Download PDF by Ina Kersten: Analytische Geometrie und Lineare Algebra 2

By Ina Kersten

ISBN-10: 393861644X

ISBN-13: 9783938616444

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Extra info for Analytische Geometrie und Lineare Algebra 2

Sample text

3. Sei B = (u1 , . . , un ) so gew¨ahlt, dass Li = Kui ist, dann gilt uj , uj =: aj = 0 f¨ ur j = 1, . . , m uj , uj = 0 f¨ ur j = m + 1, . . , n ui , uj = 0 f¨ ur alle i = j und die Behauptungen 2. und 3. 13 gilt. Bemerkung. Es sei 1 + 1 = 0 in K. Ferner sei V endlich dimensional und mit einer regul¨ aren symmetrischen Bilinearform s : V × V −→ K, (v, w) −→ v, w versehen. Dann gibt es eine orthogonale Zerlegung V = H1 ⊥ . . ⊥Hr ⊥Van mit r hyperbolischen Ebenen H1 , . . , Hr und einem anisotropen Raum Van , in dem kein Vektor isotrop ist.

1, also auch np = np , und aus N = rp np = rp np folgt dann auch rp = rp . Man kann also Summanden von Polen, die in derselben Bahn liegen zusammenfassen. Wir schreiben nun B1 , . . , Bm f¨ ur die Bahnen. Sei ri := rp , falls p ∈ Bi und ni := |Bi | = L¨ange der Bahn Bi“. Aus (1) folgt ” nun m ni (ri − 1) = 2N − 2 (2) i=1 Es ist N = ri ni f¨ ur alle i = 1, . . , m. Division der Gleichung (2) durch N ergibt: 2− (3) 2 = N m (1 − i=1 <2 1 ) ri 1/2 m 2 Es folgt m ≤ 3 , d. h. es gibt h¨ochstens drei Bahnen.

M. Dann ist f : V −→ W, ui −→ ui , vi −→ vi eine Isometrie. Der Rest folgt ebenfalls aus dem Satz. 10 Normalform schiefsymmetrischer Matrizen Wir setzen voraus, dass 1 + 1 = 0 in K ist. Satz. h. tA = −A), so ist n = 2m gerade und es gibt eine Matrix T ∈ GLn (K) so, dass t T AT = 0 −Em Em 0 gilt, wobei Em ∈ Mm×m (K) die Einheitsmatrix bezeichne. Beweis. Sei V ein n-dimensionaler K-Vektorraum und B eine Basis von V . 2 mit einer schiefsymmetrischen Bilinearform s auf V . 5 i) ist s regul¨ar.

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Analytische Geometrie und Lineare Algebra 2 by Ina Kersten


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