master > master: Formattierung
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@ -5,8 +5,8 @@ Bestimme in jedem Falle, ob φ linear ist.
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a)
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φ(x1, x2, x3) = ( 4·x1·x3 )
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( 10·x2 )
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φ(x₁, x₂, x₃) = ( 4·x₁·x₃ )
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( 10·x₂ )
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Wenn φ linear wäre, dann müsste φ(2, 0, 2) = 2·φ(1, 0, 1) gelten.
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Aber:
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@ -18,8 +18,8 @@ Also ist φ nicht linear, weil Homogenität verletzt wird.
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b)
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φ(x1, x2, x3) = ( x3² )
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( 0 )
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φ(x₁, x₂, x₃) = ( x₃² )
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( 0 )
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Wenn φ linear wäre, dann müsste φ(0, 0, 8) = 8·φ(0, 0, 1) gelten.
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Aber:
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@ -31,21 +31,21 @@ Also ist φ nicht linear, weil Homogenität verletzt wird.
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c)
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φ(x1, x2, x3) = ( x3 )
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φ(x₁, x₂, x₃) = ( x₃ )
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( 0 )
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--> linear
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d)
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φ(x1, x2, x3) = ( 0 )
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φ(x₁, x₂, x₃) = ( 0 )
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( 0 )
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--> linear
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e)
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φ(x1, x2, x3) = ( 4 )
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φ(x₁, x₂, x₃) = ( 4 )
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( 0 )
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Wenn φ linear wäre, dann müsste φ(0) = 0 gelten. [Siehe Lemma 6.1.2]
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@ -54,15 +54,15 @@ Also ist φ nicht linear.
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f)
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φ(x1, x2, x3) = ( 10·x3 )
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( -x2 + x1 )
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φ(x₁, x₂, x₃) = ( 10·x₃ )
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( -x₂ + x₁ )
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linear!
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g)
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φ(x1, x2, x3) = ( 1 - 10·x3 )
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( -x2 + x1 )
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φ(x₁, x₂, x₃) = ( 1 - 10·x₃ )
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( -x₂ + x₁ )
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Wenn φ linear wäre, dann müsste φ(0) = 0 gelten. [Siehe Lemma 6.1.2]
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Aber φ(0) = (1, 0)ᵀ.
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@ -70,7 +70,7 @@ Also ist φ nicht linear.
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h)
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φ(x1, x2, x3) = ( exp(-(7·x2 + 8·x1)) )
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φ(x₁, x₂, x₃) = ( exp(-(7·x₂ + 8·x₁)) )
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( 0 )
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Wenn φ linear wäre, dann müsste φ(0) = 0 gelten. [Siehe Lemma 6.1.2]
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@ -79,15 +79,15 @@ Also ist φ nicht linear.
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## §2. Aufgaben ähnlich zu ÜB 10-2 ##
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Seien A = (u1, u2, u3) und B = (v1, v2),
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Seien A = (u₁, u₂, u₃) und B = (v₁, v₂),
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wobei
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u1 = (3, 0, 1)ᵀ
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u2 = (0, -1, 0)ᵀ
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u3 = (4, 0, 0)ᵀ
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u₁ = (3, 0, 1)ᵀ
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u₂ = (0, -1, 0)ᵀ
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u₃ = (4, 0, 0)ᵀ
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v1 = (4, 5)ᵀ
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v2 = (0, 1)ᵀ
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v₁ = (4, 5)ᵀ
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v₂ = (0, 1)ᵀ
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Beachte:
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@ -96,33 +96,33 @@ Beachte:
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Sei nun φ : ℝ³ ⟶ ℝ² definiert durch
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φ(x1, x2, x3) = ( 4·x1 - x3 )
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( 10·x2 + x1 )
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φ(x₁, x₂, x₃) = ( 4·x₁ - x₃ )
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( 10·x₂ + x₁ )
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### Zur Linearität ###
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Seien
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(x1,x2,x3), (x1',x2',x3') ∈ ℝ³
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(x₁,x₂,x₃), (x₁',x₂',x₃') ∈ ℝ³
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c, c' ∈ ℝ
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**Zu zeigen:**
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φ(c(x1, x2, x3) +c'(x1',x2',x3')) = c·φ(x1, x2, x3) +c'·φ(x1',x2',x3')
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φ(c(x₁, x₂, x₃) +c'(x₁',x₂',x₃')) = c·φ(x₁, x₂, x₃) +c'·φ(x₁', x₂', x₃')
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Es gilt
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l. S. = φ(c(x1, x2, x3) +c'(x1',x2',x3'))
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= φ(c(x1·e1 + x2·e2 + x3·e3) +c'(x1'·e1 + x2'·e2 + x3'·e3))
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= φ((c·x1 + c'·x1)·e1 + (c·x2 + c'·x2)·e2 + (c·x3 + c'·x3)·e3)
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= φ(c·x1 + c'·x1', c·x2 + c'·x2', c·x3 + c'·x3')
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l. S. = φ(c(x₁, x₂, x₃) +c'(x₁',x₂',x₃'))
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= φ(c(x₁·e1 + x₂·e2 + x₃·e3) +c'(x₁'·e1 + x₂'·e2 + x₃'·e3))
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= φ((c·x₁ + c'·x₁)·e1 + (c·x₂ + c'·x₂)·e2 + (c·x₃ + c'·x₃)·e3)
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= φ(c·x₁ + c'·x₁', c·x₂ + c'·x₂', c·x₃ + c'·x₃')
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= ( 4·(c·x1 + c'·x1') - (c·x3 + c'·x3') )
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( 10·(c·x2 + c'·x2') + (c·x1 + c'·x1') )
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= ( 4·(c·x₁ + c'·x₁') - (c·x₃ + c'·x₃') )
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( 10·(c·x₂ + c'·x₂') + (c·x₁ + c'·x₁') )
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= ( c·(4·x1 - x3) + c'·(4·x1' - x3') )
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( c·(10·x2 + x1) + c'·(10·x2' + x1') )
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= ( c·(4·x₁ - x₃) + c'·(4·x₁' - x₃') )
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( c·(10·x₂ + x₁) + c'·(10·x₂' + x₁') )
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= c·( 4·x1 - x3 ) + c'·( 4·x1' - x3' )
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( 10·x2 + x1 ) ( 10·x2' + x1' )
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= c·( 4·x₁ - x₃ ) + c'·( 4·x₁' - x₃' )
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( 10·x₂ + x₁ ) ( 10·x₂' + x₁' )
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= r. S.
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@ -131,9 +131,9 @@ Darum ist φ linear.
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### Darstellung ###
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Zunächst beobachten wir:
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φ(x1, x2, x3) = ( 4 0 -1 ) ( x1 )
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( 1 10 0 ) ( x2 )
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( x3 )
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φ(x₁, x₂, x₃) = ( 4 0 -1 ) ( x₁ )
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( 1 10 0 ) ( x₂ )
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( x₃ )
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= C·x
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= φ_C(x) siehe [Skript, Bsp 6.2.2],
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@ -204,13 +204,14 @@ Darum gilt
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## §3. Lineare Fortsetzung von partiell definierten Funktionen ##
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Sei φ : ℝ^5 ⟶ ℝ³
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Sei φ : ℝ⁵ ⟶ ℝ³
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Seien u1, u2, u3, u4, u5 eine Basis für ℝ^5.
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Seien u₁, u₂, u₃, u₄, u₅ eine Basis für ℝ⁵.
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Seien v₁, v₂, v₃ Vektoren in ℝ³.
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Definiert werden
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φ(u1) = v1, φ(u2) = v2, φ(u4) = v3
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φ(u₁) = v₁, φ(u₂) = v₂, φ(u₄) = v₃
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**Aufgabe:** Gibt es eine lineare Abbildung, die die o. s. Gleichungen erfüllen?
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@ -218,11 +219,11 @@ Definiert werden
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**Beweis:**
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Setze
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φ(u3) := 0 (Nullvektor)
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φ(u5) := 0 (Nullvektor)
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φ(u₃) := 0 (Nullvektor)
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φ(u₅) := 0 (Nullvektor)
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Da u1, u2, u3, u4, u5 eine Basis ist,
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können wir für belieges x ∈ ℝ^5
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Da u₁, u₂, u₃, u₄, u₅ eine Basis ist,
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können wir für beliebiges x ∈ ℝ⁵
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φ(x) = ∑ c_i · φ(ui)
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