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==Теорема умножения определителей ==
rollbackEdits.php mass rollback
{{Лемма
|about = <tex>*</tex>
|statement=
<tex> \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}} </tex>
{{Лемма
|about = <tex>**</tex>
|statement=
Если <tex> {x_1} \land {x_2} \land... \land {x_p} \in {\wedge_p} </tex>, то <tex> \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} </tex>
{{Лемма
|about = <tex>***</tex>
|statement=
<tex> \mathcal{A}^{\wedge_n} z = \det \mathcal{A} \cdot z </tex>
}}
{{Теорема
|statement=
|proof =
<tex>\det (\mathcal{A} \cdot \mathcal{B}) {e_1} \land {e_2} \land... \land{e_n} = </tex><br><tex>
(\mathcal{A} \cdot \mathcal{B})^{\wedge_n}{e_1} \land {e_2} \land... \land{e_n} = ^\stackrel{\mathrm{(*)}}{=}</tex><br><tex>(\mathcal{A} \cdot \mathcal{B}) {e_1} \land (\mathcal{A} \cdot \mathcal{B}) {e_2} \land ... \land (\mathcal{A} \cdot \mathcal{B}) {e_n} = ^\stackrel{\mathrm{(def\mathcal{A} \cdot \mathcal{B})}}{=}</tex><br><tex>\mathcal{A} (\mathcal{B} {e_1}) \land \mathcal{A} (\mathcal{B} {e_2}) \land ... \land \mathcal{A} (\mathcal{B} {e_n}) = ^\stackrel{\mathrm{(**)}}{=}</tex><br><tex>\mathcal{A}^{\wedge_n}(\mathcal{B} {e_1} \land \mathcal{B} {e_2} \land ... \land \mathcal{B} {e_n})= ^\stackrel{\mathrm{(***)}}{=}</tex><br><tex>\det \mathcal{A} \cdot (\mathcal{B} {e_1} \land \mathcal{B} {e_2} \land ... \land \mathcal{B} {e_n}) = ^\stackrel{\mathrm{(***)}}{=}</tex><br><tex>
\det \mathcal{A} \cdot \mathcal{B}^{\wedge_n}({e_1} \land {e_2} \land ... \land {e_n}) = </tex><br><tex>
\det \mathcal{A} \cdot \det \mathcal{B} \cdot {e_1} \land {e_2} \land ... \land {e_n} </tex><br>
</tex>
}}
{{Утверждение
|statement=
'''Следствие 1: '''<br>
Пусть <tex>\mathcal{A} \colon X \to X</tex> обратим, т.е. <tex>(\exists \mathcal{A}^{-1}) </tex><br><tex>
\det \mathcal{A} \ne 0 </tex><br>
Тогда: <tex>\mathcal{A}^{-1} = (\det \mathcal{A})^{-1} </tex>
|proof=
<tex>\mathcal{A}^{-1} \cdot \mathcal{A} = \mathcal{I} \det (\mathcal{A}^{-1} \cdot \mathcal{A}) = </tex><br><tex>
\det \mathcal{A}^{-1} \cdot \det \mathcal{A} = \det \mathcal{I} = 1 (\det {E} = 1) </tex><br>
Итого: <tex> \det \mathcal{A}^{-1} = (\det \mathcal{A})^{-1} </tex>
}}
{{Утверждение
|statement=
'''Следствие 2: '''<br>
Пусть <tex>\tilde{A} = T^{-1}AT</tex> <br>
Тогда <tex> \det \tilde{A} = \det A </tex> <br>
<tex> \det \tilde{A} = \det T \det A \det T^{-1} = \det A </tex>
}}
