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Mathematik » Numerik & Optimierung » Matrixnorm < 1, dann A+E invertierbar
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Universität/Hochschule J Matrixnorm < 1, dann A+E invertierbar
LukasNiessen
Aktiv Letzter Besuch: im letzten Monat
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 Letzter Besuch: im letzten Monat
Dabei seit: 07.01.2015
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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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\newcommand{\hom}{\Hom} \newcommand{\dash}{\dashrightarrow} \newcommand{\schemes}{\bb{(Sch)}} \newcommand{\groups}{\bb{(Grp)}} \newcommand{\rings}{\bb{(Ring)}} \newcommand{\tx}[1]{\text{ #1 }} \newcommand{\mm}{\ff{m}} \newcommand{\zkinfsum}{\sum_{k=0}^\infty} \newcommand{\ziinfsum}{\sum_{i=0}^\infty} \newcommand{\zjinfsum}{\sum_{j=0}^\infty} \newcommand{\asum}[1]{\sum_{\a\in\N^n}#1 X^\a} \newcommand{\arr}[3]{#1\overset{#2}{\to} #3} \newcommand{\nrm}[1]{\left\|#1\right\|} \newcommand{\nr}{\nrm{-}} \newcommand{\ext}[2]{#1/{#2}} \newcommand{\lam}{\lambda} \newcommand{\a}{\alpha} \newcommand{\be}{\beta} \newcommand{\g}{\gamma} \newcommand{\de}{\delta} \newcommand{\vp}{\varphi} \newcommand{\p}{\phi} \newcommand{\bul}{\bullet} \newcommand{\t}{\tau} \newcommand{\s}{\sigma} \newcommand{\ze}{\zeta} \newcommand{\T}{\mathbb{T}} \newcommand{\tm}{\times} \newcommand{\tms}{\times\pts\times} \newcommand{\ot}{\otimes} \newcommand{\ots}{\otimes\pts\otimes} \newcommand{\pls}{+\pts +} \newcommand{\cos}{,\pts,} \newcommand{\op}{\oplus} \newcommand{\ops}{\oplus\pts\oplus} \newcommand{\cr}{\circ} \newcommand{\crs}{\circ\pts\circ} \newcommand{\sc}[1]{\mathscr{#1}} \newcommand{\scal}[2]{\sc{#1}{\!#2}} \newcommand{\ov}[2]{\begin{matrix}#1 \\ #2\end{matrix}} \newcommand{\viele}{\color{orange}{\udl{\color{black}{\sc{V}\!iele\tx{}\sc{G}\!r\overset{{}_{,,\!}}{u}\textit{ß}e}}}} \newcommand{\xst}{\color{orange}{\udl{\color{black}{X.S.T.\sim 小石头}}}} \newcommand{\gudl}[1]{\color{orange}{\udl{\color{black}{#1}}}} \newcommand{\Task}{\gudl{\sc{T}\!ask:}} \newcommand{\Exer}{\gudl{\sc{E}\!exercise:}} \newcommand{\Drinfeld}{\gudl{\sc{D}\!rinfeld:}} \newcommand{\Goss}{\gudl{\sc{G}\!oss}} \newcommand{\CK}{C/K} \newcommand{\CS}{C/S} \newcommand{\Ck}{C/k} \newcommand{\Om}{\Omega} \newcommand{\J}{\Jac_{\CS}^{g-1}} \newcommand{\Fact}{\gudl{\sc{F}\!act\colon}} \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 Letzter Besuch: im letzten Monat
Dabei seit: 30.09.2019
Mitteilungen: 152
Wohnort: Nordrhein-Westfalen, Bonn, Weststadt
Zum letzten BeitragZum nächsten BeitragZum vorigen BeitragZum erstem Beitrag  Beitrag No.3, vom Themenstarter, eingetragen 2021-03-08


Jetzt seh ich's! Danke sehr!

Grüße


-----------------
Beste Grüße, Lukas Nießen
PS: Schreibt mir gerne :-)



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