374 CHAPTER 16. VECTOR SPACES

Thus if y acts like the additive inverse, it is the additive inverse.

0x = (0+0)x = 0x+0x

Now add −0x to both sides. This gives 0 = 0x. Finally,

(−1)x+x = (−1)x+1x =(−1+1)x = 0x = 0

By the uniqueness of the additive inverse shown earlier, (−1)x =−x. ■If you are interested in considering other fields, you should have some examples other

than C, R, Q. Some of these are discussed in the following exercises. If you are happywith only considering R and C, skip these exercises. Here is an important example whichgives the typical vector space.

Example 16.1.3 Let Ω be a nonempty set and define V to be the set of functions defined onΩ. Letting a,b,c be scalars and f ,g,h functions, the vector operations are defined as

( f +g)(x) ≡ f (x)+g(x)

(a f )(x) ≡ a( f (x))

Then this is an example of a vector space.

To verify this, we check the axioms.

( f +g)(x) = f (x)+g(x) = g(x)+ f (x) = (g+ f )(x)

Since x is arbitrary, f +g = g+ f .

(( 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) . Let 0 denote the function which is given by 0(x) = 0.Then this is an additive identity because

( f +0)(x) = f (x)+0(x) = f (x)

and so f +0 = f . Let − f be the function which satisfies (− f )(x)≡− f (x) . Then

( f +(− f ))(x)≡ f (x)+(− f )(x)≡ f (x)+− f (x) = 0

Hence f +(− f ) = 0.

((a+b) f )(x)≡ (a+b) f (x) = a f (x)+b f (x)≡ (a f +b f )(x)

and so (a+b) f = a f +b f .

(a( f +g))(x)≡ a( f +g)(x)≡ a( f (x)+g(x))

= a f (x)+bg(x)≡ (a f +bg)(x)

and so a( f +g) = a f +bg.

((ab) f )(x)≡ (ab) f (x) = a(b f (x))≡ (a(b f ))(x)

so (ab f ) = a(b f ). Finally (1 f )(x)≡ 1 f (x) = f (x) so 1 f = f .Note that Rn can be considered the set of real valued functions defined on (1,2, · · · ,n).

Thus everything up till now was just a special case of this more general situation.