Page 106 - 35Linear Algebra

106 Vector Spaces

x
R
Example 63 Consider the functions f(x) = e and g(x) = e 2x in R . By taking
combinations of these two vectors we can form the plane {c 1 f +c 2 g|c 1 , c 2 ∈ R} inside
x
2x
2x
R
of R . This is a vector space; some examples of vectors in it are 4e −31e , πe −4e x
1 2x
and e .
2
A hyperplane which does not contain the origin cannot be a vector space
because it fails condition (+iv).
It is also possible to build new vector spaces from old ones using the
product of sets. Remember that if V and W are sets, then their product is
the new set
V × W = {(v, w)|v ∈ V, w ∈ W} ,
or in words, all ordered pairs of elements from V and W. In fact V × W is a
vector space if V and W are. We have actually been using this fact already:

Example 64 The real numbers R form a vector space (over R). The new vector space

R × R = {(x, y)|x ∈ R, y ∈ R}

has addition and scalar multiplication defined by

0
0
0
0
(x, y) + (x , y ) = (x + x , y + y ) and c.(x, y) = (cx, cy) .
2 {1,2}
Of course, this is just the vector space R = R .
5.1.1 Non-Examples

The solution set to a linear non-homogeneous equation is not a vector space
because it does not contain the zero vector and therefore fails (iv).

Example 65 The solution set to

1 1 x 1
=
0 0 y 0

1 0
is + c −1 c ∈ R . The vector is not in this set.
0 1 0

Do notice that if just one of the vector space rules is broken, the example is
not a vector space.
Most sets of n-vectors are not vector spaces.

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