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Пусть <tex>\mathcal{A}</tex>, <tex>\mathcal{B} \colon X \to X</tex> (автоморфизм). <br> Тогда <tex>\det (\mathcal{A} \cdot \mathcal{B}) = \det \mathcal{A} \cdot \det \mathcal{B}</tex>
|proof =
<tex>\det (\mathcal{A} \cdot \mathcal{B}) {e_1} \land {e_2} \land... \land{e_n} = (\mathcal{A} \cdot \mathcal{B})^{\wedge_pwedge_n}{e_1} \land {e_2} \land... \land{e_n}= (\mathcal{A} \cdot \mathcal{B}) {e_1} \land (\mathcal{A} \cdot \mathcal{B}) {e_2} \land ... \land (\mathcal{A} \cdot \mathcal{B}) {e_n} = \mathcal{A} (\mathcal{B} {e_1}) \land \mathcal{A} (\mathcal{B} {e_2}) \land ... \land \mathcal{A} (\mathcal{B} {e_n}) =\mathcal{A}^{\wedge_n}(\mathcal{B} {e_1} \land \mathcal{B} {e_2} \land ... \land \mathcal{B} {e_n})= \det \mathcal{A} \cdot (\mathcal{B} {e_1} \land \mathcal{B} {e_2} \land ... \land \mathcal{B} {e_n}) = \det \mathcal{A} \cdot \mathcal{B}^{\wedge_n}({e_1} \land {e_2} \land ... \land {e_n}) = \det \mathcal{A} \cdot \det \mathcal{B} \cdot {e_1} \land {e_2} \land ... \land {e_n} <br> т.е. \det (\mathcal{A} \cdot \mathcal{B}) {e_1} \land {e_2} \land... \land{e_n} = \det \mathcal{A} \cdot \det \mathcal{B} \cdot {e_1} \land {e_2} \land ... \land {e_n}</tex>
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