Introduction to Linear algebra-Strang-Solutions-Manual_ver13

Introduction to Linear algebra-Strang-Solutions-Manual_ver13

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Unformatted text preview: c /:  14 The rank of A 3I is r D 1. Changing any entry except a12 D 1 makes A diagonalizable (A will have two different eigenvalues) 15 Ak D Sƒk S 16 17 18 19 approaches zero if and only if every jj < 1; Ak ! A1 ; Ak ! 0. 1 1 2     1 1 10 1 1 10 2 ƒD and S D I ƒk ! and Sƒk S 1 ! 1 2 : steady 1 0 :2 1 1 00 2 2 state.         :9 0 3 3 3 3 3 3 ƒD , SD ; A10 D .:9/10 , A10 D .:3/10 , 2 2 0 :3 1 1 1 1 1 1        6 3 3 6 3 3 A10 D .:9/10 C .:3/10 because is the sum of C . 2 0 1 1 0 1 1          11 11 2 1 1 10 11 110 k D and A D 1 2 1 03 11 1 1 0 3k 21   2k  1 1C3 1 3k 11 . . Multiply those last three matrices to get Ak D 11 2 1 3k 1 C 3k  k   k   1 1 50 1 1 5 5k 4k k BD D . 0 1 04 0 1 0 4k  1  20 det A D .det S /.det ƒ/.det S diagonalizable. 1 / D det ƒ D 1    n . This proof works when A is 21 trace S T D .aq C bs/ C .cr C dt / is equal to .qa C rc/ C .sb C t d / D trace T S . Diagonalizable case: the trace of SƒS 1 D trace of .ƒS 22 AB BA D I is impossible since trace AB 1 /S D ƒ: sum of the ’s. t AB BAD race BA D zero ¤ trace I .   10 10 C is possible when trace .C / D 0, and E D has EE T E T E D . 11 01     1  A0 S0 ƒ0 S 0 1 23 If A D SƒS then B D D . So B has 0 2A 0S 0 2ƒ 0 S1 the additional eigenvalues 21 ; : : : ; 2n . 24 The A’s form a subspace since cA and A1 C A2 all have the same S . When S D I the A’s with those eigenvectors give the subspace of diagonal matrices. Dimension 4. 25 If A has columns x 1 ; : : : ; x n then column by column, A2 D A means every Ax i D x i . All vectors in the column space (combinations of those columns x i ) are eigenvectors with  D 1. Always the nullspace has  D 0 (A might have dependent columns, so there could be less than n eigenvectors with  D 1). Dimensions of those spaces add to n by the Fundamental Theorem, so A is diagonalizable (n independent eigenvectors altogether). 26 Two problems: The nullspace and column space can overlap, so x could be in both. There may not be r independent eigenvectors in the column space. Solutions to Exercises 63   p p p p 21 1 27 R D S ƒS D has R2 D A. B needs  D 9 and 1, trace is not real. 12     p 1 0 01 Note that can have 1 D i and i , trace 0, real square root . 10 0 1 28 AT D A gives x T AB x D .Ax /T .B x /  kAx kkB x k by the Schwarz inequality. 29 30 31 32 B T D B gives x T BAx D .B x /T .Ax /  kAx kkB x k. Add to get Heisenberg’s Uncertainty Principle when AB BA D I . Position-momentum, also time-energy. The factorizations of A and B into SƒS 1 are the same. So A D B . (This is the same as Problem 6.1.25, expressed in matrix form.) A D Sƒ1 S 1 and B D Sƒ2 S 1 . Diagonal matrices always give ƒ1 ƒ2 D ƒ2 ƒ1 . Then AB D BA from Sƒ1 S 1 Sƒ2 S 1 D S ƒ1 ƒ2 S 1 D S ƒ2 ƒ1 S 1 D Sƒ2 S 1 Sƒ1 S 1 D BA.      ab 0 b adb (a) A D has  D a and  D d : .A aI /.A d I / D 0d 0da 0 0       00 11 21 D . (b) A D has A2 D and A2 A I D 0 is true, match00...
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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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