Linear Algebra

172 Chapter 3. Maps Between Spaces
Thus the function is an isomorphism, and we can say that any n-dimensional
space is isomorphic to the n-dimensional space R n . Consequently, as noted at
the start, any two spaces with the same dimension are isomorphic. QED
2.5 Remark The parenthetical comment in that proof about the role played
by the ‘one and only one representation’ result requires some explanation. We
need to show that each vector in the domain is associated by Rep B with one
and only one vector in the codomain.
A contrasting example, where an association doesn’t have this property, is
illuminating. Consider this subset of P2, which is not a basis.
A = {1+0x +0x 2 , 0+1x +0x 2 , 0+0x +1x 2 , 1+1x +2x 2 }
Call those four polynomials �α1, ... , �α4. If, mimicing above proof, we try to
write the members of P2 as �p = c1�α1 + c2�α2 + c3�α3 + c4�α4, and associate �p
with the four-tall vector with components c1, ... , c4 then there is a problem.
For, consider �p(x) =1+x + x 2 . The set A spans the space P2, so there is at
least one four-tall vector associated with �p. But A is not linearly independent
so vectors do not have unique decompositions. In this case, both
�p(x) =1�α1 +1�α2 +1�α3 +0�α4 and �p(x) =0�α1 +0�α2 − 1�α3 +1�α4
and so there is more than one four-tall vector associated with �p.
⎛ ⎞
1
⎜
⎜1
⎟
⎝1⎠
0
and
⎛ ⎞
0
⎜ 0 ⎟
⎝−1⎠
1
If we are trying to think of this association as a function then the problem is
that, for instance, with input �p the association does not have a well-defined
output value.
Any map whose definition appears possibly ambiguous must be checked to
see that it is well-defined. For the above proof that check is Exercise 19.
That ends the proof of Theorem 2.2. We say that the isomorphism classes
are characterized by dimension because we can describe each class simply by
giving the number that is the dimension of all of the spaces in that class.
This subsection’s results give us a collection of representatives of the isomorphism
classes. ∗
2.6 Corollary A finite-dimensional vector space is isomorphic to one and only
one of the R n .
2.7 Remark 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. For a
taste of this, we will close this section by indicating how we can expand on the
proof of Lemma 2.4.
∗ More information on equivalence class representatives is in the appendix.