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| Autor |
  Matrixnorm < 1, dann A+E invertierbar |
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LukasNiessen
Aktiv 
Dabei seit: 30.09.2019
Mitteilungen: 152
Wohnort: Nordrhein-Westfalen, Bonn, Weststadt
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Hallo,
ich soll zeigen:
Sei $||.||$ eine Vektornorm auf dem $\IR^n$ und $|||.|||$ die von ihr induzierte Matrixnorm. Sei nun: Für $A \in \IR^{\text{n x n}}: |||A||| < 1$. Dann ist $E+A$ regulär.
Mein Ansatz ist:
Nach der Def. einer induzierten Matrixnorm haben wir also:
Für alle $x \in \IR^n - {0}: \frac{||Ax||}{||x||} < 1$, also $||Ax|| < ||x||$.
Man kann das dann noch umformen zu:
$||Ax||-||Ex|| < 0$
Ich habe versucht zu nutzen, dass $|||.|||$ mit $||.||$ verträglich ist, oder die umgekehrte Dreiecksungl., aber das bringt ja beides nichts, weil wir dann die Abschätzung < 0 verlieren würden.
Ich sehe aber insb. nicht wie man hier auf irgendwas von Invertierbarkeit kommen kann, zB, dass A vollen Rang haben muss, oder die Determinante nicht 0 sein kann.
Vielen Dank!
-----------------
Beste Grüße, Lukas Nießen
PS: Schreibt mir gerne :-)\(\endgroup\)
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Tirpitz
Senior 
Dabei seit: 07.01.2015
Mitteilungen: 784
 |      Beitrag No.1, eingetragen 2021-03-08
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Hallo!
Eventuell hilft der Hinweis auf eine (symbolische) geometrische Reihe weiter? Wann würde sie konvergieren?
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zippy
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Dabei seit: 24.10.2018
Mitteilungen: 2454
 | Beitrag No.2, eingetragen 2021-03-08
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\newcommand{\Factn}[1]{\gudl{\sc{F}\!act\tx{}#1\colon}}\)2021-03-08 16:57 - LukasNiessen im Themenstart schreibt:
also $||Ax|| < ||x||$. \(\endgroup\)
Und wenn jetzt $A+E$ nicht invertierbar wäre, gäbe es einen Vektor $x\neq 0$ mit $(A+E)\,x=0$ bzw. $Ax=-x$. Dann wäre aber $\|Ax\|=\|x\|$...
--zippy
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LukasNiessen
Aktiv
Dabei seit: 30.09.2019
Mitteilungen: 152
Wohnort: Nordrhein-Westfalen, Bonn, Weststadt
| Beitrag No.3, vom Themenstarter, eingetragen 2021-03-08
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Jetzt seh ich's! Danke sehr!
Grüße
-----------------
Beste Grüße, Lukas Nießen
PS: Schreibt mir gerne :-)
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LukasNiessen hat die Antworten auf ihre/seine Frage gesehen.
LukasNiessen hat selbst das Ok-Häkchen gesetzt. |  [Neues Thema]  [Druckversion] |
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