Linear Algebra, 2020a

Section I. Isomorphisms 183
(d) Produce an isomorphism from P 2 to R 3 that fits these specifications.
⎛ ⎞
1
x + x 2 ↦→ ⎝0⎠ and
⎛ ⎞
0
1 − x ↦→ ⎝1⎠
0
0
1.38 Prove that a space is n-dimensional if and only if it is isomorphic to R n . Hint.
Fix a basis B for the space and consider the map sending a vector over to its
representation with respect to B.
1.39 (Requires the subsection on Combining Subspaces, which is optional.) Let
U and W be vector spaces. Define a new vector space, consisting of the set
U × W = {(⃗u, ⃗w) | ⃗u ∈ U and ⃗w ∈ W } along with these operations.
(⃗u 1 , ⃗w 1 )+(⃗u 2 , ⃗w 2 )=(⃗u 1 + ⃗u 2 , ⃗w 1 + ⃗w 2 ) and r · (⃗u, ⃗w) =(r⃗u, r⃗w)
This is a vector space, the external direct sum of U and W.
(a) Check that it is a vector space.
(b) Find a basis for, and the dimension of, the external direct sum P 2 × R 2 .
(c) What is the relationship among dim(U), dim(W), and dim(U × W)?
(d) Suppose that U and W are subspaces of a vector space V such that V = U ⊕ W
(in this case we say that V is the internal direct sum of U and W). Show that
the map f: U × W → V given by
(⃗u, ⃗w)
f
↦−→ ⃗u + ⃗w
is an isomorphism. Thus if the internal direct sum is defined then the internal
and external direct sums are isomorphic.
I.2 Dimension Characterizes Isomorphism
In the prior subsection, after stating the definition of isomorphism, we gave some
results supporting our sense that such a map describes spaces as “the same.”
Here we will develop this intuition. When two (unequal) spaces are isomorphic
we think of them as almost equal, as equivalent. We shall make that precise by
proving that the relationship ‘is isomorphic to’ is an equivalence relation.
2.1 Lemma The inverse of an isomorphism is also an isomorphism.
Proof Suppose that V is isomorphic to W via f: V → W. An isomorphism is a
correspondence between the sets so f has an inverse function f −1 : W → V that
is also a correspondence. ∗
We will show that because f preserves linear combinations, so also does f −1 .
Suppose that ⃗w 1 , ⃗w 2 ∈ W. Because it is an isomorphism, f is onto and there
∗ More information on inverse functions is in the appendix.