Teorema rango-nulidad.
T
:
V
→
W
{\displaystyle T:V\to W}
T más la dimensión de la imagen de T es igual a la dimensión del espacio vectorial de partida V . En matemáticas , el teorema rango–nulidad es un teorema en álgebra lineal , que dice que la dimensión del dominio de una transformación lineal es la suma de su rango (dimensión de su imagen) y su nulidad (la dimensión de su núcleo o kernel).
Sean
V
{\displaystyle V}
W
{\displaystyle W}
dim
V
<
+
∞
{\displaystyle \dim V<+\infty }
T
:
V
→
W
{\displaystyle T:V\to W}
transformación lineal entonces
Rank
(
T
)
+
Null
(
T
)
=
dim
V
{\displaystyle \operatorname {Rank} (T)+\operatorname {Null} (T)=\dim V}
donde
Rank
(
T
)
:=
dim
(
Im
(
T
)
)
Null
(
T
)
:=
dim
(
Ker
(
T
)
)
{\displaystyle {\begin{aligned}\operatorname {Rank} (T)&:=\dim(\operatorname {Im} (T))\\\operatorname {Null} (T)&:=\dim(\operatorname {Ker} (T))\end{aligned}}}
es decir
dim
(
Im
(
T
)
)
+
dim
(
Ker
(
T
)
)
=
dim
V
{\displaystyle \dim(\operatorname {Im} (T))+\dim(\operatorname {Ker} (T))=\dim V}
Demostración [ editar ] Sea
T
:
V
→
W
{\displaystyle T:V\to W}
{
u
1
,
…
,
u
m
}
∈
V
{\displaystyle \{\color {green}\mathbf {u} _{1},\ldots ,\mathbf {u} _{m}\color {black}\}\in V}
base del núcleo de
T
{\displaystyle T}
Ker
T
{\displaystyle {\text{Ker}}~T}
teorema de intercambio de Steinitz , podemos extender este conjunto para formar una base de
V
{\displaystyle V}
B
=
{
u
1
,
…
,
u
m
,
w
1
,
…
,
w
n
}
⊆
V
{\displaystyle {\mathcal {B}}=\{\color {green}\mathbf {u} _{1},\ldots ,\mathbf {u} _{m}\color {black},\color {blue}\mathbf {w} _{1},\ldots ,\mathbf {w} _{n}\color {black}\}\subseteq V}
T
{\displaystyle T}
m
{\displaystyle m}
V
{\displaystyle V}
m
+
n
{\displaystyle m+n}
D
i
m
{\displaystyle Dim}
Im
T
{\displaystyle {\text{Im}}~T}
n
{\displaystyle n}
Veamos que el conjunto
{
T
(
w
1
)
,
…
,
T
(
w
n
)
}
∈
W
{\displaystyle \{\color {blue}T(\mathbf {w} _{1}),\ldots ,T(\mathbf {w} _{n})\color {black}\}\in W}
Im
T
{\displaystyle {\text{Im}}~T}
genera a
Im
T
{\displaystyle {\text{Im}}~T}
linealmente independiente .
Sea
v
{\displaystyle v}
V
{\displaystyle V}
B
{\displaystyle {\mathcal {B}}}
V
{\displaystyle V}
a
1
,
.
.
.
,
a
m
,
b
1
,
.
.
.
,
b
n
∈
K
{\displaystyle \color {green}a_{1},...,a_{m},\color {blue}b_{1},...,b_{n}\color {black}\in \mathbb {K} }
v
=
a
1
u
1
+
⋯
+
a
m
u
m
+
b
1
w
1
+
⋯
+
b
n
w
n
⇒
{\displaystyle \mathbf {v} =\color {green}a_{1}\mathbf {u} _{1}+\cdots +a_{m}\mathbf {u} _{m}\color {black}+\color {blue}b_{1}\mathbf {w} _{1}+\cdots +b_{n}\mathbf {w} _{n}\color {black}\Rightarrow }
⇒
T
(
v
)
=
a
1
T
(
u
1
)
+
⋯
+
a
m
T
(
u
m
)
+
b
1
T
(
w
1
)
+
⋯
+
b
n
T
(
w
n
)
=
b
1
T
(
w
1
)
+
⋯
+
b
n
T
(
w
n
)
{\displaystyle \Rightarrow T(\mathbf {v} )=\color {green}a_{1}T(\mathbf {u} _{1})+\cdots +a_{m}T(\mathbf {u} _{m})\color {black}+\color {blue}b_{1}T(\mathbf {w} _{1})+\cdots +b_{n}T(\mathbf {w} _{n})\color {black}=\color {blue}b_{1}T(\mathbf {w} _{1})+\cdots +b_{n}T(\mathbf {w} _{n})}
pues
u
i
∈
Ker
T
⇒
T
(
u
i
)
=
0
∀
i
∈
{
1
,
.
.
.
m
}
{\displaystyle \color {green}u_{i}\color {black}\in {\text{Ker}}~T\Rightarrow \color {green}T(u_{i})\color {black}=0\ \ \forall i\in \{1,...m\}}
Por lo tanto,
{
T
(
w
1
)
,
…
,
T
(
w
n
)
}
{\displaystyle \{\color {blue}T(\mathbf {w} _{1}),\ldots ,T(\mathbf {w} _{n})\color {black}\}}
Im
T
{\displaystyle {\text{Im}}~T}
Ahora, sólo se necesita demostrar que el conjunto
{
T
(
w
1
)
,
…
,
T
(
w
n
)
}
{\displaystyle \{\color {blue}T(\mathbf {w} _{1}),\ldots ,T(\mathbf {w} _{n})\color {black}\}}
combinación lineal de estos vectores es cero si y sólo si el coeficiente de cada vector es cero. Sean
c
1
,
.
.
.
,
c
n
∈
K
{\displaystyle \color {blue}c_{1},...,c_{n}\color {black}\in \mathbb {K} }
c
1
T
(
w
1
)
+
⋯
+
c
n
T
(
w
n
)
=
0
⇒
T
(
c
1
w
1
+
⋯
+
c
n
w
n
)
=
0
{\displaystyle \color {blue}c_{1}T(\mathbf {w} _{1})+\cdots +c_{n}T(\mathbf {w} _{n})\color {black}=\mathbf {0} \Rightarrow T(\color {blue}c_{1}\mathbf {w} _{1}+\cdots +c_{n}\mathbf {w} _{n}\color {black})=\mathbf {0} }
∴
c
1
w
1
+
⋯
+
c
n
w
n
∈
Ker
T
{\displaystyle \therefore \color {blue}c_{1}\mathbf {w} _{1}+\cdots +c_{n}\mathbf {w} _{n}\color {black}\in \operatorname {Ker} \;T}
Entonces, puesto que
{
u
i
}
i
=
1
m
{\displaystyle \color {green}\{u_{i}\}_{i=1}^{m}}
Ker
T
{\displaystyle {\text{Ker}}~T}
d
1
,
.
.
.
,
d
m
∈
K
{\displaystyle \color {green}d_{1},...,d_{m}\color {black}\in \mathbb {K} }
c
1
w
1
+
⋯
+
c
n
w
n
=
d
1
u
1
+
⋯
+
d
m
u
m
{\displaystyle \color {blue}c_{1}\mathbf {w} _{1}+\cdots +c_{n}\mathbf {w} _{n}\color {black}=\color {green}d_{1}\mathbf {u} _{1}+\cdots +d_{m}\mathbf {u} _{m}}
Pero, puesto que
B
=
{
u
1
,
…
,
u
m
,
w
1
,
…
,
w
n
}
{\displaystyle {\mathcal {B}}=\{\color {green}\mathbf {u} _{1},\ldots ,\mathbf {u} _{m}\color {black},\color {blue}\mathbf {w} _{1},\ldots ,\mathbf {w} _{n}\color {black}\}}
V
{\displaystyle V}
c
i
{\displaystyle \color {blue}c_{i}}
d
j
{\displaystyle \color {green}d_{j}}
c
i
{\displaystyle \color {blue}c_{i}}
{
T
(
w
1
)
,
…
,
T
(
w
n
)
}
{\displaystyle \{\color {blue}T(\mathbf {w} _{1}),\ldots ,T(\mathbf {w} _{n})\color {black}\}}
Im
T
{\displaystyle {\text{Im}}~T}
dimensión , esto prueba que la dimensión de
Im
T
{\displaystyle {\text{Im}}~T}
n
{\displaystyle n}
◻
{\displaystyle \quad \square }
Véase también [ editar ] Referencias [ editar ]
Datos: Q303402