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www.vutube.edu.pk SSoolluuttiioonn A Assssiiggnnm meenntt 77 Question # 1 a) show that the set S= { u1 , u 2 , u 3 } is an orthogonal basis for R 3 5 and express the vector x= 3 as a linear combination of the vectors in S 1 3 2 1 Where u1 = 3 , u 2 = 2 , u 3 = 1 0 1 4 Solution: First Consider the three possible pairs of distinct vectors, namely {u 1, u2}, {u1, u3} and {u2, u3}. u1 u2 3 2 3 2 0 1 6 6 0 0 u1 u3 31 31 0 4 3 3 0 0 u2 u3 2 1 2 1 1 4 2 2 4 0 Each pair of distinct vectors are orthogonal, and so S= {u 1, u2, u3} is an orthogonal set. Now by using this theorem which states that “If S = {u 1, u2, u3,…up} is an orthogonal set of non zero vectors in Rn, then S is linearly independent and hence is a basis for the subspace spanned by S”. So S = {u1, u2, u3} is an orthogonal basis for R3. Now the next step is to express x as a linear combination of u’s. Compute 5 3 x u1 3 3 5 3 3 3 1 0 15 9 0 24 1 0 5 2 x u2 3 2 5 2 3 2 1 1 10 6 1 3 1 1 5 1 x u3 3 1 5 1 31 1 4 5 3 4 6 1 4 3 3 u1 u1 3 3 3 3 3 3 0 0 9 9 0 18 0 0 2 2 u2 u2 2 2 2 2 2 2 1 1 4 4 1 9 1 1 1 1 u3 u3 1 1 11 11 4 4 1 1 16 18 4 4 Now by using theorem which states that, “Let {u1, u2, u3,…up} be an orthogonal basis for a subspace W of Rn. for each x in W, the weight of the linear combination. x c1u1 c2u2 c3u3 .......... c pu p Are given by cj x.u j u j .u j ( j 1, 2,3......., p) Now according to the given condition x c1u1 c2u2 c3u3 Where, c1 x.u3 x.u1 x.u2 , c2 , c3 u1.u1 u2 .u2 u3 .u3 x x.u3 x.u1 x.u2 u1 u2 u3 u1.u1 u2 .u2 u3 .u3 Now, 24 3 6 u1 u2 u3 18 9 18 4 1 1 x u1 u2 u3 3 3 3 x b). 2 4 and u= Find the orthogonal projection of y onto u. Then write y as the sum of 3 7 Let y= two orthogonal vectors, one in Span{u} and one Orthogonal to u. Solution: Compute 2 4 y u 13 3 7 4 4 u u 65 7 7 The orthogonal projection of y onto u is yˆ yu 13 1 4 4 / 5 u u uu 65 5 7 7 / 5 And the component of y orthogonal to u is 2 4 / 5 14 / 5 y yˆ 3 7 / 5 8 / 5 The sum of these two vectors is y. That is, 2 4 / 5 14 / 5 3 7 / 5 8 / 5 y yˆ ( y yˆ ) Note: If the calculations above are correct, then yˆ ( y yˆ ) will be an orthogonal set. As a check, we will compute. 4 / 5 14 / 5 56 56 yˆ ( y yˆ ) 0 7 / 5 8 / 5 25 25 Question # 2 a) Find the distance from y to w=Span { v1 , v 2 3 1 } where y= 5 1 3 1 = v1 1 , 1 1 1 = v 2 1 1 Solution: y y.v1 y.v2 v1 v2 v1.v1 v2 .v2 3 3 1 1 y v1 3 3 11 5 1 11 9 1 5 1 6 5 1 1 1 3 3 1 1 v1 v1 3 3 11 1 1 11 9 1 1 1 12, 1 1 1 1 3 1 1 1 y v2 31 1 1 5 1 1 1 3 1 5 1 6, 5 1 1 1 1 1 1 1 v2 v2 11 1 1 11 1 1 4 1 1 1 1 3 1 3 1 1 1 1 6 6 1 3 1 y 12 1 4 1 2 1 2 1 1 1 1 1 3 3 2 2 1 3 3 2 2 1 1 3 1 2 2 1 1 3 2 2 3 1 y 1 1 3 3 0 1 1 2 y-y 5 1 4 1 1 0 ^ Now ^ 2 y - y 0 4 16 0 20 y- ^ The distance from y to W is y 2 5 2 5 b) Find the least square error in the least square solution of Ax=b 1 Where A= 1 1 3 1 , b= 1 5 1 0 Solution: To use (3), compute: 1 3 1 1 1 1 1 1 3 1 1 3 3 A A 1 1 3 1 1 9 1 1 3 11 3 1 1 1 1 5 1 1 1 5 1 0 6 T A b 1 3 1 1 15 1 0 14 0 T Then the equation A A A b becomes T T 3 3 x1 6 3 11 x 14 2 Row operation can be used to solve this system, but since faster to compute. A A T 1 AT A is invertible and 2 2 , it is probably 1 11 3 24 3 3 And then to solve A Ax A b as T T xˆ AT A AT b 1 1 11 3 6 1 66 42 24 3 3 14 24 18 42 1 24 1 24 24 1 1 xˆ 1 Now we have, 3 1 3 4 1 1 1 1 0 1 1 1 1 2 1 Axˆ 1 1 Hence, 5 4 5 4 1 b Axˆ 1 0 1 0 1 0 2 0 2 2 And b Axˆ 1 1 2 2 2 2 11 4 6 The least square error is Question # 3 6 . For any x in R2 , the distance between b and the vector Ax is at least 6. Find an orthogonal basis for the subspace W of 1 3 A= 1 1 6 3 6 3 6 8 2 4 R 4 for the matrix Also make a QR-factorization of the above matrix A and check that A=QR Solution: A x1 Let , 1 3 x1 , 1 1 x2 x3 6 8 x2 , 2 4 6 3 x3 6 3 Then [ x1, x2, x3 ] is clearly linearly independent and thus is a basis for a subspace W of R 4. Now we can construct an orthogonal basis for W. Step 1: Let v1 = x1 and W1 = Span {x1} = Span {v1}. Step 2: Let v2 be the vector produced by subtracting from x 2 its projection onto the subspace W1. That is, let v2 x2 projw1 x2 x2 x2 v1 v1 v1 v1 Since v1 x1 6 1 8 3 x2 v1 6 1 8 3 2 1 4 1 6 24 2 4 36 2 1 4 1 1 1 3 3 v1 v1 1 1 3 3 11 11 1 9 1 1 12 1 1 1 1 6 1 6 1 8 3 8 3 36 v2 3 2 12 1 2 1 4 1 4 1 6 3 3 8 9 1 2 3 1 4 3 1 3 1 v2 1 1 v2 is the component of x2 orthogonal to x1, and {v1, v2} is an orthogonal basis for the subspace W2 spanned by x1 and x2. Step 3: Let v3 be the vector produced by subtracting from x 3 its projection onto the subspace W2. Use the orthogonal basis {v1, v2} to compute the projection onto W2: projW2 x3 x3 v1 x v v1 3 2 v2 v1 v1 v2 v2 6 1 3 3 x3 v1 6 1 3 3 6 1 31 6 9 6 3 6 6 1 3 1 1 1 3 3 v1 v1 1 1 3 3 11 11 1 9 1 1 12 1 1 1 1 6 3 3 1 x3 v2 6 3 31 6 1 3 1 18 3 6 3 30 6 1 3 1 3 3 1 1 v2 v2 3 3 11 11 1 1 9 1 1 1 12 1 1 1 1 1 3 3 6 30 1 12 1 12 1 1 1 1 15 1 3 5 2 1 5 1 5 1 3 3 1 51 21 21 1 1 14 7 18 4 2 6 3 4 2 v3 x3 projW2 x3 6 7 1 3 4 1 v3 6 3 3 3 2 1 Scale v3 to simplify the later computations 1 1 v3 1 v3 3 1 The columns A are the vectors x1, x2, x3 . an orthogonal basis for colA = Span [x1, x2, x3 ] has been found. 1 3 v1 , 1 1 3 1 v2 , 1 1 1 1 v3 3 1 Normalize the three vectors to obtain u1, u2, u3. 1 1 3 1 1 3 u1 v1 v1 12 1 1 1 1 12 12 12 12 3 3 1 1 1 1 u2 v2 v2 12 1 1 1 1 1 1 1 1 1 1 u3 v3 v3 12 3 3 1 1 12 12 12 12 12 12 12 12 QR Factorization: 1 3 Q 1 1 12 1 1 12 12 3 1 12 12 1 1 12 12 3 12 12 12 12 1 12 By construction, the first k columns of Q are orthonormal basis of span [x1, x2, x3 ] . From theorem of QR factorization which states; “ If A is an m n matrix with linearly independent columns, then A can be factored as A = QR, where Q is an m n matrix whose columns form an orthonormal basis for colA and R is an n n upper triangular invertible matrix with positive entries on its diagonal.” To find R, observe that QTQ = I, QT A QT QR IR R 1 12 QT 3 12 1 12 3 1 12 1 12 1 12 12 3 1 12 12 1 12 1 12 1 12 1 12 R 3 12 1 12 1 6 6 12 12 12 3 8 3 1 1 1 12 12 12 1 2 6 1 4 3 3 1 1 12 12 12 1 6 6 1 3 1 1 3 8 3 1 R 3 1 1 1 1 2 6 12 1 1 3 1 1 4 3 1 9 1 1 6 24 2 4 6 9 6 3 1 3 3 1 1 18 8 2 4 18 3 6 3 12 1 3 3 1 6 8 6 4 6 3 18 3 12 3 12 3 12 36 6 1 0 12 30 0 12 5 3 12 0 0 12 0 12 0 3 1 1 2 3 6 3 3 R 0 2 3 5 3 0 2 3 0 2 6 1 R 3 0 2 5 0 0 2 A QR 1 3 1 2 6 1 3 3 1 1 0 2 5 12 1 1 3 0 0 2 1 1 1 1 3 1 2 6 1 1 3 1 1 0 2 5 2 1 1 3 0 0 2 1 1 1 2 0 0 6 6 0 1 15 2 1 6 0 0 18 2 0 3 5 2 1 5 6 2 2 0 0 6 2 0 1 5 2 2 0 0 6 2 0 2 12 12 1 6 16 6 2 2 4 12 2 8 6 1 6 6 3 8 3 1 2 6 1 4 3 Hence it is verified that A = QR.