; in normal form its rank is $p + q$ , so [L6] gives its dimension $r = \dim V - p - q$ .

4.1step 2.1step 3.1L2

Any congruent normal form represents the same bilinear form and therefore has the same two intrinsic maxima and radical dimension. Hence its triple is the same $(p, q, r)$ , proving uniqueness.

5.1step 1.1step 4.1∎

Steps 1.1 and 4.1 prove existence and uniqueness for all finite dimensions and for degenerate as well as nondegenerate forms.

Depends on

Dependency tree · 2 levels

Sources

H. Pinkham, Linear Algebra, §7.7

Sylvester's law of inertia: every real symmetric form is congruent to diag⁡(Ip,−Iq,0r)diag(Ip​,−Iq​,0r​), and (p,q,r)(p,q,r) is unique