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Unformatted text preview: MAS 213: Linear Algebra II. Problem set #10. Solutions. Problem 1: (Problem 6.1 from [FIS].) Consider the inner product space C ([0 , 1]) equipped with the standard in tegration inner product. Let f ∈ C ([0 , 1]) be defined by f ( t ) = t and let g ∈ C ([0 , 1]) be defined by g ( t ) = e t . (i) Calculate ⟨ f, g ⟩ , ∥ f ∥ , ∥ g ∥ , and ∥ f + g ∥ . (ii) Using these calculations, check an instance of the CauchySchwarz in equality, and check an instance of the triangle inequality. Solution The calculations are: ⟨ f, g ⟩ = ∫ 1 te t dt = ∫ 1 t d dt ( e t ) dt = ∫ 1 ( d dt ( te t ) − e t ) dt = [ te t ] 1 − ∫ 1 e t dt = e − [ e t ] 1 = 1 . ∥ f ∥ = ( ⟨ f, f ⟩ ) 1 2 = (∫ 1 ( t 2 ) dt ) 1 2 = ( 1 3 [ t 3 ] 1 ) 1 2 = 1 √ 3 . 1 ∥ g ∥ = ( ⟨ g, g ⟩ ) 1 2 = (∫ 1 e 2 t dt ) 1 2 = ( [ 1 2 e 2 t ] 1 ) 1 2 = ( 1 2 ( e 2 − 1) ) 1 2 . ∥ f + g ∥ = ( ⟨ f + g, f + g ⟩ ) 1 2 = (∫ 1 ( e 2 t + 2 te t + t 2 ) dt ) 1 2 = ( e 2 − 1 2 + 2 + 1 3 ) 1 2 = ( 1 2 e 2 + 11 6 ) 1 2 . Next we’ll check that these specific calculations satisfy the CauchySchwarz inequality, which says that: ⟨ f, g ⟩ ≤ ∥ f ∥∥ g ∥ . The LHS was calculated for this instance of f and g to be 1. The RHS for this f and g is: ∥ f ∥∥ g ∥ = 1 √ 3 ∗ √ e 2 − 1 2 = √ e 2 − 1 6 . Now we can check that e 2 > 7, so e 2 − 1 > 6 so: ∥ f ∥∥ g ∥ > 1 = ⟨ f, g ⟩ . (Did we need our calculator to check that e 2 > 7? The Taylor series for e x tells us that: e 2 > 1 + 2 + 2 2 2! + 2 3 3! + 2 4 4! = 7.) 2 Finally, we’ll check that the triangle inequality is satisfied for these cal culations. It says that: ∥ f + g ∥ ≤ ∥ f ∥ + ∥ g ∥ . The square of the LHS is ( LHS ) 2 = 1 2 e 2 + 11 6 . The square of the RHS equals ( RHS ) 2 = ( 1 2 ( e 2 − 1) ) + 2 √ 1 6 ( e 2 − 1) + 1 3 = 1 2 e 2 − 1 6 + 2 √ 1 6 ( e 2 − 1) > 1 2 e 2 − 1 6 + 2 = ( LHS ) 2 . So we conclude that for this f and g , the triangle inequality behaves just as expected. 3 Problem 2: Let T be a linear operator on an inner product space V satisfying the property that ∥ T ( v ) ∥ = ∥ v ∥ for all v ∈ V . Prove that an operator with this property is 11. Solution According to a standard theorem, it is enough for us to show that the null space of T is { } . So let v be a vector in the null space of T . For this v : 0 = ∥ ∥ = ∥ T ( v ) ∥ = ∥ v ∥ . Axiom 4 of an inner product space tells us that the only vector with zero norm is the zero vector. So v = 0. Problem 3: In each of the following parts you are given a possible definition of an inner product space. In each part either prove that the given definition does indeed define an inner product space, or state (with justification) which axioms fail....
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 Fall '10
 ANDREWKRICKER

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