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Unit 28: Positive Forms and More on Forms
28.2 More on Forms Notes
Theorem 3: Let f be a form on a real or complex vector space V and { , ..., } a basis for the finite
1 r
dimensional subspace W of V. Let M be the r r matrix with entries
M = f( , )
jk k j
and W’ the set of all vectors in V such that f( , ) for all in W. Then W’ is subspace of V, and
W W’ = {0} if and only if M is invertible. When this is the case, V = W + W’.
Proof: If and are vectors in W’ and c is a scalar, then for every in W
f( , c + ) = c f( , ) + f( , )
= 0.
Hence, W’ is a subspace of V.
r r
Now suppose = x x k and that = y j j . Then
k 1 j 1
r
f( , ) = , y M x
jk k
, j k
= y M jk x . k
j
k j
It follows from this that W W’ {0} if and only if the homogeneous system
r
y M jk = 0, 1 k r
j
j 1
has a non-trivial solution (y ..., y ). Hence W W‘ {0} if and only if M* is invertible. But the
1 r
invertibility of M* is equivalent to the invertibility of M.
Suppose that M is invertible and let
–1
–1
A = (M*) = (M )*
r
g ( ) = A jk ( f k , )
j
k 1
Then
g (c + ) = kn ( f k , c )
j
k
= c A jk ( f k , ) A jk ( f k , )
k k
= cg ( ) + g ( )
j j
Hence, each g is a linear function on V. Thus we may define a linear operator E on V by setting
j
r
E = g j ( ) j
j 1
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