Introduction to Linear algebra-Strang-Solutions-Manual_ver13

# Solutions to exercises 3 elimination on b1 a 83 b d 1

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Unformatted text preview: " # "# " # "# 9 12 0 :6 :8 0 T :6 , v3 D 0 . A A D 12 16 0 has  D 25; 0; 0 and v1 D :8 , v2 D 0 00 0 0 1 Here A D Œ 3 4 0  has rank 1 and AAT D Œ 25  and 1 D 5 is the only singular value in † D Œ 5 0 0 . 1 AT A D 2 3 4 5 6 7 8 9 10  Solutions to Exercises 11 12 13 14 15 16 17 18 19 20 21 81 "#" # " # :2 :12 :36 :48 0 A D Œ 1  Œ 5 0 0 V T and ACDV 0 D :16 ; AC AD :48 :64 0 I AACDŒ 1  0 0 0 00 The zero matrix has no pivots or singular values. Then † D same 2 by 3 zero matrix and the pseudoinverse is the 3 by 2 zero matrix. If det A D 0 then rank.A/ < n; thus rank.AC / < n and det AC D 0. A must be symmetric and positive deﬁnite, if † D ƒ and U D V D eigenvector matrix Q is orthogonal. (a) AT A is singular (b) This x C in the row space does give AT Ax C D AT b (c) If .1; 1/ in the nullspace of A is added to x C , we get another solution to AT Ab D AT b. x But this b is longer than x C because the added part is orthogonal to x C in the row x space. x C in the row space of A is perpendicular to b x C in the nullspace of AT A D x nullspace of A. The right triangle has c 2 D a2 C b 2 . AAC p D p, AAC e D 0, AC Ax r D x r , AC Ax n D 0.   :36 :48 AC D V †C U T is 1 Œ :6 :8  D Œ :12 :16  and AC A D Œ 1  and AAC D D 5 :48 :64 projection. L is determined by `21 . Each eigenvector in S is determined by one number. The counts are 1 C 3 for LU , 1 C 2 C 1 for LDU , 1 C 3 for QR, 1 C 2 C 1 for U †V T , 2 C 2 C 0 for SƒS 1 . LDLT and QƒQT are determined by 1 C 2 C 0 numbers because A is symmetric. P Column times row multiplication gives A D U †V T D i ui vT and also AC D i P1 CT T C V† U D i vi ui . Multiplying A A and using orthogonality of each ui to all P C C other 1vi vT . Similarly AAC D i P ujT leaves theTprojection matrix A A: A A D 1ui ui from V V D I . 22 Keep only the r by r corner †r of † (the rest is all zero). Then A D U †V T has the Tb T b required form A D U M1 †r M2 V T with an invertible M D M1 †r M2 in the middle.       0Au Av u The singular values of A are 23 D D . v v eigenvalues of this block matrix. AT 0 AT u Problem Set 8.1, page 418 " # c1 C c2 c2 0 c2 c2 C c3 c3 1 Det AT C0 A0 D is by direct calculation. Set c4 D 0 to 0 0 c3 c3 C c4 ﬁnd det AT C1 A1 D c1 c2 c3 . 1 3" " #2 1 # c1 100 111 5011D 2 .AT C1 A1 / 1 D 1 1 0 4 c2 1 1 1 111 001 c 21 33 1 1 c1 c1 c1 5. 4c 1 c 1 C c 1 c1 1 C c2 1 1 1 2 1 1 1 1 1 1 c1 c1 C c2 c1 C c2 C c3 Solutions to Exercises 82 3 The rows of the free-free matrix in equation (9) add to Œ 0 0 0  so the right side needs f1 C f2 C f3 D 0. f D . 1; 0; 1/ gives c2 u1 c2 u2 D 1; c3 u2 c3 u3 D 1; 0 D 0. Then uparticular D . c2 1 c3 1 ; c3 1 ; 0/. Add any multiple of unullspace D .1; 1; 1/.     Z Z d du du 1 4 c.x/ dx D c.x/ D 0 (bdry cond) so we need f .x/ dx D 0. dx dx dx 0 Zx Z1 dy 5 D f .x/ gives y.x/ D C f .t /dt . Then y.1/ D 0 gives C D f .t /dt dx 0 0 Z1 and y.x/ D f .t /dt . If the load is f .x/ D 1 then the displacement is y.x/ D 1 x . x as columns of AT times c ’s times rows of A1 . The ﬁrst...
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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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