problemset11solutions - MAS 213: Linear Algebra II. Problem...

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Unformatted text preview: 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 , , 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 , 1 , 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))(1 , , 0) + ( x · (0 , 1 , 0))(0 , 1 , 0) + ( x · (0 , , 1))(0 , , 1) = (1) ∗ (1 , 1 , 2) + (1) ∗ (0 , 1 , 0) + (2) ∗ (0 , , 1) . 1 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 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...
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This note was uploaded on 12/08/2010 for the course SPMS MAS213 taught by Professor Andrewkricker during the Fall '10 term at Nanyang Technological University.

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problemset11solutions - MAS 213: Linear Algebra II. Problem...

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