ExampleConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-13

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A rank-two alternating form on $R^{3}$ has one symplectic pair and a one-dimensional radical

Example

On $R^{3}$ , the form with matrix

$A = (\begin{array}{ccc} 0 & 1 & 0 \\ - 1 & 0 & 0 \\ 0 & 0 & 0 \end{array})$

has the symplectic pair $(e_{1}, e_{2})$ , radical $span {e_{3}}$ , and rank $2$ .

Facts & Assumptions

Given: The displayed matrix in the standard basis.

[L1]

The standard coordinate vectors form a basis of $R^{3}$ (The standard list $e : n \to F^{n}$ with $e_{i} (i) = 1_{F}$ and $e_{i} (j) = 0_{F}$ for $j \neq i$ is an ordered basis of $F^{n}$ ; hence $\dim_{F} F^{n} = n$ , and $F^{0}$ is the zero space with basis $\emptyset$ and dimension $0$ ).

[L2]

An alternating form has a basis of symplectic pairs followed by a radical basis, with two rank units per pair (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).

Verification

technique · direct matrix inspection

1.1

The matrix has zero diagonal and is skew-symmetric, so $B (v, v) = 0$ by direct expansion. Also $B (e_{1}, e_{2}) = 1$ and $B (e_{2}, e_{1}) = - 1$ .

L1algebra

1.2

Multiplication by $A$ kills exactly the vectors $(0, 0, z)$ , so the radical is $span {e_{3}}$ ; the first two columns are independent, giving rank $2$ .

L1algebra

2.1

Thus $(e_{1}, e_{2})$ is one symplectic pair followed by the radical basis $(e_{3})$ , exactly the normal form of [L2].

step 1.1step 1.2L2∎

Depends on

Used by

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Sources

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

A rank-two alternating form on R3R3 has one symplectic pair and a one-dimensional radical