Analytische Geometrie und Lineare Algebra 1 by Ina Kersten

By Ina Kersten

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Example text

Lemma. ahle v1 , . . , vs ∈ V so, dass Sei {u1 , . . , ur } eine Basis von kern(f ). W¨ f (v1 ), . . , f (vs ) eine Basis von bild(f ) bilden. Dann ist B := {u1 , . . , ur , v1 , . . , vs } eine Basis von V . Beweis. Unabh¨ angigkeit Sei 0 = λ1 u1 + · · · + λr ur + µ1 v1 + · · · µs vs mit λi , µj ∈ K ∀i, j f =⇒ 0 = f (0) = µ1 f (v1 ) + · · · + µs f (vs ), da u1 , . . , ur ∈ kern(f ), =⇒ µ1 = · · · = µs = 0, da f (v1 ), . . f (vs ) linear unabh¨angig sind =⇒ λ1 = · · · = λr = 0, da u1 , .

Bn1 ⎞ ⎞ b1 .. ⎟ . ⎠ · · · bn ··· c1 .. ⎟ .

Man pr¨ ufe, ob U := { f ∈ V | f (x) = f (−x) ∀ x ∈ ❘ } ein Untervektorraum von V ist. Aufgabe 9. Man untersuche, welche der folgenden vier Mengen Untervektorr¨aume von ❘2 sind: U1 = { (x, y) ∈ ❘2 | y = x2 } U2 = { (x, y) ∈ ❘2 | x y } U3 = { (x, y) ∈ ❘2 | y = 2x } U4 = { (x, y) ∈ ❘2 | xy 0 } Aufgabe 10. Man stelle den Vektor w ∈ v1 , v2 , v3 dar: ❘3 jeweils als Linearkombination der Vektoren a) w = (3, 2, 1) , v1 = (1, 0, 1) , v2 = (7, 3, 1) , v3 = (4, 3, −1) b) w = (−8, 17, −14) , v1 = (2, 1, 0) , v2 = (3, 0, 5) , v3 = (−1, 4, −1) .

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