Linear Algebra, 2020a

Section I. Isomorphisms 187
In the proof we express elements ⃗v of the domain space as combinations
of members of the basis B and then associate ⃗v with the column vector of
coefficients. That there is at least one expansion of each ⃗v holds because B is a
basis and so spans the space.
The worry that there is no more than one associated member of the codomain
is subtler. A contrasting example, where an association fails this unique output
requirement, illuminates the issue. Let the domain be P 2 and consider a set that
is not a basis (it is not linearly independent, although it does span the space).
A = {1 + 0x + 0x 2 ,0+ 1x + 0x 2 ,0+ 0x + 1x 2 ,1+ 1x + 2x 2 }
Call those polynomials ⃗α 1 ,...,⃗α 4 . In contrast to the situation when the set
is a basis, here there can be more than one expression of a domain vector in
terms of members of the set. For instance, consider ⃗v = 1 + x + x 2 . Here are
two different expansions.
⃗v = 1⃗α 1 + 1⃗α 2 + 1⃗α 3 + 0⃗α 4 ⃗v = 0⃗α 1 + 0⃗α 2 − 1⃗α 3 + 1⃗α 4
So this input vector ⃗v is associated with more than one column.
⎛ ⎞ ⎛ ⎞
1 0
1
0
⎜ ⎟ ⎜ ⎟
⎝1⎠
⎝−1⎠
0 1
Thus, with A the association is not well-defined. (The issue is that A is not
linearly independent; to show uniqueness Theorem Two.III.1.12’s proof uses only
linear independence.)
In general, any time that we define a function we must check that output
values are well-defined. Most of the time that condition is perfectly obvious but
in the above proof it needs verification. See Exercise 22.
2.8 Corollary Each finite-dimensional vector space is isomorphic to one and only
one of the R n .
This gives us a collection of representatives of the isomorphism classes.
All finite dimensional
vector spaces:
⋆ R 0 ⋆ R 3
⋆ R 2
...
⋆ R 1
One representative
per class
The proofs above pack many ideas into a small space. Through the rest of
this chapter we’ll consider these ideas again, and fill them out. As a taste of
this we will expand here on the proof of Lemma 2.5.