Andrei Antonenko
March 10, 2003
1 Image and kernel
Last lecture we studied image and kernel of a linear function. Now we will prove one of the
properties of image and kernel. First let’s consider kernel.
Let
f
:
V
→
U
be a linear function, and let its kernel be Ker
f
— set of all elements
v
from
V
which map to
0
. Then we can state the following properties of it.
Existence of zero.
The zero vector
0
belongs to kernel of
f
, since
f
(
0
) =
0
— maps to
0
, so
0
is in kernel.
Summation.
Let vectors
v
and
u
belong to kernel, so,
f
(
v
) =
0
and
f
(
u
) =
0
. Then
f
(
v
+
u
) =
f
(
v
) +
f
(
u
) =
0
,
and thus
u
+
v
belongs to Ker
f
.
Multiplication by a scalar.
Let vector
v
belongs to the kernel of
f
. Then we know that
f
(
v
) =
0
. Now for any constant
k
we have:
f
(
kv
) =
kf
(
v
) =
k
·
0
=
0
,
thus
kv
belongs to Ker
f
.
So, we proved the following theorem:
Theorem 1.1.
The kernel of linear function
f
:
V
→
U
is a vector subspace in
V
.
Example 1.2.
Consider the projection function
f
(
x,y,z
) = (
x,y,
0)
. It’s kernel consists of
vectors of the form
(0
,
0
,c
)
for any constant
c
. Geometrically speaking, this is a
z
axis in the
3dimensional space. This is a vector subspace.
Now let’s consider the image. Let
f
:
V
→
U
be a linear function, and it’s image Im
f
is
the set of all vectors from
U
where we can get by applying a function to vectors from
V
. We’ll
state some properties of it.
1
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 Spring '03
 Andant
 Linear Algebra, Vector Space, Linear function, Ker

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