6—Vector Spaces 128the numbers{ai}are called thecomponentsof~ain the specified basis. Note that you don’t have to
talk about orthogonality or unit vectors or any other properties of the basis vectors save that they span
the space and they’re independent.
Example 1 is the prototype for the subject, and the basis usually chosen is the one designated
ˆx,yˆ, (andzˆfor three dimensions). Another notation for this isˆı,ˆ,kˆ— I’ll usexˆ-yˆ. In any case, the
two (or three) arrows are at right angles to each other.
In example 5, the simplest choice of basis is
~e 1 = ( 1 0 0 ... 0 )
~e 2 = ( 0 1 0 ... 0 )
..
.
~en= ( 0 0 0 ... 1 ) (6.1)
In example 6, if the domain of the functions is from−∞to+∞, a possible basis is the set of
functions
ψn(x) =xne−x
(^2) / 2
.
The major distinction between this and the previous cases is that the dimension here is infinite. There
is a basis vector corresponding to each non-negative integer. It’s not obvious that this is a basis, but
it’s true.
If two vectors are equal to each other and you express them in the same basis, the corresponding
components must be equal.
∑iai~ei=
∑
ibi~ei =⇒ ai=bi for alli (6.2)
Suppose you have the relation between two functions of time
A−Bω+γt=βt (6.3)
that is, that the twofunctionsare the same, think of this in terms of vectors: on the vector space of
polynomials inta basis is
~e 0 = 1, ~e 1 =t, ~e 2 =t^2 , etc.
Translate the preceding equation into this notation.
(A−Bω)~e 0 +γ~e 1 =β~e 1 (6.4)
For this to be valid the corresponding components must match:
A−Bω= 0, and γ=β
Differential Equations
When you encounter differential equations such as
m
d^2 x
dt^2
+b
dx
dt
+kx= 0, or γ
d^3 x
dt^3
+kt^2
dx
dt
+αe−βtx= 0, (6.5)
the sets of solutions to each of these equations form vector spaces. All you have to do is to check the
axioms, and because of the theorem in section6.3you don’t even have to do all of that. The solutions