Linear Algebra

Section II. Homomorphisms 189
To show that BR spans the rangespace, consider h(�v) ∈ R(h) and write �v
as a linear combination �v = c1 � β1 + ···+ cn � βn of members of BV . This gives
h(�v) =h(c1 � β1+···+cn � βn) =c1h( � β1)+···+ckh( � βk)+ck+1h( � βk+1)+···+cnh( � βn)
and since � β1, ... , � βk are in the nullspace, we have that h(�v) =�0 +···+ �0 +
ck+1h( � βk+1)+···+ cnh( � βn). Thus, h(�v) is a linear combination of members of
BR, andsoBR spans the space. QED
2.14 Example Where h: R 3 → R 4 is
⎛
⎝ x
⎞
y⎠
z
h
⎛ ⎞
x
⎜
↦−→ ⎜0
⎟
⎝y⎠
0
we have that the rangespace and nullspace are
⎛ ⎞
a
⎜
R(h) ={ ⎜0⎟
�
⎟ �
⎝b⎠
a, b ∈ R}
0
and
⎛ ⎞
0
N (h) ={ ⎝0⎠
z
� � z ∈ R}
and so the rank of h is two while the nullity is one.
2.15 Example If t: R → R is the linear transformation x ↦→ −4x, then the
range is R(t) =R 1 , and so the rank of t is one and the nullity is zero.
2.16 Corollary The rank of a linear map is less than or equal to the dimension
of the domain. Equality holds if and only if the nullity of the map is zero.
We know that there an isomorphism exists between two spaces if and only
if their dimensions are equal. Here we see that for a homomorphism to exist,
the dimension of the range must be less than or equal to the dimension of the
domain. For instance, there is no homomorphism from R 2 onto R 3 —there are
many homomorphisms from R 2 into R 3 , but none has a range that is all of
three-space.
The rangespace of a linear map can be of dimension strictly less than the
dimension of the domain (an example is that the derivative transformation on
P3 has a domain of dimension four but a range of dimension three). Thus, under
a homomorphism, linearly independent sets in the domain may map to linearly
dependent sets in the range (for instance, the derivative sends {1,x,x 2 ,x 3 } to
{0, 1, 2x, 3x 2 }). That is, under a homomorphism, independence may be lost. In
contrast, dependence is preserved.
2.17 Lemma Under a linear map, the image of a linearly dependent set is
linearly dependent.
Proof. Suppose that c1�v1 + ··· + cn�vn = �0V , with some ci nonzero. Then,
because h(c1�v1 +···+cn�vn) =c1h(�v1)+···+cnh(�vn) and because h(�0V )=�0W ,
we have that c1h(�v1)+···+ cnh(�vn) =�0W with some nonzero ci. QED