A First Course in Linear Algebra, 2017a

460 Vector Spaces
B + A = B
[ ] 0 0 0
=
1 0 0
It follows that A + B ≠ B + A. Therefore addition as defined for V is not commutative and V fails
this axiom. Hence V is not a vector space.
Consider another example of a vector space.
Example 9.8: Vector Space of Functions
Let S be a nonempty set and define F S to be the set of real functions defined on S. Inotherwords,
we write F S : S ↦→ R. Letting a,b,c be scalars and f ,g,h functions, the vector operations are defined
as
Show that F S is a vector space.
( f + g)(x) = f (x)+g(x)
(af)(x) = a( f (x))
♠
Solution. To verify that F S is a vector space, we must prove the axioms beginning with those for addition.
Let f ,g,h be functions in F S .
• First we check that addition is closed. For functions f ,g defined on the set S, their sum given by
( f + g)(x)= f (x)+g(x)
is again a function defined on S. Hence this sum is in F S and F S is closed under addition.
• Secondly, we check the commutative law of addition:
Since x is arbitrary, f + g = g + f .
• Next we check the associative law of addition:
( f + g)(x)= f (x)+g(x)=g(x)+ f (x)=(g + f )(x)
(( f + g)+h)(x)=(f + g)(x)+h(x)=(f (x)+g(x)) + h(x)
= f (x)+(g(x)+h(x)) = ( f (x)+(g + h)(x)) = ( f +(g + h))(x)
and so ( f + g)+h = f +(g + h).
• Next we check for an additive identity. Let 0 denote the function which is given by 0(x)=0. Then
this is an additive identity because
and so f + 0 = f .
( f + 0)(x)= f (x)+0(x)= f (x)