Introduction to Linear algebra-Strang-Solutions-Manual_ver13

# A tridiagonal matrix a has bidiagonal factors l and u

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Unformatted text preview: 0 1 1 140 41 0 4 4D4 1 4 0 1 1 D LDLT . 011 004 011 4 001 2 32 32 3 aaaa 1 a a a a a ¤ 0 All of the b a b a b a7 b ¤ a multipliers 6a b b b 7 61 1 76 . Need 4 a b c c 5 D 4 1 1 1 54 c b c b5 c ¤ b are `ij D 1 abcd 1111 dc d ¤ c for this A 2 32 32 3 arrr 1 a r r r a¤0 b r s r s r7 b¤r 6a b s s 7 61 1 76 . Need 4a b c t 5 D 41 1 1 54 c s t s5 c¤s abcd 1111 dt d ¤t         5 2 24 2 2 10 . gives x D xD . Then gives c D cD 3 3 01 3 11 41       2 24 2 24 D c. xD . Forward to xD Ax D b is LU x D 3 01 11 8 17 "# # "# " "# # "# " 3 4 111 4 4 100 1 1 0 c D 5 gives c D 1 . Then 0 1 1 x D 1 gives x D 0 . 1 1 001 1 6 111 Those are the forward elimination and back substitution steps for # "# #" " 4 111 1 1 1 xD 5 . Ax D 1 1 6 1 111 (b) I goes to L 1 (c) LU goes to U . Elimination multiply by L 1 ! 18 (a) Multiply LDU D L1 D1 U1 by inverses to get L1 1 LD D D1 U1 U 1 . The left side is lower triangular, the right side is upper triangular ) both sides are diagonal. (b) L; U; L1 ; U1 have diagonal 1’s so D D D1 . Then L1 1 L and U1 U 1 are both I . # " # # " " #" a 1 110 a a 0 b b 1 1 D LI U I a a C b D (same L) 19 1 1 0 b bCc c 1 011 (same U ). A tridiagonal matrix A has bidiagonal factors L and U . 20 A tridiagonal T has 2 nonzeros in the pivot row and only one nonzero below the pivot (one operation to ﬁnd ` and then one for the new pivot!). T D bidiagonal L times bidiagonal U . 17 (a) L goes to I Solutions to Exercises 23 21 For the ﬁrst matrix A; L keeps the 3 lower zeros at the start of rows. But U may not 22 23 24 25 26 have the upper zero where A24 D 0. For the second matrix B; L keeps the bottom left zero at the start of row 4. U keeps the upper right zero at the start of column 4. One zero in A and two zeros in B are ﬁlled in. " # " # " # 531 420 200 Eliminating upwards, 3 3 1 ! 2 2 0 ! 2 2 0 D L. We reach 111 111 111 a lower triangular L, and the multipliers are in an upper triangular U . A D UL with " # 111 UD 0 1 1 . 001 The 2 by 2 upper submatrix A2 has the ﬁrst two pivots 5; 9. Reason: Elimination on A starts in the upper left corner with elimination on A2 . The upper left blocks all factor at the same time as A: Ak is Lk Uk . The i; j entry of L 1 is j= i for i  j . And Li i 1 is .1 i /= i below the diagonal .K 1 /ij D j.n i C 1/=.n C 1/ for i  j (and symmetric): .n C 1/K 1 looks good. Problem Set 2.7, page 115  1 1 AD 9  1 AD c      0 19 1 0 T 1 has A D ;A D ; .A 1 /T D .AT / 3 03 3 1=3    10 c c has AT D A and A 1 D 2 D .A 1 /T . 0 1 cc 1   1 3 D ; 0 1=3 2 .AB/T is not AT B T except when AB D BA. Transpose that to ﬁnd: B T AT D AT B T . 1T / D .B 1 A 1 /T D .A 1 /T .B 1 /T . This is also .AT / 1 .B T / 1 . (b) If U is upper triangular, so is U 1 : then .U 1 /T is lower triangular.   01 AD has A2 D 0. The diagonal of AT A has dot products of columns of A with 00 themselves. If AT A D 0, zero dot products ) zero columns ) A D zero matrix.    "0#  123...
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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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