A First Course in Linear Algebra, 2017a

512 Vector Spaces
9.8 The Kernel And Image Of A Linear Map
Outcomes
A. Describe the kernel and image of a linear transformation.
B. Use the kernel and image to determine if a linear transformation is one to one or onto.
Here we consider the case where the linear map is not necessarily an isomorphism. First here is a
definition of what is meant by the image and kernel of a linear transformation.
Definition 9.77: Kernel and Image
Let V and W be vector spaces and let T : V → W be a linear transformation. Then the image of T
denoted as im(T ) is defined to be the set
{T (⃗v) :⃗v ∈ V }
In words, it consists of all vectors in W which equal T (⃗v) for some ⃗v ∈ V. The kernel, ker(T ),
consists of all⃗v ∈ V such that T (⃗v)=⃗0. Thatis,
{
}
ker(T )= ⃗v ∈ V : T (⃗v)=⃗0
Then in fact, both im(T ) and ker(T ) are subspaces of W and V respectively.
Proposition 9.78: Kernel and Image as Subspaces
Let V,W be vector spaces and let T : V → W be a linear transformation. Then ker(T ) ⊆ V and
im(T ) ⊆ W . In fact, they are both subspaces.
Proof. First consider ker(T ). It is necessary to show that if ⃗v 1 ,⃗v 2 are vectors in ker(T ) and if a,b are
scalars, then a⃗v 1 + b⃗v 2 is also in ker(T ). But
T (a⃗v 1 + b⃗v 2 )=aT(⃗v 1 )+bT(⃗v 2 )=a⃗0 + b⃗0 =⃗0
Thus ker(T ) is a subspace of V.
Next suppose T (⃗v 1 ),T(⃗v 2 ) are two vectors in im(T ).Thenifa,b are scalars,
aT(⃗v 2 )+bT(⃗v 2 )=T (a⃗v 1 + b⃗v 2 )
and this last vector is in im(T ) by definition.
♠
Consider the following example.