TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-13

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Every alternating form on a finite-dimensional space has a basis of symplectic pairs followed by a basis of its radical; in particular its rank is even

Statement

Let $B$ be an alternating bilinear form on a finite-dimensional vector space $V$ . There is a basis

$e_{1}, f_{1}, \dots, e_{r}, f_{r}, z_{1}, \dots, z_{s}$

such that $B (e_{i}, f_{i}) = 1$ , $B (f_{i}, e_{i}) = - 1$ , every other pairing of distinct listed blocks is zero, and $z_{1}, \dots, z_{s}$ is a basis of $rad B$ . Thus the matrix is a direct sum of $r$ blocks $(\begin{array}{cc} 0 & 1 \\ - 1 & 0 \end{array})$ and an $s \times s$ zero block, so $rank B = 2 r$ is even.

Facts & Assumptions

Given: A finite-dimensional $F$ -vector space $V$ and an alternating bilinear form $B$ .

[L1]

Alternation means $B (v, v) = 0$ for every $v$ (Bilinear forms, and symmetric, skew-symmetric, and alternating bilinear forms), and every alternating form is skew-symmetric (Alternating forms are skew-symmetric; the converse holds when $char F \neq 2$ , while in characteristic $2$ alternating forms are symmetric).

[L2]

The radical consists of vectors pairing to zero with every vector, and the rank is the rank of the associated map into the dual (The matrix, left and right radicals, rank, and nondegeneracy of a bilinear form on a finite-dimensional space).

[L3]

Subspaces of finite-dimensional spaces are finite-dimensional and admit bases (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 $U = V$ ); a basis is an independent spanning set, with the empty basis for zero space (Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis).

Proof

technique · induction on $n=\dim V$

1.1

If $n = 0$ , the empty basis has the asserted form. If $B = 0$ , any basis of $V = rad B$ works with $r = 0$ .

baseL2L3

1.2

Assume $B \neq 0$ and the result below dimension $n$ . Choose $e, f$ with $B (e, f) \neq 0$ and rescale $f$ so $B (e, f) = 1$ ; [L1] gives $B (f, e) = - 1$ .

ihL1givenchoose

2.1

Put $W = span {e, f}$ and $W^{⊥} = {v : B (v, e) = B (v, f) = 0}$ . For every $v \in V$ , the vector $z = v - B (v, f) e + B (v, e) f$ lies in $W^{⊥}$ , so $V = W + W^{⊥}$ . If $a e + b f \in W^{⊥}$ , pairing with $e$ and $f$ gives $b = 0 = a$ , hence $V = W \oplus W^{⊥}$ .

step 1.2L1algebra

3.1

Since $e \notin W^{⊥}$ , this subspace is proper and [L3] gives $\dim W^{⊥} < n$ . By induction its restricted alternating form has symplectic pairs followed by a basis of its radical. Adjoining $e, f$ gives the displayed basis of $V$ ; because $W$ is nondegenerate and orthogonal to $W^{⊥}$ , the remaining radical is exactly the radical of the whole form.

step 1.2step 2.1ihL2L3

4.1

In that basis the associated map has one invertible rank-two block per pair and is zero on the radical block, so its rank is $2 r$ . This remains valid in characteristic $2$ , where $- 1 = 1$ .

step 3.1L1L2algebra

5.1

The base cases and induction step establish the normal form and even-rank conclusion in every finite dimension, including odd-dimensional and degenerate forms.

step 1.1step 4.1discharge-induction∎

Depends on

Used by

Symplectic vector space Definition
A rank-two alternating form on ℝ³ has one symplectic pair and a one-dimensional radical Example
Symplectic vector spaces have even dimension Proposition
Darboux theorem Theorem
Nondegeneracy is equivalent to a nonvanishing top wedge Theorem

Dependency tree · two levels

35 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

K. Conrad, Bilinear Forms, §6 (standard reference, not scraped)