Page 345 - DMTH502_LINEAR_ALGEBRA

Unit 31: Skew-symmetric Bilinear Forms

Theorem 1: Let V be an n-dimensional vector space over a subfield of the complex numbers, and Notes
let f be a skew-symmetric bilinear form on V. Then the rank r of f is even, and if r = 2k there is an
ordered basis for V in which the matrix of f is the direct sum of the (n – r) (n – r) zero matrix and
k copies of the 2 2 matrix

0 1
1 0

Proof: Let , , ... , be vectors satisfying conditions (a), (b), and (c) above. Let { , ..., } be
1 1 k k i s
any ordered basis for the subspace W . Then
0
= { , , , , ..., , , ..., }
1 1 2 2 k k 1 s
is an ordered basis for V. From (a), (b), and (c) it is clear that the matrix of f in the ordered basis
is the direct sum of the (n – 2k) (n – 2k) zero matrix and k copies of the 2 2 matrix

0 1
...(2)
1 0
Furthermore, it is clear that the rank of this matrix, and hence the rank of f, is 2k.
One consequence of the above is that if f is a non-degenerate, skew-symmetric bilinear form on
V, then the dimension of V must be even. If dim V = 2k, there will be an ordered basis { , , ...,
1 1
, } for V such that
k k
1, i j
f( , ) =
i j
1, i j
f( , ) = f( , ) = 0
i j i j
The matrix of f in this ordered basis is the direct sum of k copies of the 2 2 skew-symmetric
matrix (2).

Self Assessment

n
1. Let f be a symmetric bilinear form on c and g a skew symmetric bilinear form on c .
n
Suppose f + g = 0. Show that f = 0, g = 0.
2. Let V be an n-dimensional vector space over a subfield F of C. Prove that
(a) The equation
1 1
(Pf) ( , ) = f( , ) – f( , ) defines
2 2
a linear operator P on L (V, V, F)
2
(b) P = P, i.e. P is a projection

31.2 Summary

 A bilinear form f on V is called skew-symmetric if f( , ) = –f( , )
 The space L(V, V, F) of the bilinear forms is the direct sum of the sub-space of symmetric
forms and the subspace of skew-symmetric forms.
 In an n-dimensional vector space over a subfield of the complex numbers, the skew
symmetric bilinear form f has an even rank r = 2k, k being an integer.

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