Page 275 - DMTH502_LINEAR_ALGEBRA

Unit 25: Linear Functional and Adjoints of Inner Product Space

Notes
2
for all , and [ ( )/ ] .
f
Example 1: We should give one example showing that Theorem 1 is not true without the
assumption that V is finite dimensional. Let V be the vector space of polynomials over the field
of complex numbers, with the inner product

1
t
(f g) = f ( ) ( ) .
dt
g
t
0
This inner product can also be defined algebraically. If f = a x and g = b x , then
k
k
k k
j
(f g) = a b .
j k
j k 1
. j k
Let z be a fixed complex number, and let L be the linear functional ‘evaluation at z’:
L(f) = f(z).
Is there a polynomial g such that (f g) = L(f) for every f? The answer is no; for suppose we have
1
t
t
f(z) = f ( ) ( ) dt
g
0
for every f. Let h = x z, so that for any f we have (hf) (z) = 0. Then
1
t
f
t
t
g
0 = h ( ) ( ) ( )dt
0
for all f. In particular this holds when f hg so that
1
h ( ) 2 g ( ) 2 dt = 0
t
t
0
and so hg = 0. Since h 0, it must be that g = 0. But L is not the zero functional; hence, no such g
exists.
One can generalize the example somewhat, to the case where L is a linear combination of point
evaluations. Suppose we select fixed complex numbers z , ...., z and scalars c , ...., c and let
1 n 1 n
L(f) = c f(z ) + ... +c f (z ).
1 1 n n
Then L is a linear functional on V, but there is no g with L(f) = (f g), unless c = c = .... = c = 0. Just
1 2 n
repeat the above argument with h = (x z ) ... (x z ) in the Example 1.
1 n
We turn now to the concept of the adjoint of a linear operator.
25.2 Adjoint of Linear Operators

Theorem 2: For any linear operator T on a finite-dimensional inner product space V, there exists
a unique linear operator T* on V such that

(T ) = ( T* ) ...(2)
for all in V.
Proof: Let be any vector in V. Then (T ) is a linear functional on V. By Theorem 1 there
is a unique vector in V such that (T ) = ( ) for every in V. Let T* denote the mapping

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