Page 146 - DMTH502_LINEAR_ALGEBRA

Linear Algebra

Notes Now one may look at the system of equations from the ‘dual’ point of view. That is, suppose that
n
we are given m vectors in F .
(A ,...,A )
1 1 i in
and we wish to find the annihilator of the subspace spanned by these vectors. Since a typical
n
linear functional on F has the form
x
x
f ( ,... ) c x ... c x
1 1
1
n n
n
the condition that f be in this annihilator is that
n
A c 0, i 1,...,m
ij j
j 1
that is, that ( ,....,c 1 c n ) be a solution of the system AX = 0. From this point of view, row-reduction
gives us a systematic method of finding the annihilator of the subspace spanned by a given finite
set of vectors in F .
n

4
Example 6: Here are three linear functionals on R :
x
f ( ,x ,x ,x ) = x 2x 2x x
1 1 2 3 4 1 2 3 4
x
f 2 ( ,x 2 ,x 3 ,x 4 ) = 2x 2 x 4
1
f 3 ( ,x 2 ,x 3 ,x 4 ) = 2x 1 4x 2 3 .
x
x
1
4
The subspace which they annihilate may be found explicitly by finding the row-reduced echelon
form of the matrix
1 2 2 1
A = 0 2 0 1
2 0 4 3

A short calculation, shows that A goes over 2R as

1 0 2 0
R = 0 1 0 0
0 0 0 1

Therefore, the linear functionals
x
g 1 ( ,x 2 ,x 3 ,x 4 ) = x 1 2x 3
1
g 2 ( ,x 2 ,x 3 ,x 4 ) = x 2
x
1
x
g 3 ( ,x 2 ,x 3 ,x 4 ) = x 4
1
4
span the same subspace of (R )* and annihilate the same subspace of R as do f f 2 , . The
,
4
f
1
3
subspace annihilated consists of the vectors with
x = –2x
1 3
x = x = 0
2 4
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