Solutions Manual Introduction to Linear PDF

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INTRODUCTION TO LINEAR ALGEBRA Fourth Edition MANUAL FOR INSTRUCTORS

Gilbert Strang Massachusetts Institute of Technology math.mit.edu/linearalgebra web.mit.edu/18.06 video lectures: ocw.mit.edu math.mit.edu/gs www.wellesleycambridge.com email: [email protected] Wellesley - Cambridge Press Box 812060 Wellesley, Massachusetts 02482

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Solutions to Exercises

2

Problem Set 1.1, page 8 The combinations give (a) a line in R3

(b) a plane in R3

(c) all of R3 .

v C w D .2; 3/ and v  w D .6; 1/ will be the diagonals of the parallelogram with v and w as two sides going out from .0; 0/. This problem gives the diagonals v C w and v  w of the parallelogram and asks for the sides: The opposite of Problem 2. In this example v D .3; 3/ and w D .2; 2/. 3v C w D .7; 5/ and cv C d w D .2c C d ; c C 2d /. u Cv D .2; 3; 1/ and u Cv Cw D .0; 0; 0/ and 2u C2v Cw D . add first answers/ D .2; 3; 1/. The vectors u; v ; w are in the same plane because a combination gives .0; 0; 0/. Stated another way: u D v  w is in the plane of v and w. The components of every cv C d w add to zero. c D 3 and d D 9 give .3; 3; 6/. The nine combinations c.2; 1/ C d .0; 1/ with c D 0; 1; 2 and d D .0; 1; 2/ will lie on a lattice. If we took all whole numbers c and d , the lattice would lie over the whole plane. The other diagonal is v  w (or else w  v). Adding diagonals gives 2v (or 2w). The fourth corner can be .4; 4/ or .4; 0/ or .2; 2/. Three possible parallelograms! i  j D .1; 1; 0/ is in the base (x-y plane). i C j C k D .1; 1; 1/ is the opposite corner from .0; 0; 0/. Points in the cube have 0  x  1, 0  y  1, 0  z  1. Four more corners .1; 1; 0/; .1; 0; 1/; .0; 1; 1/; .1; 1; 1/. The center point is .21; 12 ; 21 /. Centers of faces are . 21 ; 21 ; 0/; . 12 ; 12 ; 1/ and .0; 21 ; 21 /; .1; 12 ; 21 / and . 21 ; 0; 12 /; . 12 ; 1; 12 /. A four-dimensional cube has 24 D 16 corners and 2  4 D 8 three-dimensional faces and 24 two-dimensional faces and 32 edges in Worked Example . ı Sum D zero vector. p Sum D 2:00 vector D 8:00 vector. 2:00 is 30 from horizontal D .cos 6 ; sin 6 / D . 3=2; 1=2/. Moving the origin to 6:00 adds j D .0; 1/ to every vector. So the sum of twelve vectors changes from 0 to 12j D .0; 12/. 1 3 The point v C w is three-fourths of the way to v starting from w. The vector 4 4 1 1 1 1 v C w is halfway to u D v C w. The vector v C w is 2u (the far corner of the 2 4 4 2 parallelogram).

All combinations with c C d D 1 are on the line that passes through v and w. The point V D v C 2w is on that line but it is beyond w. All vectors cv C cw are on the line passing through .0; 0/ and u D 21v C 21 w. That line continues out beyond v C w and back beyond .0; 0/. With c  0, half of this line is removed, leaving a ray that starts at .0; 0/. The combinations cv C d w with 0  c  1 and 0  d  1 fill the parallelogram with sides v and w. For example, if v D .1; 0/ and w D .0; 1/ then cv C d w fills the unit square. With c  0 and d  0 we get the infinite “cone” or “wedge” between v and w. For example, if v D .1; 0/ and w D .0; 1/, then the cone is the whole quadrant x  0, y  0. Question: What if w D v? The cone opens to a half-space.

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Solutions to Exercises

50 " " # " # " # 1 0 0 1 You can see why q 1 D 0 , q 2 D 0 , q 3 D 1 . A D 0 0 1 0 0 QR.

0 0 1

0 1 0

# 1 2 4 0 3 6 D 0 0 5

#"

(a) One basis for the subspace S of solutions to x1 C x2 C x3  x4 D 0 is v1 D .1; 1; 0; 0/, v 2 D .1; 0; 1; 0/, v 3 D .1; 0; 0; 1/ (b) Since S contains solutions to .1; 1; 1; 1/Tx D 0, a basis for S ? is .1; 1; 1; 1/ (c) Split .1; 1; 1; 1/ D b1 C b2 by projection on S ? and S : b2 D . 12 ; 21; 12 ;  21 / and b1 D . 21 ; 12 ; 21 ; 23 /. This question   breakdown  R22 D 0 when   shows 2by 2 formulas  A is sinfor QR;  1 1 1 1 5 3 1 2 1 1 1 2 1  D p . Singular  p gular. 1 1 D p 1 1 1 1 2 2 1 5 0 1 5   1 2 2 p . The Gram-Schmidt process breaks down when ad  bc D 0. 2 0 0 T c  .q2TC /q 2 D BT B because q 2 D k B k and the extra q 1 in C  is orthogonal to q 2 . B B B

When a and b are not orthogonal, the projections onto these lines do not add to the projection onto the plane of a and b. We must use the orthogonal A and B (or orthonormal q 1 and q 2 ) to be allowed to add 1D projections. There are mn multiplications in (11) and21m2 n multiplications in each part of (12). q 1 D 31.2; 2; 1/, q 2 D 31 .2; 1; 2/, q 3 D 13 .1; 2; 2/.

The columns of the wavelet matrix W are orthonormal. Then W 1 D W T . See Section 7.2 for more about wavelets : a useful orthonormal basis with many zeros. (b) The pro(a) c D 21 normalizes all the orthogonal columns to have unit length 1 T T jection .a b=a a/a of b D .1; 1; 1; 1/ onto the first column is p1 D 2 .1; 1; 1; 1/. (Check e D 0.) To project onto the plane, add p2 D 12 .1; 1; 1; 1/ to get .0; 0; 1; 1/. " #   1 0 0 1 0 0 1 across plane y C z D 0. Q1 D reflects across x axis, Q2 D 0 0 1 0 1 0

Orthogonal and lower triangular ) ˙1 on the main diagonal and zeros elsewhere. (a) Qu D .I  2uuT /u D u  2uuT u. This is u, provided that uT u equals 1 (b) Qv D .I  2uuT /v D u  2uuT v D u, provided that uT v D 0.

Starting from A D .1; 1; 0; 0/, the orthogonal (not orthonormal) vectors B D .1; 1; 2; 0/ and C D .1; 1; 1; 3/ and D D .1; 1; 1; 1/ are in the directions of q 2 ; q 3 ; q 4 . The 4 by 4 and 5 by 5 matrices with integer orthogonal columns 2 (not orthogonal3rows, 2 3 1 1 1 1 1 1 1 17 6 since not orthonormal Q!) are 4 A B C D 5 D 4 and 0 2 1 15 0 0 3 1 3 2 1 1 1 1 1 1 1 1 17 6 1 7 6 1 1 17 6 0 2 4 0 0 3 1 15 0 0 0 4 1

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Solutions to Exercises

90

U U 1 D I : Back substitution needs 12 j 2 multiplications on column j , using the j by j upper left block. Then 12 .12 C 22 C    C n2 /  12 .31n3 / D total to find U 1 .           1 0 0 1 2 2 2 2 1 0 ; and L D D U with P D ! ! :5 1 1 0 0 1 1 0 2 2 # " # " # " # " 2 2 0 2 2 0 2 2 0 2 2 0 2 0 ! 0 2 0 D U with A ! 1 0 1 ! 0 1 1 ! 0 0 1 1 0 2 0 0 2 0 0 0 1 # " # " 1 0 0 0 1 0 1 0 . P D 0 0 1 and L D 0 1 0 0 :5 :5 1 2 3 1 1 0 0 6 1 1 1 07 has cofactors C13 D C31 D C24 D C42 D 1 and C14 D C41 D AD4 0 1 1 15 0 0 1 1 1. A1 is a full matrix! With 16-digit floating point arithmetic the errors kx  x computed k for " D 103 , 106 , 109 , 1012, 10 15 are of order 1016 , 1011 , 107 , 104 , 103 .      p p 1 3 1 1 10 14 1 1 p p (a) cos  D 1= 10, sin  D 3= 10, R D 10 3 1 3 8. 5 D 10 0 (b) A haseigenvalues  in row  1 of Q: either  of the unit eigenvectors  4 and 2. Put one 1 3 1 1 2 4 1 1 1 QD p and QAQ1 D or Q D p10 and QAQ D 2 1 3 1 0 4 1   4 4 . 0 2

When A is multiplied by a plane rotation Qij , this changes the 2n (not n2 ) entries in rows i and j . Then multiplying on the right by .Qij /1 D .Qij /T changes the 2n entries in columns i and j . Qij A uses 4n multiplications (2 for each entry in rows i and j /. By factoring out cos  , the entries 1 and ˙ tan  need only 2n multiplications, which leads to 32 n3 for QR. 1 cos  or The .2; 1/ entry of Q21A pis 3 . sin  C 2 cos /. This ispzero if sin  D 2p tan  D 2. Then the 2; 1; 5 right triangle has sin  D 2= 5 and cos  D 1= 5.

Every 3 by 3 rotation with det Q D C1 is the product of 3 plane rotations. This problem shows how elimination is more expensive (the nonzero multipliers are counted by nnz(L) and nnz(LL)) when we spoil the tridiagonal K by a random permutation. If on the other hand we start with a poorly ordered matrix K, an improved ordering is found by the code symamd discussed in this section. The “red-black ordering” puts rows and columns 1 to 10 in the odd-even order 1; 3; 5; 7, 9; 2; 4; 6; 8; 10. When K is the 1; 2; 1 tridiagonal matrix, odd points are connected

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