Page 334 - DMTH502_LINEAR_ALGEBRA

Linear Algebra

Notes Now the functions f defined by
ij
f ( , ) = L ( )L ( )
ij i j
are bilinear forms. If
= x + ... + x and = y + ... + y
1 1 n n 1 1 n n
then
f ( , ) = x y
ij i j
Let f be any bilinear form on V and let A be the matrix of f in the ordered basis . Then

f( , ) = A x y
ij i j
, i j
which simply says that

f = A f
ij ij
, i j
2
It is now clear that the n forms f comprise a basis for L(V, V, F).
ij
One can rephrase the proof of the corollary as follows. The bilinear from f has as its matrix in the
ij
i,j
ordered basis the matrix ‘unit’ E , whose only non-zero entry is a 1 in now i and column j. Since
these matrix units comprise a basis for the space of n n matrices, the forms f comprise a basis
ij
for the space of bilinear forms.
The concept of the matrix of a bilinear form in an ordered basis is similar to that of the matrix of
a linear operator in an ordered basis. Just as for linear operators, we shall be interested in what
happens to the matrix representing a bilinear form, as we change from one ordered basis to
another. So, suppose = { , ... } and = { , ..., } are two ordered bases for V and that f is
1 n 1 n
a bilinear form on V. How are the matrices [f] and [f] , related? Well, let P be the (invertible)
n n matrix such that
[ ] = P[ ]
for all in V. In other words, define P by
n
= P ij i
j
i 1
For any vectors , in V
t
f( , ) = [ ] [f] [ ]
t
= (P[ ] ) [ f] P[ ]
t
t
= [ ] (P [ f] P)[ ] .
By the definition and uniqueness of the matrix representing f in the ordered basis , we must
have
t
[f] = P [f] P. ...(4)

2
Example 3: Let V be the vector space R . Let f be the bilinear form defined on = (x , x )
1 2
and = (y , y ) by
1 2
f( , ) = x y + x y + x y + x y
1 1 1 2 2 1 2 2

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