problemset10solutions - MAS 213 Linear Algebra II Problem...

Info icon This preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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
Image of page 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
Image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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 .
Image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern