## Analytische Geometrie und Lineare Algebra 2 by Ina Kersten

By Ina Kersten

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Additional info for Analytische Geometrie und Lineare Algebra 2

Example text

Nun sind wir in der Situation der ersten beiden F¨alle und k¨ onnen die Gleichung durch Bewegungen auf Normalform bringen. 4 Quadriken Eine Quadrik ist die L¨osungsmenge einer quadratischen Gleichung n n aii x2i + f (x1 , . . 3 beschriebenen Methode die Quadriken in Dimension n klassifizieren. F¨ ur n = 3 ergibt sich folgendes: Jede nicht ausgeartete Quadrik in ❘3 kann durch Bewegungen von ❘3 auf eine der folgenden Normalformen gebracht werden. i) Ellipsoid : b1 y12 + b2 y22 + b3 y32 − 1 = 0 ii) einschaliges Hyperboloid : b1 y12 + b2 y22 − b3 y32 − 1 = 0 iii) zweischaliges Hyperboloid : b1 y12 − b2 y22 − b3 y32 − 1 = 0 iv) elliptisches Paraboloid : b1 y12 + b2 y22 − y3 = 0 v) hyperbolisches Paraboloid : b1 y12 − b2 y22 − y3 = 0 wobei jeweils b1 , b2 , b3 > 0 sind.

0 ... 0 wobei B ∈ GLr (K) mit r = rang MB (s) . Beweis. 8 Sei C = (ϕ1 , . . , ϕn ) die durch ϕi (vj ) = surjektiv. 1 f¨ ur i = j 0 f¨ ur i = j definierte Basis von HomK (V, K). 4, dass MCB ( ) = (aij ) ist. Es ist vi , vj = (vj )(vi ) = a1j ϕ1 (vi ) + · · · + anj ϕn (vi ) = aij nach Definition von ϕi Also gilt MB (s) = MCB ( ). Da rang MCB ( ) = dimK bild( ) ist (vgl. 13), folgt nun surjektiv ⇐⇒ rang MB (s) = n und damit i). 4 und bestimme eine Basis B = (v1 , . . , vn ) von V so, dass (v1 , .

2 Schiefsymmetrische Bilinearformen 49 Umgekehrt gibt es zu jeder symmetrischen Matrix A ∈ Mn×n (K) genau eine symmetrische Bilinearform s : V × V −→ K so, dass MB (s) = A gilt, vgl. 3 Basiswechsel: Seien B und B Basen von V . 2 so gilt MB (s) = t T MB (s) T Schiefsymmetrische Bilinearformen Definition. Eine schiefsymmetrische Bilinearform ist eine Abbildung s : V × V −→ K, (v, w) −→ v, w , mit den Eigenschaften: F¨ ur alle u, v, w ∈ V und µ ∈ K gelten 1. 1 angesprochenen Punkte f¨ trische Bilinearformen s : V × V −→ K, (v, w) −→ v, w ?