Download e-book for kindle: Algebra Volume 2, 2nd Edition by P. M. Cohn

By P. M. Cohn

ISBN-10: 047192234X

ISBN-13: 9780471922346

The most emphasis of this revised algebra textbook is on fields, jewelry and modules. The textual content contains new chapters at the consultant thought of finite teams, coding thought and algebraic language concept. units, lattices, different types and graphs are brought firstly of the textual content. The textual content, which has been rewritten with the purpose of constructing the topic more straightforward to know, includes simplified proofs and plenty of new illustrations and workouts.

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Additional resources for Algebra Volume 2, 2nd Edition

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Alors det A 0 0 B = det(A) det(B). 2. REGLE DE CALCUL POUR LES DETERMINANTS 53 Remarque. En particulier on a  λ1 0 0 . . 0  0 λ2 0 . . 0  .  ...   .  = λ1 · λ2 · · · λ n det  .. . . ..   ..  0 0 . . 0 λn  pour λi ∈ K. D´emonstration. Par r´ecurrence sur n. Exercice. 8. Pour tout A ∈ Mn (K) on a det(At ) = det(A). Ici At est la matrice transpos´e de A. D´emonstration. On admet la d´emonstration qui est un peu p´enible. Mais on peut v´erifier comme exercice le th`eor`eme dans le cas n = 2, 3 avec les formules explicites donn´ees ci-dessus.

Remarque. Les formules s’´ecrivent plus simples sous la forme ∇ f (g, h) = (∇f )(g, h) ∇g . ∇h Remarque. Un cas sp´ecial important est le cas d’une composition f (g, h) avec des fonctions g(x) et h(x) d’une seule variable. En les consid`ere comme fonction en deux variable en posant g˜(x, y) = g(x) et analogue pour h. On a ∂˜ g (x, y) = g (x), et donc ainsi ∂x d ∂f ∂f f (g, h) = (g, h) g (x) + (g, h) h (x). dt ∂x ∂y Exemple. Pour calculer la d´eriv´ee de t2 φ(t) = sin(s) ds log t on pose y f (x, y) = sin(s) ds, x et donc 1 d f (log t, t2 ) = − sin(log t) · + sin(t2 ) · 2t.

On a |g (x)| = |f (x, φ(x)) − f (x, ψ(x))| ≤ L |φ(x) − ψ(x)| = L |g(x)|, avec une constante L. En plus, g(x0 ) = 0. Le th´eor`eme est maintenant une cons´equence du lemme suivant. Lemma. Soit G une fonction d´erivable sur un intervalle ouvert I. On suppose qu’il existe une constante L telle que |g (x)| ≤ L |g(x)| pour tout x ∈ I, et un x0 ∈ I tel que g(x0 ) = 0. Alors g ≡ 0. 2. VARIATION DE LA CONSTANTE Demonstration. On suppose que g n’est pas identiquement 0. Soit > 0 tel que Lε < 1. Car g est continue est poss`ede au moins un z´ero (par exemple x0 ) il existe un x1 tel que g(x1 ) = 0, mais g(x1 − ε) = 0 (exercice).

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Algebra Volume 2, 2nd Edition by P. M. Cohn


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