Linear Algebra

114 Chapter 2. Vector Spaces
1.5 Definition For any Rn ,
⎛ ⎞
1
⎜
⎜0
⎟
En = 〈 ⎜.
⎝.
⎟
. ⎠
0
,
⎛ ⎞
0
⎜
⎜1
⎟
⎜.
⎝.
⎟
. ⎠
0
,...,
⎛ ⎞
0
⎜
⎜0
⎟
⎜.
⎝.
⎟
. ⎠
1
〉
is the standard (or natural) basis. We denote these vectors by �e1,...,�en.
Note that the symbol ‘�e1’ means something different in a discussion of R 3 than
it means in a discussion of R 2 . (Calculus books call R 2 ’s standard basis vectors
�ı and �j instead of �e1 and �e2, and they call R 3 ’s standard basis vectors �ı, �j, and
� k instead of �e1, �e2, and�e3.)
1.6 Example We can give bases for spaces other than just those comprised of
column vectors. For instance, consider the space {a · cos θ + b · sin θ � � a, b ∈ R}
of function of the real variable θ. This is a natural basis
〈1 · cos θ +0· sin θ, 0 · cos θ +1· sin θ〉 = 〈cos θ, sin θ〉
while, another, more generic, basis is 〈cos θ − sin θ, 2cosθ +3sinθ〉. Verfication
that these two are bases is Exercise 22.
1.7 Example A natural basis for the vector space of cubic polynomials P3 is
〈1,x,x 2 ,x 3 〉. Two other bases for this space are 〈x 3 , 3x 2 , 6x, 6〉 and 〈1, 1+x, 1+
x + x 2 , 1+x + x 2 + x 3 〉. Checking that these are linearly independent and span
the space is easy.
1.8 Example The trivial space {�0} has only one basis, the empty one 〈〉.
1.9 Example The space of finite degree polynomials has a basis with infinitely
many elements 〈1,x,x 2 ,...〉.
1.10 Example We have seen bases before. For instance, we have described
the solution set of homogeneous systems such as this one
by paramatrizing.
x + y − w =0
z + w =0
⎛ ⎞ ⎛ ⎞
−1 1
⎜
{ ⎜ 1 ⎟ ⎜
⎟
⎝ 0 ⎠ y + ⎜ 0 ⎟
⎝−1⎠
0 1
w � � y, w ∈ R}
That is, we have described the vector space of solutions as the span of a twoelement
set. We can easily check that this two-vector set is also linearly independent.
Thus the solution set is a subspace of R 4 with a two-element basis.