Page 294 - 35Linear Algebra

294 Kernel, Range, Nullity, Rank

Proof. The proof of this theorem is Review Exercise 2.

linear
Theorem 16.2.3. If L: V −−−→ W then ker L is a subspace of V .

Proof. Notice that if L(v) = 0 and L(u) = 0, then for any constants c, d,
L(cu+dv) = 0. Then by the subspace theorem, the kernel of L is a subspace
of V .

3
Example 150 Let L: R → R be the linear transformation defined by L(x, y, z) =
3
(x + y + z). Then ker L consists of all vectors (x, y, z) ∈ R such that x + y + z = 0.
Therefore, the set
3
V = {(x, y, z) ∈ R | x + y + z = 0}
3
is a subspace of R .
When L : V → V , the above theorem has an interpretation in terms of
the eigenspaces of L. Suppose L has a zero eigenvalue. Then the associated
eigenspace consists of all vectors v such that Lv = 0v = 0; the 0-eigenspace
of L is exactly the kernel of L.
In the example where L(x, y) = (x + y, x + 2y, y), the map L is clearly
2
3
not surjective, since L maps R to a plane through the origin in R . But any
plane through the origin is a subspace. In general notice that if w = L(v)
0
0
and w = L(v ), then for any constants c, d, linearity of L ensures that
0
0
cw + dw = L(cv + dv ) .
Now the subspace theorem strikes again, and we have the following theorem:

Theorem 16.2.4. If L: V → W is linear then the range L(V ) is a subspace
of W.

Example 151 Let L(x, y) = (x + y, x + 2y, y). The range of L is a plane through
3
the origin and thus a subspace of R . Indeed the matrix of L in the standard basis is
 
1 1
1 2 .
 
0 1
The columns of this matrix encode the possible outputs of the function L because

     
1 1 1 1
x
L(x, y) =  1 2  = x   + y   .
2
1
y
0 1 0 1
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