Diagonalization, Rank, and Sylvester's Law

Diagonalizing by completing the square

The first structural theorem: over any field of characteristic ≠ 2, every symmetric bilinear form admits an orthogonal basis — a basis in which the Gram matrix is diagonal. Equivalently every symmetric matrix is congruent to a diagonal one, and every quadratic form is equivalent to some ⟨d₁, …, dₙ⟩. The proof is just completing the square, done by an algorithm you already know from school but now run over a general field.

1. If some diagonal entry q(eᵢ) = B(eᵢ, eᵢ) is nonzero, reorder so it is the first variable x₁; use it as a pivot.
2. Gather every term containing x₁ and complete the square: a₁(x₁ + …)² where the “…” is a linear combination of x₂, …, xₙ. Substitute y₁ = x₁ + …, removing all cross terms with x₁.
3. What remains is a quadratic form in x₂, …, xₙ alone; recurse on it.
4. Edge case: if every diagonal entry is 0 but some B(eᵢ, eⱼ) ≠ 0, first substitute xᵢ ↦ xᵢ + xⱼ to create a nonzero square (this needs char ≠ 2), then proceed.

Diagonalize q(x1, x2, x3) = x1^2 + 2 x1 x2 + 2 x2 x3 over Q.

Pivot on x1:  x1^2 + 2 x1 x2 = (x1 + x2)^2 - x2^2.
  Set y1 = x1 + x2.   q = y1^2 - x2^2 + 2 x2 x3.

Now handle the rest:  -x2^2 + 2 x2 x3 = -(x2^2 - 2 x2 x3)
                    = -(x2 - x3)^2 + x3^2.
  Set y2 = x2 - x3,  y3 = x3.

  q = y1^2 - y2^2 + y3^2.        Diagonal form  <1, -1, 1>.

Rank = 3 (three nonzero entries), so q is nondegenerate.
Over R: one minus sign, two plus signs -> signature (2, 1), s = 2 - 1 = 1.
The change of variables is invertible, so this is a CONGRUENCE,
not an orthonormal (eigenvalue) diagonalization -- no square roots needed.

Rank: the field-independent invariant

Once diagonalized as ⟨d₁, …, dₙ⟩, the number of nonzero dᵢ is the rank of the form — the rank of its Gram matrix, which congruence cannot change because P is invertible. The form is nondegenerate exactly when the rank equals dim V, i.e. all dᵢ ≠ 0. Rank is the *one* invariant that works over every field: over an algebraically closed field like ℂ it is the *only* invariant, since there every nonzero dᵢ becomes 1 after scaling by 1/(√dᵢ)² — so over ℂ a form is determined entirely by its rank, and ⟨d₁, …, dᵣ, 0, …⟩ ≅ ⟨1, …, 1, 0, …⟩.

Sylvester's law of inertia

Over ℝ we can scale each nonzero dᵢ to ±1 (divide by (√|dᵢ|)²), so every real form is congruent to ⟨1, …, 1, −1, …, −1, 0, …, 0⟩ with p ones, m minus-ones, and z zeros. The diagonalization is wildly non-unique, but [[sylvester-law-of-inertia|Sylvester's law of inertia]] says the triple (p, m, z) is not: any two diagonalizations of the same real form have the same number of positive, negative, and zero entries. The pair (p, m) is the [[signature-of-a-form|signature]]; often one reports the single number s = p − m.

The clean proof of uniqueness of p: it equals the maximal dimension of a subspace on which q is strictly positive. If U is a subspace where q > 0 (dimension p, the span of the “+” basis vectors) and W is where q ≤ 0 (dimension m + z, the rest), then U ∩ W = {0} because a vector in both has q both > 0 and ≤ 0. Hence dim U + dim W ≤ n, forcing p to be maximal and basis-independent. The same argument pins m. This dimension characterization is *intrinsic* — no basis mentioned — which is exactly why the count cannot depend on the chosen diagonalization.