Page 171 - Matrix Analysis & Applied Linear Algebra

166 Chapter 4 Vector Spaces

Solution: By definition, S spans V if and only if for each b ∈V there exist
scalars α i such that

 
α 1

 α 2 
.  = Ax.
b = α 1 a 1 + α 2 a 2 + ··· + α n a n = a 1 | a 2 | ··· | a n

.
 . 
α n
Note: This simple observation often is quite helpful. For example, to test whether
3
or not S = {(111 ) , (1 −1 −1) , (311 )} spans , place these
rows as columns in a matrix A, and ask, “Is the system
     
1 1 3 x 1 b 1
1 −11 =
   x 2   b 2 
1 −11 x 3 b 3
3
consistent for every b ∈ ?” Recall from (2.3.4) that Ax = b is consis-
tent if and only if rank[A|b]= rank (A). In this case, rank (A)=2, but
rank[A|b] = 3 for some b ’s (e.g., b 1 =0,b 2 =1,b 3 =0), so S doesn’t span
3

. On the other hand, S = {(111 ) , (1 −1 −1) , (312 )} is a
3
spanning set for because
 
1 1 3
A =  1 −11 
1 −12
is nonsingular, so Ax = b is consistent for all b (the solution is x = A −1 b ).

As shown below, it’s possible to “add” two subspaces to generate another.

Sum of Subspaces

If X and Y are subspaces of a vector space V, then the sum of X
and Y is defined to be the set of all possible sums of vectors from X
with vectors from Y. That is,

X + Y = {x + y | x ∈X and y ∈Y}.

• The sum X + Y is again a subspace of V. (4.1.1)
• If S X , S Y span X, Y, then S X ∪S Y spans X + Y. (4.1.2)