Introduction to Linear algebra-Strang-Solutions-Manual_ver13

# Qd p p i 1 0 cos i sin 2 2i1 p p p 1 1c 3 1ci 1p i 1 0

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Unformatted text preview: ination x k C1 . The solution to Problem 28 in this Fourth Edition is straightforward and important. Since H D Q 1 AQ D QT AQ is symmetric if A D AT , and since H has only one lower diagonal by construction, then H has only one upper diagonal: H is tridiagonal and all the recursions in Arnoldi’s method have only 3 terms (Problem 29). Solutions to Exercises 95 1 AQ is similar to A, so H has the same eigenvalues as A (at the end of Arnoldi). When Arnoldi stops sooner because the matrix size is large, the eigenvalues of Hk (called Ritz values) are close to eigenvalues of A. This is an important way to compute approximations to  for large matrices. 30 In principle the conjugate gradient method converges in 100 (or 99) steps to the exact solution x . But it is slower than elimination and its all-important property is to give good approximations to x much sooner. (Stopping elimination part way leaves you nothing.) The problem asks how close x 10 and x 20 are to x 100 , which equals x except for roundoff errors. 29 H D Q Problem Set 10.1, page 498 2 C 2i , 2 cos  and products 5, 2 i , 1. Note .e i /.e p i p 1 2 In polar form these are 5e , 5e 2i , p e i  , 5. 1 (a)(b)(c) have sums 4, 5 1 , and 100. The angles are 10 D 2 , jz w j  5. 3 p 1 C 23 i ; w 12 D 1. 2 i i 3 The absolute values are r D 10, 100, 4 jz  w j D 6, jz C w j  5, jz=w j 5 a C ib D p 3 2 C 1i, 2 1 2 C p 3 i, 2 i, i / D 1.  , 2 ,  and 2 . 6 1=z has absolute value 1=r and angle  ; .1=r/e times re equals 1.         a b c ac b d real part 1 3 1 10 7 D is the matrix b ad bc C ad imaginary part 3 1 3 0 form of .1 C 3i /.1 3i / D 10.     A1 A2 x1 b1 8 D gives complex matrix D vector multiplication .A1 C A2 A1 x2 b2 iA2 /.x 1 C i x 2 / D b1 C i b2 .  9 2 C i ; .2 C i /.1 C i / D 1 C 3i ; e 10 z C z is real; z 11 12 13 14 15 i =2 D i; e i D 1; 1i 1Ci D i ; . i /103 D i . z is pure imaginary; z z is positive; z=z has absolute value 1.   ab 01 includes aI (which just adds a to the eigenvalues and b . So the ba 10 eigenvectors are x 1 D .1; i / and x 2 D .1; i /. The eigenvalues are 1 D a C bi and 2 D a b i . We see x 1 D x 2 and 1 D 2 as expected for real matrices with complex eigenvalues. p (a) When a D b D d D 1 the square root becomes 4c ;  is complex if c < 0 (b)  D 0 and  D a C d when ad D bc (c) the ’s can be real and different. 2 Complex ’s when .a Cd / < 4.ad b c/; write .a Cd /2 4.ad b c/ as .a d /2 C4bc which is positive when bc > 0. det.P I / D 4 1 D 0 has  D 1, 1, i , i with eigenvectors .1; 1; 1; 1/ and .1; 1; 1; 1/ and .1; i; 1; i / and .1; i ; 1; i / D columns of Fourier matrix. The 6 by 6 cyclic shift P has det.P6 I / D 6 1 D 0. Then  D 1, w , w 2 , w 3 , w 4 , w 5 with w D e 2i=6 . These are the six solutions to b D 1 as in Figure 10.3 (The sixth roots of 1).   Solutions to Exercises 96 16 The symmetric block matrix has real eigenvalues; so i  is real and  is pure imaginary. 17 (a) 2e i=3 , 4e 2i=3 50e  i=2 (c) 7e 3i=2 , 49e 3i .D 49/ (d) . 18 r D 1, angle  2 19 a C i b D 1, i , 20 1, e (b) e 2i , e 4i 2i=3  ; multiply by e i to get e i=2 D i . 1 1, i , ˙ p ˙ 2 i p. 2 The root w D w 4i=3 1 De ,e are cube roots of 1. The cube roots of Altogether six roots of z 6 D 1. 2i=8 1 are p p is 1= 2 50e  i=4 p i = 2. 1, e i=3 , e  i=3 . 21 cos 3 D ReŒ.cos  Ci sin  /3  D cos3  3 cos  sin2  ; sin 3 D 3 cos2  sin  sin3  . 22 If the conjugate z D 1=z then jz j2 D 1 and z is any point e i on the unit circle. 23 e i is at angle  D 1 on the unit circle; ji e j D 1e ; Inﬁnitely many i e D e i.=2C2 n/e . 24 (a) Unit circle (b) Spiral in to e 2 (c) Circle continuing around to angle  D 2 2 . Problem Set 10.2, page 506 1 kuk D uH v). p p 9 D 3, kvk D 3, uH v D 3i C 2, vH u D " 2 0 0 2 1Ci 1Ci 2 AH A D 1i1i 2 share the eigenvalues 4 and 2. # H and AA D  3i C 2 (this is the conjugate of  31 are Hermitian matrices. They 13 3 z D multiple of .1 C i; 1 C i; 2/; Az D 0 gives zH AH D 0H so z (not z!) is orthogonal to all columns of AH (using complex inner product zH times columns of AH ). 4 The four fundamental subspaces are now C.A/, N.A/, C.AH /, N.AH /. AH and not AT . 5 (a) .AH A/H D AH AHH D AH A again (b) If AH Az D 0 then .zH AH /.Az/ D 0. This is kAzk D 0 so Az D 0. The nullspaces of A and 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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