An introduction to real and complex manifolds. by Giuliano Sorani

By Giuliano Sorani

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Xn+d fUr aile Xl>'" 'Xn+l E V (Schranken-Lemma). , an E B paarweise verschieden, dann gilt nicht E(ab' .. , an). Gibt es dann eine endliche Menge A mit V = (A), so kann man aile Ergebnisse aus 5 bis 7 analog herleiten. Die Eigenschaft E entspricht der Aussage "sind linear abhlingig", (A) ist das Analogon zu SpanA. Mit dieser Verallgemeinerung beweist man in der Algebra die Existenz von sogenannten Transzendenz-Basen. 30 1. Vektorraume § 5. Anwendungen I. Die reellen Zahlen als Vektorraum tiber

3 gilt hier (3) dim Span(ab ... , an) = Rang{al' ... , an} = max {k E IN: es gibt k linear unabhangige Elemente unter den aI, ... ,an}. Als Beispiel ein Gleichungssystem von 4 Gleichungen in den 4 Unbekannten + ~2 ~1 + ~2 4~1 + 8~2 ~I (4) ~I + + ~4 = 0, +~3+ ~4=0, + ~3 ~3 + ~4 = 0, + 4~4 = 0. ~}, ... , ~4: Dieses System hat die Form (I), wenn man definiert Hier sind a}, a2, a3 linear unabhangig, aber a}, az, a3, a4 linear abhangig. Es folgt Rang{a},a2,a3,a4} = 3 und dimL(a},az,a3,a4) = 1. Man braucht also nur ein uEL(a},az, a3,a4), U 1= 0, zu finden und weiBdann, daBjede L6sungvon (4) die Form lJ(umit § 6.

Ein (i) feb;) = g(b;) fur i = I, ... ,n. (ii) f = g. D Beweis. Folgt aus (H). Existeoz-Satz. 1st b l , ... , b" eine Basis von V, so gibt es zujeder Wahl von Vektoren a'J' ... ,a~ aus V' einen Homomorphismus J: V --+ V' mit feb;) = a; fur i = I, ... , n. Beweis. Man definiert f(x) := IXla; + ... + IXna~, falls x = IXlb l + ... + IXnb n. 5 ist f wohldefiniert und ein Homomorphismus der Vektorraume. D 3. Kern Dod Bild. IstJ: V --+ V' ein Homomorphismus der Vektorraume und U eine Teilmenge von V, so setzt man f( U) : = {f( x): x E U} und definiert Kern und BUd durch Kernf:= {XE V:f(x) = Bildf:=f(V):= {f(x): O}, XE V}.

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