Оператор
A
:
X
→
Y
{\displaystyle A:X\rightarrow Y}
A
(
x
→
+
y
→
)
=
A
(
x
→
)
+
A
(
y
→
)
;
{\displaystyle A({\vec {x}}+{\vec {y}})=A({\vec {x}})+A({\vec {y}});}
A
(
λ
x
→
)
=
λ
A
(
x
→
)
.
{\displaystyle A(\lambda {\vec {x}})=\lambda A({\vec {x}}).}
Θ
:
∀
x
→
:
Θ
x
→
=
0
→
y
{\displaystyle \Theta :\forall {\vec {x}}:\Theta {\vec {x}}={\vec {0}}_{y}}
I
:
X
→
X
;
∀
x
→
:
I
x
→
=
x
→
{\displaystyle I:X\rightarrow X;\forall {\vec {x}}:I{\vec {x}}={\vec {x}}}
(
A
1
+
A
2
)
x
→
=
d
e
f
A
1
x
→
+
A
2
x
→
{\displaystyle (A_{1}+A_{2}){\vec {x}}{\overset {def}{=}}A_{1}{\vec {x}}+A_{2}{\vec {x}}}
A
2
x
→
=
d
e
f
A
(
A
(
x
→
)
)
;
A
k
x
→
=
A
(
A
k
−
1
(
x
→
)
)
{\displaystyle A^{2}{\vec {x}}{\overset {def}{=}}A(A({\vec {x}}));A^{k}{\vec {x}}=A(A^{k-1}({\vec {x}}))}
A
:
X
→
Y
,
B
:
Y
→
Z
;
(
A
B
)
x
→
=
d
e
f
B
(
A
x
→
)
{\displaystyle A:X\rightarrow Y,B:Y\rightarrow Z;(AB){\vec {x}}{\overset {def}{=}}B(A{\vec {x}})}
править
A
:
X
→
Y
,
d
i
m
(
X
)
=
m
,
d
i
m
(
Y
)
=
n
.
{\displaystyle A:X\rightarrow Y,dim(X)=m,dim(Y)=n.}
Возьмём базис
e
→
1
,
…
,
e
→
m
{\displaystyle {\vec {e}}_{1},\dots ,{\vec {e}}_{m}}
f
→
1
,
…
,
f
→
n
{\displaystyle {\vec {f}}_{1},\dots ,{\vec {f}}_{n}}
Покажем, что если известны результаты действия оператора А на базис, то оператор А полностью определён:
A
(
e
→
i
)
=
h
→
i
∈
Y
,
i
=
1
…
m
.
{\displaystyle A({\vec {e}}_{i})={\vec {h}}_{i}\in Y,i=1\dots m.}
∀
x
→
∈
X
:
x
→
=
∑
j
=
1
m
x
j
e
→
j
⇒
A
(
x
→
)
=
A
(
∑
j
=
1
m
x
j
e
→
j
)
=
∑
j
=
1
m
x
j
A
(
e
→
j
)
=
∑
j
=
1
m
x
j
h
→
j
.
{\displaystyle \forall {\vec {x}}\in X:{\vec {x}}=\sum \limits _{j=1}^{m}x_{j}{\vec {e}}_{j}\Rightarrow A({\vec {x}})=A(\sum \limits _{j=1}^{m}x_{j}{\vec {e}}_{j})=\sum \limits _{j=1}^{m}x_{j}A({\vec {e}}_{j})=\sum \limits _{j=1}^{m}x_{j}{\vec {h}}_{j}.}
Матрица оператора
A
e
f
=
|
|
a
i
j
|
|
n
×
m
=
(
a
11
a
12
…
a
1
m
…
…
…
…
a
n
1
a
n
2
…
a
n
m
)
{\displaystyle A_{ef}=||a_{ij}||_{n\times m}=\left({\begin{smallmatrix}a_{11}&a_{12}&\dots &a_{1m}\\\dots &\dots &\dots &\dots \\a_{n1}&a_{n2}&\dots &a_{nm}\end{smallmatrix}}\right)}
A
e
→
1
=
a
11
f
→
1
+
⋯
+
a
n
1
f
→
n
{\displaystyle A{\vec {e}}_{1}=a_{11}{\vec {f}}_{1}+\dots +a_{n1}{\vec {f}}_{n}}
…
{\displaystyle \dots }
A
e
→
m
=
a
1
m
f
→
1
+
⋯
+
a
n
m
f
→
n
{\displaystyle A{\vec {e}}_{m}=a_{1m}{\vec {f}}_{1}+\dots +a_{nm}{\vec {f}}_{n}}
Утверждение. Если матрица
B
=
|
|
b
i
j
|
|
m
×
n
{\displaystyle B=||b_{ij}||_{m\times n}}
Утверждение. Если оператор
C
=
A
+
B
.
{\displaystyle C=A+B.}
(
A
:
L
→
M
;
B
:
L
→
M
;
C
:
L
→
M
)
{\displaystyle \left(A:L\rightarrow M;B:L\rightarrow M;C:L\rightarrow M\right)}
C
e
f
=
A
e
f
+
B
e
f
{\displaystyle C_{ef}=A_{ef}+B_{ef}}
Доказательство.
(
A
e
f
+
B
e
f
)
(
x
1
⋮
x
n
)
=
A
e
f
(
x
1
⋮
x
n
)
+
B
e
f
(
x
1
⋮
x
n
)
=
A
(
x
→
)
+
B
(
x
→
)
=
(
A
+
B
)
(
x
→
)
=
C
(
x
→
)
=
C
e
f
(
x
1
⋮
x
n
)
{\displaystyle \left(A_{ef}+B_{ef}\right)\left({\begin{smallmatrix}x_{1}\\\vdots \\x_{n}\end{smallmatrix}}\right)=A_{ef}\left({\begin{smallmatrix}x_{1}\\\vdots \\x_{n}\end{smallmatrix}}\right)+B_{ef}\left({\begin{smallmatrix}x_{1}\\\vdots \\x_{n}\end{smallmatrix}}\right)=A({\vec {x}})+B({\vec {x}})=(A+B)({\vec {x}})=C({\vec {x}})=C_{ef}\left({\begin{smallmatrix}x_{1}\\\vdots \\x_{n}\end{smallmatrix}}\right)}
A
:
L
→
M
{\displaystyle A:L\rightarrow M}
Если А - изоморфизм, то:
y
→
→
x
→
,
∀
y
→
∈
M
∃
!
x
→
:
A
x
→
=
y
→
⇒
{\displaystyle {\vec {y}}\rightarrow {\vec {x}},\forall {\vec {y}}\in M\exists !{\vec {x}}:A{\vec {x}}={\vec {y}}\Rightarrow }
A
−
1
:
A
−
1
(
y
→
)
=
x
→
.
{\displaystyle A^{-1}:A^{-1}({\vec {y}})={\vec {x}}.}
Покажем, что
A
−
1
{\displaystyle A^{-1}}
1)
A
−
1
(
y
→
)
→
x
→
;
A
−
1
(
y
→
1
)
→
x
→
1
{\displaystyle A^{-1}({\vec {y}})\rightarrow {\vec {x}};A^{-1}({\vec {y}}_{1})\rightarrow {\vec {x}}_{1}}
A
−
1
(
y
→
+
y
→
1
)
=
x
→
+
x
→
1
=
A
−
1
(
y
→
)
+
A
−
1
(
y
→
1
)
,
(
т.к.
A
(
x
→
+
x
→
1
)
=
A
(
x
→
)
+
A
(
x
→
1
)
=
y
→
+
y
→
1
)
.
{\displaystyle A^{-1}({\vec {y}}+{\vec {y}}_{1})={\vec {x}}+{\vec {x}}_{1}=A^{-1}({\vec {y}})+A^{-1}({\vec {y}}_{1}),({\text{т.к.}}A({\vec {x}}+{\vec {x}}_{1})=A({\vec {x}})+A({\vec {x}}_{1})={\vec {y}}+{\vec {y}}_{1}).}
2)
A
−
1
(
λ
y
→
)
=
λ
x
→
=
λ
A
−
1
(
y
→
)
,
(
т.к.
A
(
λ
x
→
)
=
λ
A
(
x
→
)
=
λ
y
→
)
.
{\displaystyle A^{-1}(\lambda {\vec {y}})=\lambda {\vec {x}}=\lambda A^{-1}({\vec {y}}),({\text{т.к.}}A(\lambda {\vec {x}})=\lambda A({\vec {x}})=\lambda {\vec {y}}).}
Условие обратимости:
A
:
L
→
M
{\displaystyle A:L\rightarrow M}
⇔
{\displaystyle \Leftrightarrow }
Матрица обратного оператора
A
:
L
→
M
{\displaystyle A:L\rightarrow M}
A
{\displaystyle A}
c
{\displaystyle c}
|
|
A
x
|
|
≥
c
|
|
x
|
|
{\displaystyle ||Ax||\geq c||x||}
x
{\displaystyle x}
(
n
=
m
)
{\displaystyle (n=m)}
∃
A
−
1
:
M
→
L
{\displaystyle \exists A^{-1}:M\rightarrow L}
Возьмём
e
→
1
,
…
,
e
→
n
{\displaystyle {\vec {e}}_{1},\dots ,{\vec {e}}_{n}}
f
→
1
,
…
,
f
→
n
{\displaystyle {\vec {f}}_{1},\dots ,{\vec {f}}_{n}}
∀
x
→
=
(
x
1
⋮
x
n
)
=
∑
i
=
1
n
x
i
e
→
i
;
∀
y
→
=
A
(
x
→
)
=
(
y
1
⋮
y
n
)
=
∑
i
=
1
n
y
i
f
→
i
{\displaystyle \forall {\vec {x}}={\begin{pmatrix}x_{1}\\\vdots \\x_{n}\end{pmatrix}}=\sum _{i=1}^{n}x_{i}{\vec {e}}_{i};\forall {\vec {y}}=A({\vec {x}})={\begin{pmatrix}y_{1}\\\vdots \\y_{n}\end{pmatrix}}=\sum _{i=1}^{n}y_{i}{\vec {f}}_{i}}
Y
=
A
e
f
X
⇔
A
e
f
−
1
Y
=
A
e
f
−
1
A
e
f
X
⇔
A
e
f
−
1
Y
=
X
⇒
{\displaystyle Y=A_{ef}X\Leftrightarrow A_{ef}^{-1}Y=A_{ef}^{-1}A_{ef}X\Leftrightarrow A_{ef}^{-1}Y=X\Rightarrow }
A
e
f
−
1
{\displaystyle A_{ef}^{-1}}
A
−
1
⇒
A
e
f
−
1
{\displaystyle A^{-1}\Rightarrow A_{ef}^{-1}}
