Page 335 - DMTH502_LINEAR_ALGEBRA

Unit 30: Bilinear Forms and Symmetric Bilinear Forms

Now Notes
1 1 y 1
f( , ) = [x , x ]
1 2
1 1 y 2
and so the matrix of f in the standard ordered basis = { , } is
1 2
1 1
[f] =
1 1
Let = { , } be the ordered basis defined by = (1, –1), = (1, 1). In this case, the matrix P
1 2 1 2
which changes coordinates from to is
1 1
P =
1 1
Thus
t
[f] = P [f] P
1 1 1 1 1 1
=
1 1 1 1 1 1
1 1 0 2
=
1 1 0 2

0 0
=
0 4
What this means is that if we express the vectors and by means of their coordinates in the
basis , say
= x + x , = y + x
1 1 2 2 1 1 2 2
then
f( , ) = 4x y
2 2
One consequence of the change of basis formula (4) is the following: If A and B are n × n matrices
which represent the same bilinear form on V in (possibly) different ordered bases, then A and B
have the same rank. For, if P is an invertible n n matrix and B = P AP, it is evident that A and B
t
have the same rank. This makes it possible to define the rank of a bilinear form on V as the rank
of any matrix which represents the form in an ordered basis for V.

It is desirable to give a more intrinsic definition of the rank of a bilinear form. This can be done
as follows: Suppose F is a bilinear form on the vector space V. If we fix a vector in V, then
f( , ) is linear as a function of . In this way, each fixed determines a linear functional on V;
let us denote this linear functional by L ( ). To repeat, if is a vector in V, then L ( ) is the linear
f f
functional on V whose value on any vector is f( , ). This gives us a transformation L ( )
f
form V into the dual space V*. Since
f(c , + , ) = cf( , ) + f( , )
1 2 1 2
we see that
L (c , + ) = cL ( ) + L ( )
f 1 2 f 1 f 2
that is L is a linear transformation from V into V*.
f

LOVELY PROFESSIONAL UNIVERSITY 329