Let
K
{\displaystyle K}
be a field. We now consider
K
{\displaystyle K}
as a vector space over itself.
From school we already know the vector space
R
3
{\displaystyle \mathbb {R} ^{3}}
over the field
R
{\displaystyle \mathbb {R} }
. The vectors in
R
3
{\displaystyle \mathbb {R} ^{3}}
have the form
(
x
,
y
,
z
)
T
{\displaystyle (x,y,z)^{T}}
with
x
,
y
,
z
∈
R
{\displaystyle x,y,z\in \mathbb {R} }
. We can consider the vectors in a 3-dimensional coordinate system. Since
R
3
{\displaystyle \mathbb {R} ^{3}}
is a vector space, we can add and scale vectors.
We also know the vector space
R
2
{\displaystyle \mathbb {R} ^{2}}
. The vectors in
R
2
{\displaystyle \mathbb {R} ^{2}}
have the form
(
x
,
y
)
T
{\displaystyle (x,y)^{T}}
with
x
,
y
∈
R
{\displaystyle x,y\in \mathbb {R} }
. We can get
R
2
{\displaystyle \mathbb {R} ^{2}}
from
R
3
{\displaystyle \mathbb {R} ^{3}}
by deleting one of the coordinates
x
,
y
,
z
{\displaystyle x,y,z}
(e.g., the last one). Illustratively, we then go from the 3-dimensional coordinate system to the
x
y
{\displaystyle xy}
plane. So when omitting a coordinate from
R
3
{\displaystyle \mathbb {R} ^{3}}
, the vector space structure is conserved. What happens if we delete another coordinate?
For example, if we omit the second coordinate of
(
x
,
y
)
{\displaystyle (x,y)}
, only
x
{\displaystyle x}
remains and we get an element in
R
{\displaystyle \mathbb {R} }
. Illustratively, we thus go from the
x
y
{\displaystyle xy}
plane to the
x
{\displaystyle x}
axis. Again, when deleting a coordinate, the vector space structure should not be broken.
We can add and scale the elements in
R
{\displaystyle \mathbb {R} }
(just like vectors), because for all
x
,
y
∈
R
{\displaystyle x,y\in \mathbb {R} }
we have
x
+
y
∈
R
{\displaystyle x+y\in \mathbb {R} }
and for all
λ
∈
R
{\displaystyle \lambda \in \mathbb {R} }
and
x
∈
R
{\displaystyle x\in \mathbb {R} }
it holds that
λ
⋅
x
∈
R
{\displaystyle \lambda \cdot x\in \mathbb {R} }
.
Addition of the vectors
1
{\displaystyle 1}
and
2
{\displaystyle 2}
on the real line
Scalar multiplication of the vector
2
{\displaystyle {\sqrt {2}}}
with the scalar
1.5
{\displaystyle 1.5}
on the real line
Now our field
R
{\displaystyle \mathbb {R} }
should be an
R
{\displaystyle \mathbb {R} }
-vector space.
Visually, this vector space is the number line.
We can apply this idea to an arbitrary field
K
{\displaystyle K}
, since also in an arbitrary
K
{\displaystyle K}
we can add elements and multiply them by scalars in
K
{\displaystyle K}
. Therefore, we conjecture that
K
{\displaystyle K}
is a
K
{\displaystyle K}
-vector space.
Definition of the vector space structure
Bearbeiten
Let
(
K
,
+
,
⋅
)
{\displaystyle (K,+,\cdot )}
be a field.
Then we can define an addition and a scalar multiplication.
The field is a vector space over itself
Bearbeiten
Theorem (
K
{\displaystyle K}
is a vector space)
(
K
,
⊞
,
⊡
)
{\displaystyle (K,\boxplus ,\boxdot )}
is a
K
{\displaystyle K}
-vector space .
How to get to the proof? (
K
{\displaystyle K}
is a vector space)
We proceed as in the article Proofs for vector spaces .
Proof (
K
{\displaystyle K}
is a vector space)
So now we have to establish the eight Vector space axioms .
Proof step: Associativity of addition
Let
x
,
y
,
z
∈
K
{\displaystyle x,y,z\in K}
.
Then:
(
x
⊞
y
)
⊞
z
=
↓
definition of
⊞
=
(
x
+
y
)
⊞
z
↓
definition of
⊞
=
(
x
+
y
)
+
z
↓
associativity of addition in
K
=
x
+
(
y
+
z
)
↓
definition of
⊞
=
x
⊞
(
y
⊞
z
)
.
{\displaystyle {\begin{aligned}(x\boxplus y)\boxplus z&=\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxplus \right.}\\[0.3em]&=(x+y)\boxplus z\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxplus \right.}\\[0.3em]&=(x+y)+z\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{associativity of addition in }}K\right.}\\[0.3em]&=x+(y+z)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxplus \right.}\\[0.3em]&=x\boxplus (y\boxplus z).\end{aligned}}}
This shows the associativity of the addition.
Proof step: Commutativity of addition
Let
x
,
y
∈
K
{\displaystyle x,y\in K}
.
Then:
x
⊞
y
=
↓
definition of
⊞
=
x
+
y
↓
commutativity of the addition in
K
=
y
+
x
↓
definition of
⊞
=
y
⊞
x
.
{\displaystyle {\begin{aligned}x\boxplus y&=\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxplus \right.}\\[0.3em]&=x+y\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{commutativity of the addition in }}K\right.}\\[0.3em]&=y+x\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxplus \right.}\\[0.3em]&=y\boxplus x.\end{aligned}}}
This shows the commutativity of the addition.
Proof step: Neutral element of addition
Proof step: Inverse with respect to addition
Proof step: Scalar distributive law
Let
λ
,
μ
∈
K
{\displaystyle \lambda ,\mu \in K}
and
x
∈
K
{\displaystyle x\in K}
.
Then:
(
λ
+
μ
)
⊡
x
=
↓
definition of
⊡
=
(
λ
+
μ
)
⋅
x
↓
distributive law in
K
=
λ
⋅
x
+
μ
⋅
x
↓
definition of
⊞
=
(
λ
⋅
x
)
⊞
(
μ
⋅
x
)
↓
definition of
⊡
=
(
λ
⊡
x
)
⊞
(
μ
⊡
x
)
.
{\displaystyle {\begin{aligned}(\lambda +\mu )\boxdot x&=\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxdot \right.}\\[0.3em]&=(\lambda +\mu )\cdot x\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{distributive law in }}K\right.}\\[0.3em]&=\lambda \cdot x+\mu \cdot x\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxplus \right.}\\[0.3em]&=(\lambda \cdot x)\boxplus (\mu \cdot x)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxdot \right.}\\[0.3em]&=(\lambda \boxdot x)\boxplus (\mu \boxdot x).\end{aligned}}}
Thus the scalar distributive law is shown.
Proof step: Vectorial distributive law
Let
λ
∈
K
{\displaystyle \lambda \in K}
and
x
,
y
∈
K
{\displaystyle x,y\in K}
.
Then:
λ
⊡
(
x
⊞
y
)
=
↓
definition of
⊞
=
λ
⊡
(
x
+
y
)
↓
definition of
⊡
=
λ
⋅
(
x
+
y
)
↓
distributive law in
K
=
λ
⋅
x
+
λ
⋅
y
↓
definition of
⊞
=
(
λ
⋅
x
)
⊞
(
λ
⋅
y
)
↓
Definition of
⊡
=
(
λ
⊡
x
)
⊞
(
λ
⊡
y
)
.
{\displaystyle {\begin{aligned}\lambda \boxdot (x\boxplus y)&=\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxplus \right.}\\[0.3em]&=\lambda \boxdot (x+y)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxdot \right.}\\[0.3em]&=\lambda \cdot (x+y)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{distributive law in }}K\right.}\\[0.3em]&=\lambda \cdot x+\lambda \cdot y\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxplus \right.}\\[0.3em]&=(\lambda \cdot x)\boxplus (\lambda \cdot y)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{Definition of }}\boxdot \right.}\\[0.3em]&=(\lambda \boxdot x)\boxplus (\lambda \boxdot y).\end{aligned}}}
Thus the vectorial distributive law is shown.
Proof step: Assoziativität bezüglich Multiplikation
Let
λ
,
μ
∈
K
{\displaystyle \lambda ,\mu \in K}
and
x
∈
K
{\displaystyle x\in K}
.
Then:
(
λ
⋅
μ
)
⊡
x
=
↓
definition of
⊡
=
(
λ
⋅
μ
)
⋅
x
↓
associative law for multiplication in
K
=
λ
⋅
(
μ
⋅
x
)
↓
definition of
⊡
=
λ
⊡
(
μ
⋅
x
)
↓
definition of
⊡
=
λ
⊡
(
μ
⊡
x
)
.
{\displaystyle {\begin{aligned}(\lambda \cdot \mu )\boxdot x&=\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxdot \right.}\\[0.3em]&=(\lambda \cdot \mu )\cdot x\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{associative law for multiplication in }}K\right.}\\[0.3em]&=\lambda \cdot (\mu \cdot x)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxdot \right.}\\[0.3em]&=\lambda \boxdot (\mu \cdot x)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxdot \right.}\\[0.3em]&=\lambda \boxdot (\mu \boxdot x).\end{aligned}}}
This shows the associative law for multiplication.
Proof step: Unitary law
Let
x
∈
K
{\displaystyle x\in K}
.
Then:
1
K
⊡
x
=
1
K
⋅
x
=
x
.
{\displaystyle 1_{K}\boxdot x=1_{K}\cdot x=x.}
Thus we have shown the unitary law.
With this we have established all eight vector space axioms and thus
(
K
,
⊞
,
⊡
)
{\displaystyle (K,\boxplus ,\boxdot )}
is a
K
{\displaystyle K}
vector space.