Review Note
Last Update: 01/26/2025 07:45 PM
Current Deck: Mathématiques::Classiques::Thomas lefèvre
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Inégalité de Hadamard
Soit \(A = M_{\mathcal B_{can} }(C_1, \cdots, C_n)\). Alors \({{c1::|\det(A)| \leqslant \displaystyle \prod_{i=1}^n \|C_i\|}}\).
Démonstration :
Si \(A\) n'est pas inversible, \({{c2::\det(A) = 0}}\).
Sinon :\[|\det(A)| = {{c2::|\det(O)||\det(T)|}} = {{c2::\displaystyle \prod_{i=1}^n |\langle C_i, V_i \rangle|}} \leqslant {{c2::\displaystyle \displaystyle \prod_{i=1}^n \|C_i\|\|V_i\|}}\]
Soit \(A = M_{\mathcal B_{can} }(C_1, \cdots, C_n)\). Alors \({{c1::|\det(A)| \leqslant \displaystyle \prod_{i=1}^n \|C_i\|}}\).
Démonstration :
Si \(A\) n'est pas inversible, \({{c2::\det(A) = 0}}\).
Sinon :\[|\det(A)| = {{c2::|\det(O)||\det(T)|}} = {{c2::\displaystyle \prod_{i=1}^n |\langle C_i, V_i \rangle|}} \leqslant {{c2::\displaystyle \displaystyle \prod_{i=1}^n \|C_i\|\|V_i\|}}\]
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