problemset10solutions

# problemset10solutions - MAS 213 Linear Algebra II Problem...

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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 Cauchy-Schwarz in- equality, and check an instance of the triangle inequality. Solution The calculations are: f, g = 1 0 te t dt = 1 0 t d dt ( e t ) dt = 1 0 ( d dt ( te t ) e t ) dt = [ te t ] 1 0 1 0 e t dt = e [ e t ] 1 0 = 1 . f = ( f, f ) 1 2 = (∫ 1 0 ( t 2 ) dt ) 1 2 = ( 1 3 [ t 3 ] 1 0 ) 1 2 = 1 3 . 1

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g = ( g, g ) 1 2 = (∫ 1 0 e 2 t dt ) 1 2 = ( [ 1 2 e 2 t ] 1 0 ) 1 2 = ( 1 2 ( e 2 1) ) 1 2 . f + g = ( f + g, f + g ) 1 2 = (∫ 1 0 ( 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 Cauchy-Schwarz 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

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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 1-1. Solution According to a standard theorem, it is enough for us to show that the null space of T is { 0 } . So let v be a vector in the null space of T . For this v : 0 = 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. Solution to Part (i). In this case the underlying vector space is P n ( R ), and the candidate inner product is: a 0 + a 1 x + . . . + a n x n , b 0 + b 1 x + . . . + b n x n = n i =0 ( i + 1) a i b i . This definition does indeed define an inner product structure. In the proof, we’ll let a ( x ) , b ( x ) and c ( x ) be the three arbitrary polynomials from P n ( R ): a = a 0 + . . . + a n x n , b = b 0 + . . . + b n x n , c = c 0 + . . . + c n x n .
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