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Two real symmetric bilinear forms are congruent if and only if they have the same inertia

Statement

Two real symmetric bilinear forms on vector spaces of the same finite dimension are congruent if and only if they have the same inertia $(p, q, r)$ .

Facts & Assumptions

Given: Real symmetric bilinear forms $B$ and $C$ in the same finite dimension.

[L1]

Sylvester's law gives each form a unique normal form $diag (I_{p}, - I_{q}, 0_{r})$ (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).

[L2]

A basis change acts on a bilinear-form matrix by congruence $A \mapsto P^{T} A P$ (A basis change by $P$ changes the matrix of a bilinear form from $A$ to $P^{T} A P$ ).

Proof

technique · direct, proving both implications

1.1

If $B$ and $C$ are congruent, [L2] says they are two matrix representations of the same form after an invertible coordinate identification. The uniqueness clause of [L1] therefore gives them the same inertia.

L1L2given

1.2

Conversely, suppose both have inertia $(p, q, r)$ . By [L1], choose bases in which both matrices equal $D = diag (I_{p}, - I_{q}, 0_{r})$ . The linear map sending the first chosen basis to the second is an isomorphism and carries one form to the other, so the forms are congruent; equivalently, compose the two invertible change-of-basis matrices in [L2].

L1L2choose

2.1

Steps 1.1 and 1.2 prove both directions, including the zero-dimensional and degenerate cases.

step 1.1step 1.2∎

Depends on

Used by

Same complexification with different killing form signatures Counterexample

Dependency tree · two levels

13 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

H. Pinkham, Linear Algebra, §7.7 (standard reference, not scraped)