problemset11solutions - MAS 213 Linear Algebra II Problem...

Info icon This preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
MAS 213: Linear Algebra II. Problem set #11. Solutions . Problem 1: (Problem 6.2.2 from [FIS], various parts.) In the following parts you are given an inner product space V , a set of vectors S V , and a distinguished vector x . In each case find an orthonormal basis for span( S ), then use the inner product to compute the co-ordinate vector of the given vector x with respect to the basis that you have just determined. Check your answer. Solution to part (i). In this part the vector space is R 3 with the standard inner product, the given set of vectors is S = { (1 , 0 , 1) , (0 , 1 , 1) , (1 , 3 , 3) } , and the distinguished vector is x = (1 , 1 , 2). This part is easier than it looks. Note that the vectors in this set are linearly independent, so, because there are three of them, their span is all of R 3 . We are asked to find an orthonormal basis for their span, which is R 3 , so we might as well take the standard orthonormal basis β = { (1 , 0 , 0) , (0 , 1 , 0) , (0 , 0 , 1) } . The co-ordinate vector of x = (1 , 1 , 2) with respect to this basis is obvi- ously itself. The inner product calculation gives that answer too: x = ( x · (1 , 0 , 0))(1 , 0 , 0) + ( x · (0 , 1 , 0))(0 , 1 , 0) + ( x · (0 , 0 , 1))(0 , 0 , 1) = (1) (1 , 1 , 2) + (1) (0 , 1 , 0) + (2) (0 , 0 , 1) . 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
Solution to part (ii). In the case the inner product space is V = M 2 × 2 ( R ) with the standard trace inner product, the set S = { v 1 , v 2 , v 3 } is S = {[ 3 5 1 1 ] , [ 1 9 5 1 ] , [ 7 17 2 6 ]} , and x = [ 1 27 4 8 ] . The first step is to apply the Gram-Schmidt procedure to turn this set into an orthogonal set. The procedure starts: w 1 = v 1 = [ 3 5 1 1 ] w 2 = v 2 v 2 , w 1 w 1 , w 1 = [ 1 9 5 1 ] tr ([ 3 1 5 1 ] [ 1 9 5 1 ]) tr ([ 3 1 5 1 ] [ 3 5 1 1 ]) [ 3 5 1 1 ] = [ 1 9 5 1 ] 36 36 [ 3 5 1 1 ] = [ 4 4 6 2 ] w 3 = v 3 v 3 , w 1 w 1 , w 1 w 1 v 3 , w 2 w 2 , w 2 w 2 = [ 7 17 2 6 ] [ 3 1 5 1 ] [ 7 17 2 6 ] [ 3 1 5 1 ] [ 3 5 1 1 ] [ 3 5 1 1 ] [ 4 6 4 2 ] [ 7 17 2 6 ] [ 4 6 4 2 ] [ 4 4 6 2 ] [ 4 4 6 2 ] . 2
Image of page 2
This expression gives: w 3 = [ 7 17 2 6 ] ( 72 36 ) [ 3 5 1 1 ] ( 72 72 ) [ 4 4 6 2 ] = [ 9 3 6 6 ] . That completes the Gram-Schmidt procedure. The second step is to normalize these vectors to give them unit length. We proceed: e 1 = 1 w 1 w 1 = 1 6 [ 3 5 1 1 ] e 2 = 1 w 2 w 2 = 1 6 2 [ 4 4 6 2 ] e 3 = 1 w 3 w 3 = 1 9 2 [ 9 3 6 6 ] . The next part of the problem is to find the co-ordinate vector of the given matrix x with respect to this orthonormal basis β = { e 1 , e 2 , e 3 } . Because this basis is orthonormal, we can calculate this co-ordinate vector just by taking the inner products of x with the various vectors in the basis: [ x ] β = x, e 1 x, e 2 x, e 3 . The numbers we need are calculated: x, e 1 = tr ( 1 6 [ 3 1 5 1 ] [ 1 27 4 8 ]) = 1 6 ((3) ( 1) + ( 1) ( 4) + (5) (27) + (1) (8)) = 24 .
Image of page 3

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

View Full Document Right Arrow Icon
Image of page 4
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