Page 201 - 35Linear Algebra

9.2 Building Subspaces 201

Hence, thanks to the subspace theorem, the set of all vectors in U that are mapped
to the zero vector is a subspace of V . It is called the kernel of L:

kerL := {u ∈ U|L(u) = 0} ⊂ U.

Note that finding a kernel means finding a solution to a homogeneous linear equation.

Example 113 (The image of a linear map).
Suppose L : U → V is a linear map between vector spaces. Then if

0
0
v = L(u) and v = L(u ) ,
linearity tells us that

0
0
0
αv + βv = αL(u) + βL(u ) = L(αu + βu ) .
Hence, calling once again on the subspace theorem, the set of all vectors in V that
are obtained as outputs of the map L is a subspace. It is called the image of L:

imL := {L(u) | u ∈ U} ⊂ V.

Example 114 (An eigenspace of a linear map).

Suppose L : V → V is a linear map and V is a vector space. Then if

L(u) = λu and L(v) = λv ,

linearity tells us that

L(αu + βv) = αL(u) + βL(v) = αL(u) + βL(v) = αλu + βλv = λ(αu + βv) .

Hence, again by subspace theorem, the set of all vectors in V that obey the eigenvector
equation L(v) = λv is a subspace of V . It is called an eigenspace

V λ := {v ∈ V |L(v) = λv}.

For most scalars λ, the only solution to L(v) = λv will be v = 0, which yields the
trivial subspace {0}. When there are nontrivial solutions to L(v) = λv, the number λ
is called an eigenvalue, and carries essential information about the map L.

Kernels, images and eigenspaces are discussed in great depth in chap-
ters 16 and 12.

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