Introduction to Linear algebra-Strang-Solutions-Manual_ver13

# To be in both of those perpendicular spaces ax must

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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)...
View Full Document

## 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.

Ask a homework question - tutors are online