Page 360 - A Course in Linear Algebra with Applications

344 Chapter Nine: Advanced Topics

will still be a basis of V. Also important is the remark that this
substitution will not nullify what has already been achieved;
the reason is that when i > 2

/
/(zi,Zi + czi) = (zi,z i ) + c/(z 1 ,z 1 ) = 0 .
We have now reached the point where

/(zi,z 2 ) = 1 = -/(z 2 ,zi) and (zi,z,) = 0 = /(z 2 ,Zi),
/

for all i > 2. Now we rename our first two basis elements,
writing Ui = zi and vi = z 2 .
So far the matrix representing / has the form

/ O i l 0 0\
- 1 0 1 0 0
I
V 0 0 I B J

where B is a skew-symmetric matrix with n — 2 rows and
columns. We can now repeat the argument just given for the
subspace with basis {Z3,..., z n }; it follows by induction on n
that there is a basis for this subspace with respect to which
/ is represented by a matrix of the required form. Indeed
let u 2 ,...,u f c , v 2 ,...,Vfc,wi,...,w n _2A; be this basis. By
adjoining ui and vi, we obtain a basis of V with respect to
which / is represented by a matrix of the required form.

Example 9.3.4
Find the canonical form of the skew-symmetric matrix

/ 0 0 2
A= \ 0 0 - 1
\ - 2 1 0

We need to carry out the procedure indicated in the proof
3
of the theorem. Let {E±, E 2, E 3} be the standard basis of R .