Page 359 - A Course in Linear Algebra with Applications

9.3: Bilinear Forms 343

Corollary 9.3.7
A skew-symmetric n x n matrix A over R or C is congruent
to a matrix M of the above form.

This is because the bilinear form / given by f(X, Y) =
X T AY is skew-symmetric and hence is represented with re-
spect to a suitable basis by a matrix of type M; thus A must
be congruent to M.

Proof of 9.3.6
j
z
/
Let i , . . . , z n be any basis of V. If (ZJ, Zj) = 0 for all i and ,
then (u, v) = 0 for all vectors u and v, so that / is the zero
/
bilinear form and it is represented by the zero matrix. This is
the case k = 0. So assume that (ZJ, Zj) is not zero for some i
/
and . Since the basis can be reordered, we may suppose that
j
/
/(zi, z 2 ) = a ^ 0. Then (a _ 1 zi, z 2 ) = a _ 1 /(zi 5 z 2 ) = 1.
_ 1
Now replace zi by a zi; the effect is to make (zi, z 2 ) = 1,
/
/
and of course (z 2 , zi) = —1 since / is skew-symmetric.
Next put bi = /(zi,Zj) where i > 2. Then
-
-
/(zi.zi 6z 2 ) = /(zi,Zi) 6/(zi,z 2 ) = 6 - 6 = 0.
This suggests that we modify the basis further by replacing Zj
by Zj — 6z 2 for i > 2; notice that this does not disturb linear
independence, so we still have a basis of V. The effect of this
substitution is make

3
/(zi,Zj) = 0 for i = ,.. ,n.
.
Next we have to address the possibility that (z 2 , z;) may
/
be non-zero when i > 2; let c = (z 2 , Zj). Then
/

/ ( z 2 , Z j + CZi) = / ( z 2 , Z j ) + c / ( z 2 , Z i ) = C + C ( - 1 ) = 0 .

This suggests that the next step should be to replace Zj by
Zj + czi where i > 2; again we need to observe that Zi,..., z n