Материал из Викиконспекты
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Строка 49: |
| | |statement= | | |statement= |
| | '''Следствие 2: '''<br> | | '''Следствие 2: '''<br> |
| − | Пусть {<tex>{e_i}</tex>}<tex> \to </tex>{<tex>{e_i}</tex>} <br> | + | Пусть <tex>\tilde{A} = T^{-1}AT</tex> <br> |
| − | Пусть <tex>A = T^{-1}AT</tex> <br>
| + | Тогда <tex> \det \tilde{A} = \det A </tex> <br> |
| − | Тогда <tex> \det A = \det A </tex> <br> | + | <tex> \det \tilde{A} = \det T \det A \det T^{-1} = \det A </tex> |
| − | <tex> \det A = \det T \det A \det T^{-1} = \det A </tex> | |
| | }} | | }} |
Версия 03:20, 15 июня 2013
| Лемма (*): |
[math] \mathcal{A}^{\wedge_p} {e_{i_1}} \land {e_{i_2}} \land ... \land {e_{i_p}} \stackrel{\mathrm{def}}{=} \mathcal{A}{e_{i_1}} \land \mathcal{A}{e_{i_2}} \land... \land \mathcal{A}{e_{i_p}} [/math] |
| Лемма (**): |
Если [math] {x_1} \land {x_2} \land... \land {x_p} \in {\wedge_p} [/math], то [math] \mathcal{A}^{\wedge_p} {x_1} \land {x_2} \land ... \land {x_p} = \mathcal{A}{x_1} \land \mathcal{A}{x_2} \land... \land \mathcal{A}{x_p} [/math] |
| Лемма (***): |
[math] \mathcal{A}^{\wedge_n} z = \det \mathcal{A} \cdot z [/math] |
Теорема умножения определителей
| Теорема: |
Пусть [math]\mathcal{A}[/math], [math]\mathcal{B} \colon X \to X[/math] (автоморфизм).
Тогда [math]\det (\mathcal{A} \cdot \mathcal{B}) = \det \mathcal{A} \cdot \det \mathcal{B}[/math] |
| Доказательство: |
| [math]\triangleright[/math] |
|
[math]\det (\mathcal{A} \cdot \mathcal{B}) {e_1} \land {e_2} \land... \land{e_n} = [/math]
[math]
(\mathcal{A} \cdot \mathcal{B})^{\wedge_n}{e_1} \land {e_2} \land... \land{e_n} = ^{(*)}[/math]
[math]
(\mathcal{A} \cdot \mathcal{B}) {e_1} \land (\mathcal{A} \cdot \mathcal{B}) {e_2} \land ... \land (\mathcal{A} \cdot \mathcal{B}) {e_n} = ^{(def\mathcal{A} \cdot \mathcal{B})}[/math]
[math]
\mathcal{A} (\mathcal{B} {e_1}) \land \mathcal{A} (\mathcal{B} {e_2}) \land ... \land \mathcal{A} (\mathcal{B} {e_n}) = ^{(**)}[/math]
[math]
\mathcal{A}^{\wedge_n}(\mathcal{B} {e_1} \land \mathcal{B} {e_2} \land ... \land \mathcal{B} {e_n})= ^{(***)}[/math]
[math]
\det \mathcal{A} \cdot (\mathcal{B} {e_1} \land \mathcal{B} {e_2} \land ... \land \mathcal{B} {e_n}) = ^{(***)}[/math]
[math]
\det \mathcal{A} \cdot \mathcal{B}^{\wedge_n}({e_1} \land {e_2} \land ... \land {e_n}) = [/math]
[math]
\det \mathcal{A} \cdot \det \mathcal{B} \cdot {e_1} \land {e_2} \land ... \land {e_n} [/math]
т.е. [math] \det (\mathcal{A} \cdot \mathcal{B}) {e_1} \land {e_2} \land... \land{e_n} = [/math]
[math]
\det \mathcal{A} \cdot \det \mathcal{B} \cdot {e_1} \land {e_2} \land ... \land {e_n}
[/math] |
| [math]\triangleleft[/math] |
| Утверждение: |
Следствие 1:
Пусть [math]\mathcal{A} \colon X \to X[/math] обратим, т.е. [math](\exists \mathcal{A}^{-1}) [/math]
[math]
\det \mathcal{A} \ne 0 [/math]
Тогда: [math]\mathcal{A}^{-1} = (\det \mathcal{A})^{-1} [/math] |
| [math]\triangleright[/math] |
|
[math]\mathcal{A}^{-1} \cdot \mathcal{A} = \mathcal{I} \det (\mathcal{A}^{-1} \cdot \mathcal{A}) = [/math]
[math]
\det \mathcal{A}^{-1} \cdot \det \mathcal{A} = \det \mathcal{I} = 1 (\det {E} = 1) [/math]
Итого: [math] \det \mathcal{A}^{-1} = 1/(\det \mathcal{A}) [/math] |
| [math]\triangleleft[/math] |
| Утверждение: |
Следствие 2:
Пусть [math]\tilde{A} = T^{-1}AT[/math]
Тогда [math] \det \tilde{A} = \det A [/math]
[math] \det \tilde{A} = \det T \det A \det T^{-1} = \det A [/math] |
