Page 147 - DMTH502_LINEAR_ALGEBRA

Unit 10: Linear Functionals

Notes
Example 7: Let W be the subspace of R which is spanned by the vectors
4
= (2, –2, 3, 4, –1) = (0, 0, –1, –2, 3)
1 3
= (–1, 1, 2, 5, 2) =(1, –1, 2, 3, 0).
2 4
0
How does one describe W , the annihilator of W? Let us form the 4 × 5 matrix A with row vectors
, , , , and find the row-reduced echelon matrix R which is row-equivalent of A:
1 2 3 4
2 2 3 4 1 1 1 0 1 0
1 1 2 5 2 0 0 1 2 0
A R
0 0 1 2 3 0 0 0 0 1
1 1 2 3 0 0 0 0 0 0

If f is a linear functional on R :
5
5
x
f ( ,....,x ) = c x
1 5 j j
j 1
then f is in W° if and only if ( ) 0,f i i 1, 2, 3, 4, i.e., if and only if

5
A c = 0, 1 i 4.
ij j
j 1
This is equivalent to

5
R c = 0, 1 i 3
ij j
j 1
or

c – c – c = 0
1 2 4
c + 2c = 0
3 4
c = 0
5
We obtain all such linear functionals f by assigning arbitrary values to c and c , say c = a and
2 4 2
c = b, and then finding the corresponding c = a + b, c = – 2b, c = 0. So W° consists of all linear
4 1 3 5
functionals f of the form
x
f ( ,x 2 ,x 3 ,x 4 ,x 5 ) (a b )x 1 ax 2 2bx 3 bx 4
1
The dimension of W* is 2 the basis (f , f ) for W* can be found by first taking a = 1, b = 0 and then
1 2
a = 0 and b = 1:
f 1 ( ,x 2 ,x 3 ,x 4 ,x 5 ) = x 1 x 2
x
1
f ( ,x ,x ,x ,x ) = x 2x x
x
2 1 2 3 4 5 1 3 4
The above general f in W* is f a f 1 b 2 . f

LOVELY PROFESSIONAL UNIVERSITY 141