Linear Algebra

188 Chapter 3. Maps Between Spaces
because it contains �0V , as S contains �0W . To show that it is closed under
combinations, let �v1 and �v2 be elements of the inverse image, so that h(�v1) and
h(�v2) are members of S. Then c1�v1 + c2�v2 is also in the inverse image because
under h it is sent h(c1�v1 +c2�v2) =c1h(�v1)+c2h(�v2) to a member of the subspace
S. QED
2.10 Definition The nullspace or kernel of a linear map h: V → W is
N (h) ={�v ∈ V � � h(�v) =�0W } = h −1 (�0W ).
The dimension of the nullspace is the map’s nullity.
2.11 Example The map from Example 2.3 has this nullspace N (d/dx) =
{a0 +0x +0x 2 +0x 3 � � a0 ∈ R}.
2.12 Example The map from Example 2.4 has this nullspace.
� �
a b ��
N (h) ={
a, b ∈ R}
0 −(a + b)/2
Now for the second insight from the above pictures. In Example 2.5, each
of the vertical lines is squashed down to a single point—π, in passing from the
domain to the range, takes all of these one-dimensional vertical lines and “zeroes
them out”, leaving the range one dimension smaller than the domain. Similarly,
in Example 2.6, the two-dimensional domain is mapped to a one-dimensional
range by breaking the domain into lines (here, they are diagonal lines), and
compressing each of those lines to a single member of the range. Finally, in
Example 2.7, the domain breaks into planes which get “zeroed out”, and so the
map starts with a three-dimensional domain but ends with a one-dimensional
range—this map “subtracts” two from the dimension. (Notice that, in this
third example, the codomain is two-dimensional but the range of the map is
only one-dimensional, and it is the dimension of the range that is of interest.)
2.13 Theorem A linear map’s rank plus its nullity equals the dimension of its
domain.
Proof. Let h: V → W be linear and let BN = 〈 � β1,... , � βk〉 be a basis for
the nullspace. Extend that to a basis BV = 〈 � β1,..., � βk, � βk+1,..., � βn〉 for the
entire domain. We shall show that BR = 〈h( � βk+1),...,h( � βn)〉 is a basis for the
rangespace. Then counting the size of these bases gives the result.
To see that BR is linearly independent, consider the equation ck+1h( � βk+1)+
···+ cnh( � βn) =�0W . This gives that h(ck+1 � βk+1 + ···+ cn � βn) =�0W and so
ck+1 � βk+1 +···+cn � βn is in the nullspace of h. AsBN is a basis for this nullspace,
there are scalars c1,...,ck ∈ R satisfying this relationship.
c1 � β1 + ···+ ck � βk = ck+1 � βk+1 + ···+ cn � βn
But BV is a basis for V so each scalar equals zero. Therefore BR is linearly
independent.