Linear Algebra

190 Chapter 3. Maps Between Spaces
When is independence not lost? One obvious sufficient condition is when
the homomorphism is an isomorphism (this condition is also necessary; see
Exercise 34.) We finish our comparison of homomorphisms and isomorphisms by
observing that a one-to-one homomorphism is an isomorphism from its domain
onto its range.
2.18 Definition A linear map that is one-to-one is nonsingular.
(In the next section we will see the connection between this use of ‘nonsingular’
for maps and its familiar use for matrices.)
2.19 Example This nonsingular homomorphism ι: R 2 → R 3
⎛
� �
x ι
↦−→ ⎝
y
x
⎞
y⎠
0
gives the obvious correspondence between R 2 and the xy-plane inside of R 3 .
We will close this section by adapting some results about isomorphisms to
this setting.
2.20 Theorem In an n-dimensional vector space V , then these
(1) h is nonsingular, that is, one-to-one
(2) h has a linear inverse
(3) N (h) ={�0 }, that is, nullity(h) =0
(4) rank(h) =n
(5) if 〈 � β1,..., � βn〉 is a basis for V then 〈h( � β1),...,h( � βn)〉 is a basis for R(h)
are equivalent statements about a linear map h: V → W .
Proof. We will first show that (1) ⇐⇒ (2). We will then show that (1) =⇒
(3) =⇒ (4) =⇒ (5) =⇒ (2).
For (1) =⇒ (2), suppose that the linear map h is one-to-one, and so has an
inverse. The domain of that inverse is the range of h and so a linear combination
of two members of that domain has the form c1h(�v1) +c2h(�v2). On that
combination, the inverse h −1 gives this.
h −1 (c1h(�v1)+c2h(�v2)) = h −1 (h(c1�v1 + c2�v2))
= h −1 ◦ h (c1�v1 + c2�v2)
= c1�v1 + c2�v2
= c1h −1 ◦ h (�v1)c2h −1 ◦ h (�v2)
= c1 · h −1 (h(�v1)) + c2 · h −1 (h(�v2))
Thus the inverse of a one-to-one linear map is automatically linear. But this also
gives the (1) =⇒ (2) implication, because the inverse itself must be one-to-one.
Of the remaining implications, (1) =⇒ (3) holds because any homomorphism
maps �0V to �0W , but a one-to-one map sends at most one member of V
to �0W .