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VI. Egenvektorer og diagonalisering – 2. Diagonalisering 466 og giver potensen A k = BΛ k B −1 2.23. Nyt eksempel, potens ☞ [LA] 10 Diagonalisering Eksempel 2 - fortsat 11 5 −6 12 −6 5 1 3 2 0 = 3 2 4 0 35 −8 6 3 −2 1931 −1266 = 2532 −1656 2.24. Advarsel! ☞ [LA] 10 Diagonalisering Eksempel 3 (Pas på) Betragt matricen Matricen A kan ikke diagonaliseres. λ = 3 er eneste egenværdi. A = 3 1 0 3
VI. Egenvektorer og diagonalisering – 2. Diagonalisering 467 Egenrummet er nulrum for matricen A − 3I3 = 0 1 0 0 2.25. Advarsel! ☞ [LA] 10 Diagonalisering Eksempel 3 - fortsat Egenrummet er E3 = span(e1) En matrix B, hvis søjler er egenvektorer B = har determinant 0 og dermed ikke invertibel. Altså kan A ikke diagonaliseres. b1 b2 0 0 2.26. Opgave Opgave 2 Det oplyses, at matricen A givet ved ☞ Matematik Alfa 1, August 2002 ⎡ −1 −3 ⎤ −3 A = ⎣ 3 5 3 ⎦ −3 −3 −1
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CALCULUS "SLIDES" TIL CALCULUS 1 +
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3. Generelle områder 201 4. Koordi
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Forord "Slides" til forelæsningern
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I. Differentiation - 1. Kontinuitet
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I. Differentiation - 7. Lagrangemet
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II. Integration - 1. Dobbelt integr
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III. Potensrækker - 4. Taylorpolyn
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1. Grafiske/numeriske metoder IV Di
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IV. Differentialligninger - 1. Graf
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1. Vektorer og matricer V Matricer
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V. Matricer - 1. Vektorer og matric
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V. Matricer - 2. Lineære afbildnin
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V. Matricer - 3. Lineære ligninger
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VIII. Appendiks - 1. Polære koordi
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Litteratur • Arnold, Ordinary dif
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0-række/søjle, 423 n-te afsnitssu
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Identitetsmatricen, 383 Identitetsm
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