Introduction to Linear algebra-Strang-Solutions-Manual_ver13

To be in both of those perpendicular spaces ax must

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Unformatted text preview: projections onto orthogonal vectors. This is important. The projections of .1; 1/ onto .1; 0/ and .1; 2/ are p1 D .1; 0/ and p2 D .0:6; 1:2/. Then p1 C p2 ¤ b. Since A is invertible, P D A.AT A/ 1 AT D AA 1 .AT / 1 AT D I : project on all of R2 .       0:2 0:4 0:2 10 0:2 This is not a1 D .1; 0/ P2 D , P2 a1 D , P1 D , P1 P2 a1 D . 0:4 0:8 0:4 00 0 No, P1 P2 ¤ .P1 P2 /2 . 11 (a) p D A.AT A/ " 1T A b D .2; 3; 0/, e D .0; 0; 4/, AT e D 0 (b) p D .4; 4; 6/, e D 0. # 100 0 1 0 D projection matrix onto the column space of A (the xy plane) 000 " # 0:5 0:5 0 Projection matrix onto the second column space. P2 D 0:5 0:5 0 = Certainly .P2 /2 D P2 . 0 0 1 2 3 2 3 23 23 100 1000 1 1 60 1 07 60 1 0 07 627 627 13 A D 4 , P D square matrix D 4 , p D P 4 5 D 4 5. 0 0 15 0 0 1 05 3 3 000 0000 4 0 14 The projection of this b onto the column space of A is b itself when b is in that space. " # "# 58 4 0 1 8 17 2 and b D P b D p D 2 . But P is not necessarily I . P D 21 4 2 20 4 12 P1 D 15 2A has the same column space as A. b for 2A is half of b for A. x x 16 1 .1; 2; 2 2 1/ C 3 .1; 0; 1/ D .2; 1; 1/. So b is in the plane. Projection shows P b D b. 2 P /2 D .I P /.I P / D I P I I P C P 2 D I P . When P projects onto the column space, I P projects onto the left nullspace. (a) I P is the projection matrix onto .1; 1/ in the perpendicular direction to .1; 1/ (b) I P projects onto the plane x C y C z D 0 perpendicular to .1; 1; 1/. " # 5=6 1=6 1=3 For any basis vectors in the plane x y 2z D 0, 1=6 5=6 1=3 . say .1; 1; 0/ and .2; 0; 1/, the matrix P is 1=3 1=3 1=3 "# " # " # 1 1=6 1=6 1=3 5=6 1=6 1=3 ee T D 1 , Q D eTe 1=6 1=6 1=3 , I Q D 1=6 5=6 1=3 . eD 2 1=3 1=3 2=3 1=3 1=3 1=3 2 A.AT A/ 1 AT D A.AT A/ 1 .AT A/.AT A/ 1 AT D A.AT A/ 1 AT . So P 2 D P . P b is in the column space (where P projects). Then its projection P .P b/ is P b. P T D .A.AT A/ 1 AT /T D A..AT A/ 1 /T AT D A.AT A/ 1 AT D P . (AT A is symmetric!) If A is invertible then its column space is all of Rn . So P D I and e D 0. The nullspace of AT is orthogonal to the column space C .A/. So if AT b D 0, the projection of b onto C .A/ should be p D 0. Check P b D A.AT A/ 1 AT b D A.AT A/ 1 0. 17 If P D P then .I 18 19 20 21 22 23 24 Solutions to Exercises 46 25 The column space of P will be S . Then r D dimension of S D n. 26 A 1 exists since the rank is r D m. Multiply A2 D A by A T T 1 to get A D I . 27 If A Ax D 0 then Ax is in the nullspace of A . But Ax is always in the column space of A. To be in both of those perpendicular spaces, Ax must be zero. So A and AT A have the same nullspace. 28 P 2 D P D P T give P T P D P . Then the .2; 2/ entry of P equals the .2; 2/ entry of P T P which is the length squared of column 2. 29 A D B T has independent columns, so AT A (which is BB T ) must be invertible.    aaT 1 9 12 3 so PC D T D . 4 aa 25 12 25 (b) The row space is the line through v D .1; 2; 2/ and PR D vvT =vT v. Always PC A D A (columns of A project to themselves) and APR D A. Then PC APR D A ! 30 (a)...
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This note was uploaded on 09/25/2012 for the course PHY 103 taught by Professor Minki during the Spring '12 term at Korea Advanced Institute of Science and Technology.

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