problemset11solutions - MAS 213 Linear Algebra II Problem...

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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
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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
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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 .
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  • Fall '10
  • ANDREWKRICKER
  • Linear Algebra, inner product, Inner product space, Orthonormal basis

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