Page 143 - DMTH502_LINEAR_ALGEBRA

Unit 10: Linear Functionals

so that the unique expression for as a linear combination of the , is Notes
i
n
= f ( ) .
i i
i 1
Equation (7) provides us with a nice way of describing what the dual basis is. It says if
 1 , 2 ..., n is an ordered basis for V and * f 1 ,..., f n is the dual basis, then f is precisely
i
the function which assigns to each vector in V the ith coordinate of relative to the ordered
basis . Thus we may also call the f the coordinate functions for . The formula (6), when
i
combined with tells us the following:
If f is in V*, and we let ( )f i a i , then when

= x 1 1 ... x n n

we have
f(x) = a x ... a x . ...(8)
1 1
n n
In other words, if we choose an ordered basis  for V and describe each vector in V by its n-tuple
of coordinates (x ,....x ) relative to , then every linear functional on V has the form. This is the
1 n
natural generalization of Example 1, which is the special case V = F and  = { , ..., }.
n
1 n
Example 5: Let V be the vector space of all polynomial functions from R into R which
have degree less than or equal to 2. Let t , t and t be any three distinct real numbers, and let
1 2 3
L (p) = p(t )
i i
Then L , L and L are linear functionals on V. These functionals are linearly independent; for,
1 2 3
suppose

L = c L c L c L
3 3
1 1
2 2
If L = 0, i.e., if L(p) = 0 for each p in V, then applying L to the particular polynomial ‘functions’ 1,
2
x, x , we obtain
c 1 c 2 c 3 = 0
t c t c t c
3 3 = 0
1 1
2 2
2
2
2
t c t c t c = 0
2 2
1 1
3 3
From this it follows that c 1 c 2 c 3 0, because (as a short computation shows) the matrix
1 1 1
t 1 t 2 t 3
t 2 1 t 2 2 t 3 2
is invertible when t , t and t are distinct. Now the L are independent and since V has dimension
1 2 3 i
3, these functional from a basis for V*. What is the basis for V, of which this is the dual? Such a
basis {p , p , p } for V must satisfy
1 2 3
L (p ) =
i i ij
or p (t ) = .
j i ij

LOVELY PROFESSIONAL UNIVERSITY 137