Page 274 - DMTH502_LINEAR_ALGEBRA

Linear Algebra

Notes This function f is a linear functional on V, because by its very definition, ( ) is linear as a function
of . If V is finite-dimensional, every linear functional on V arises in this way from some .
Theorem 1: Let V be a finite-dimensional inner product space, and f a linear functional on V. Then
there exists a unique vector in V such that f( ) = ( ) for all in V.
Proof: Let { , , ...., } be an orthonormal basis for V. Put
1 2 n
n
= ( f j ) j ...(1)
j 1

and let f be the linear functional defined by
f ( ) = ( ).
Then

f ( ) = k ( f j ) j ( f k )
k
j
Since this is true for each , is follows that f = f . Now suppose is a vector in V such that ( )
k
= ( ) for all . Then ( ) = 0 and . Thus there is exactly one vector determining
the linear functional f in the stated manner.

The proof of this theorem can be reworded slightly, in terms of the representation of linear
functionals in a basis. If we choose on orthonormal basis { , ...., ) for V, the inner product of
1 n
= x + ... + x and = y + ...+ y will be
1 1 n n 1 1 n n
( ) = x y ... x y .
1 1
n n
If f is any linear functional on V, then f has the form
f( ) = c x + ... +c x
1 1 n n
for some fixed scalars c , ...., c determined by the basis. Of course c = f ( ). If we wish to find a
1 n j j
vector in V such that ( ) = f ( ) for all , then clearly the coordinates y of must satisfy
j
y i c j or y i ( f j ). Accordingly,

= ( f 1 ) 1 ... ( f n ) n
is the desired vector.
Some further comments are in order. The proof of Theorem 1 that we have given is admirably
brief, but it fails to emphasize the essential geometric fact that lies in the orthogonal complement
of the null space of f. Let W be the null space of f. Then V = W + W , and f is completely determined
by its values on W . In fact, if P is the orthogonal projection of V on W , then
f( ) = f (P )

for all in V. Suppose f 0. Then f is of rank 1 and dim (W ) = 1. If is any non-zero vector in W ,
it follows that
( )
P = 2

for all in V. Thus
f ( )
f( ) = ( ).
2

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