Page 297 - DMTH502_LINEAR_ALGEBRA

Unit 27: Introduction and Forms on Inner Product Spaces

27.2 Forms on Inner Product Spaces Notes

If T is a linear operator on a finite-dimensional inner product space V the function f defined on
V × V by

f( , ) = (T | )
may be regarded as a kind of substitute of T. Many questions about T are equivalent to questions
concerning f. In fact, it is easy to see that f determines T. For if = { , ..., } is an orthonormal
1 n
basis for V, then the entries of the matrix of T in are given by
A = f( , )
jk k j
It is important to understand why f determines T from a more abstract point of view. The crucial
properties of f are described in the following definition.
Definition: A (sesquilinear) form on a real or complex vector space V is a function f on V V with
values in the field of scalars such that
(a) f(c + , ) = cf( , ) + f( , )
(b) f( + c , ) = c f( , ) + f( , )
for all , , in V and all scalars c.

Thus, a sesquilinear form is a function on V V such that f( , ) is a linear function of for fixed
and a conjugate-linear function of for fixed . In the real case, f( , ) is linear as a function of
each argument; in other words, f is a bilinear form. In the complex case, the sesquilinear form f
is not bilinear unless f = 0. In the remainder of this chapter, we shall omit the adjective ‘sesquilinear’
unless it seems important to include it.
If f and g are forms on V and c is a scalar, it is easy to check that cf + g is also a form. From this it
follows that any linear combination of forms on V is again a form. Thus the set of all forms on
V is a subspace of the vector space of all scalar-valued functions on V V.
Theorem 1: Let V be a finite-dimensional inner product space and f a form on V. Then there is a
unique linear operator T on V such that
f( , ) = (T | )
for all , , in V and the map f T is an isomorphism of the space of forms onto L(V, V).
Proof: Fix a vector in V. Then a f( , ) is a linear function on V. By theorem 6 in unit 26 there
is a unique vector in V such that f( , ) = ( | ) for every . We define a function U from V into
V by setting U = . Then
f( |c + ) = ( |U(c + ))
= cf ( , ) f ( , )

= ( |c U ) ( |U )
= ( |c U U )

for all , , in V and all scalars c. Thus U is a linear operator on V and T = U* is an operator such
that f( , ) = (T | ) for all and . If we also have f( , ) = (T’ | ), then
(T – T’ | ) = 0

for all and ; so T = T’ for all . Thus for each form f there is a unique linear operator T such
f
that
f( , ) = (T | )
j

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