Page 361 - A Course in Linear Algebra with Applications

9.3: Bilinear Forms 345

The matrix A determines a skew-symmetric bilinear form /
with the properties f{E 1,E 3) = 2 = -f(E 3, E x), f{E 3, E 2) =
1 = -f(E 2,E 3), f(E 1,E 2) = 0 = f(E 2,E 1).
The first step is to reorder the basis as {Ei,E 3,E 2};
this is necessary since f(E\,E 2) = 0 whereas f(Ei,E 3) ^
0. Now replace {Ei,E 3,E 2} by {±Ei,E 3,E 2}, noting that
f(\E uE 3) = 1 = - / ( £ 3 , | £ i ) . Next f(E 3,E 2) = 1, so we
replace E 2 by

E 2 + f(E 3,E 2)-Ei = -Ei + E 2.

Note that f{\E u \E X + E 2) = 0 = f(E 3, \E X + E 2).
The procedure is now complete. The bilinear form is
represented with respect to the new ordered basis

by the matrix
/ ( 0 1 0
M = - 1 0 0
I 0 0 0
which is in canonical form. The change of basis from
{^Ei,E 3, | E \ + E 2} to the standard ordered basis is repre-
sented by the matrix

1/2 0 1/2'
5 = | 0 0 1
0 1 0

T
The reader should now verify that S AS equals M, the canon-
ical form of A, as predicted by the proof of 9.3.6.