Page 109 - DMTH502_LINEAR_ALGEBRA

Unit 6: Computation Concerning Subspaces

 = (0 0 1 4 0) Notes
2
 = (0 0 0 0 1)
3
of R
(c) The row-space W consists of all vectors of the form

 = c  + c  + c 
1 1 2 2 3 3
= (c , 2c , c , 3c + 4c , c )
1 1 2 1 2 3
where c , c , c are arbitrary scalars. Thus (b , b , b , b , b ) is in W if and only if
1 2 3 1 2 3 4 5
(b , b , b , b , b ) = b  + b  + b 
1 2 3 4 5 1 1 3 2 5 3
which is true if and only if
b = 2b
2 1
b = 3b + 4b .
4 1 3
These equations are instances of the general system (3) in unit 5, and using them we may
tell at a glance whether a given vector lies in W. Thus (–5, –10, 1, –11, 20) is a linear
combination of the rows of A, but (1, 2, 3, 4, 5) is not.

(d) The coordinate matrix of the vector (b , 2b , b , 3b + 4b , b ) in the basis { ,  ,  } is
1 1 3 1 3 5 1 2 3
evidently
b  1 
 
 b 3 
5
 b  
(e) There are many ways to write the vectors in W as linear combinations of the rows of A.
 = (b , 2b , b , 3b + 4b , b )
1 1 3 1 3 5
= [b , b , b , 0, 0] . R
1 3 5
= [b , b , b , 0, 0] . PA
1 3 5

 1 0 0 0 0 
 1  1 0 0 0 
 

= [ ,b b b , 0, 0] 0 0 0 0 1 A
,
1 3 5
 
  1 1 1 0 0 
  3 1 0 1  1 
 
= [b + b , –b , 0, 0, b ] . A
1 3 3 5
In particular, with  = (–5, –10, 1, –11, 20) we have

 1 2 0 3 0
 1 2  1  1 0 
 
 = ( 4, 1, 0, 0, 20)  0 0 1 4 0


 
 2 4 1 10 1 
 0 0 0 0 1 
 

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