Introduction to Linear algebra-Strang-Solutions-Manual_ver13

# Solutions to exercises 76 4 a st v d v b st v1

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Unformatted text preview: 32 immediately says that An D zero matrix. The key example is a single n by n Jordan block (with n 1 ones above the diagonal): Check directly that J n D zero matrix. 23 Certainly Q1 R1 is similar to R1 Q1 D Q1 1 .Q1 R1 /Q1 . Then A1 D Q1 R1 c s 2 I is similar to A2 D R1 Q1 c s 2 I: 24 A could have eigenvalues  D 2 and  D 1 (A could be diagonal). Then A 1 has the 2 same two eigenvalues (and is similar to A). 22 If all roots are  D 0, this means that det.A Problem Set 6.7, page 371  1 A D U †V T D u1  u2  1 0  v1 v2 T 32 3 3 2p 1 3 5 4 50 0 5 4 1 25 00 2 1 3 1 Dp p 10 5 2 4  12 2 This A D is a 2 by 2 matrix of rank 1. Its row space has basis v1 , its nullspace 36 has basis v2 , its column space has basis u1 , its left nullspace has basis u2 :   11 1 2 Row space p Nullspace p 2 1 5 5   1 1 1 3 T Column space p ; N .A / p : 3 1 10 10 3 If A has rank 1 then so does AT A. The only nonzero eigenvalue of AT A is its trace, 2 2 which is the sum of all aij . (Each diagonal entry of AT A is the sum of aij down one column, so the trace is the sum down all columns.) Then 1 D square root of this sum, 2 2 and 1 D this sum of all aij . p p   5 But A is 3C 5 2 3 21 T T 2 4 A A D AA D has eigenvalues 1 D , 2 D . 11 indeﬁnite 2 2 p p 1 D .1 C 5/=2 D 1 .A/; 2 D . 5 1/=2 D 2 .A/; u1 D v1 but u2 D v2 . 5 A proof that eigshow ﬁnds the SVD. When V 1 D .1; 0/; V 2 D .0; 1/ the demo ﬁnds AV 1 and AV 2 at some angle  . A 90ı turn by the mouse to V 2 ; V 1 ﬁnds AV 2 and AV 1 at the angle   . Somewhere between, the constantly orthogonal v1 and v2 must produce Av1 and Av2 at angle =2. Those orthogonal directions give u1 and u2 . p    p  21 1=p2 1=p2 2 2 6 AAT D has 1 D 3 with u1 D and 2 D 1 with u2 D . 12 1= 2 1= 2 p3 2 2 p3 &quot; # 1=p6 1= 2 110 2 2 AT A D 1 2 1 has 1 D 3 with v1 D 4 2=p6 5, 2 D 1 with v2 D 4 0p 5; 011 1= 2 1= 6 p3 2   p  1=p3 110 300 and v3 D 4 1=p3 5. Then D Œ u1 u2  Œ v1 v2 v3 T . 011 0 10 1= 3 Solutions to Exercises 75 p 3 and 2 D 1 in †. The smallest change to rank 1 is to make  2 D 0. In the factorization 7 The matrix A in Problem 6 had 1 D A D U †V T D u1 1 vT C u2 2 vT 1 2 this change 2 ! 0 will leave the closest rank–1 matrix as u1 1 vT . See Problem 14 1 for the general case of this problem. 8 The number max .A 1 /max .A/ is the same as max .A/=min .A/. This is certainly  1. It equals 1 if all  ’s are equal, and A D U †V T is a multiple of an orthogonal matrix. The ratio max =min is the important condition number of A studied in Section 9:2. 9 A D U V T since all j D 1, which means that † D I . 10 A rank–1 matrix with Av D 12u would have u in its column space, so A D uwT for some vector w. I intended (but didn’t say) that w is a multiple of the unit vector v D 1 .1; 1; 1; 1/ in the problem. Then A D 12uvT to get Av D 12u when vT v D 1. 2 11 If A has orthogonal columns w1 ; : : : ; wn of lengths 1 ; : : : ; n , then AT A will be di2 2 agonal with entries 1 ; : : : ; n . So the  ’s are deﬁnitely the singular values of A (as expected). The eigenvalues of t...
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