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1 MATH 434 - Numerical Linear Algebra , Fall 2004
SOLUTION 1. Chapter 6. OF#26 HOMEW(a). We knoORw KAx (^7) = b and will perturb the vector b by δb, where
A =
^00 ..^44454444 00 ..^44444445 −−..^22222222 −. 2222 −. 2222 0. 1112
b =
^00 ..^66676667 −. 3332
, x = ^11 1
, δb = ^1010 −−^44 10 −^4
and let δx be the perturbation on x. Since Aδx = δb, using Matlab (see appendix), we have and also cond(A) = 10000, and now w^ δex w^ =an^ [t .to^3334 verify^.^3334 1.^3333 ]t cond^ ‖δ(Ab‖)‖b‖ ≤^ ‖ ‖δxx‖‖ ≤^ cond(A)^ ·^ ‖ ‖δbb‖‖ which is obtained from Theorem 6.6.1 of the textbook. Putting the Matlab values we get
7321104 ·· 101 −^4 ≤ 11 .. (^41427321) ≤ 104 · 1. 7321101 −^4 ⇔ 1. 7321 · 10 − (^8) ≤ 0. 8165 ≤ 1. 7321 and the result is verified.
Chapter 6. #28 (a). Clearly Cond(A−^1 ) = ‖A−^1 ‖‖(A−^1 )−^1 ‖ = ‖A‖‖A−^1 ‖ = Cond(A)
Chapter (i) Since 6. I (^) =#28 A · (^) A(b).− (^1) we have
and so Cond(A) ≥ 1.^1 =^ ‖I‖^ =^ ‖A^ ·^ A−^1 ‖^ ≤^ ‖A‖‖A−^1 ‖^ =^ Cond(A) (ii) Applying the definition Cond(AtA) = ‖AtA‖‖(AtA)−^1 ‖ = ‖AtA‖‖A−^1 (A−^1 )t‖ = ‖A‖^2 ‖A−^1 ‖^2 = [‖A‖‖A−^1 ‖]^2 = Cond(A)^2 c exercisehapter 628 #29 b (ii)(a). w (^) eLet get O be an orthogonal matrix, we know that OtO = I, and applying the result of the 1 = Cond(I) ⇒ = CCondond((OO)t O=) 1 = Cond(O)^2 since chapter Cond 6 #29(O) (^) (b)≥ 0. ⇒ Now suppose Cond(A) = ‖A‖ 2 ‖A− (^1) ‖ = 1. Define We ha (^) vRe = A/‖A‖ 2 , which immediately implies ‖R‖ 2 = 1. R−^1 = ‖A‖ 2 A−^1 ⇒ ‖R−^1 ‖ 2 = ‖A‖ 2 ‖A−^1 ‖ 2 = 1 and also, for any vector x with the same size of the matrix A, ‖x‖ 2 = ‖R−^1 Rx‖ 2 ≤ ‖R−^1 ‖ 2 ‖Rx‖ 2 ≤ ‖R‖ 2 ‖x‖ 2 = ‖x‖ 2
and we conclude that ‖Rx‖ 2 = ‖x‖ 2 , which means ‖Rx‖^22 = ‖x‖^22 and so
⇒^ ( Rxxt()RtRxtR −= Ix)txx =⇔ 0 x fortR tallRx x^ = and^ xtx so RtR − I = 0 ⇒ RtR = I and (^) ⇐so RSupp is orthogonalose A = α (^) Oand with A =O ‖tOA ‖= 2 R I (^) ,= α α&=R 0.. By definition
Cond(A) = ‖A‖‖A−^1 ‖ = ‖αO‖‖α−^1 O−^1 ‖ = |α| · (^) |^1 α| · ‖O‖‖O−^1 ‖ = Cond(O) = 1. which completes the proof.
- ‖U (^) ‖Chapter∞‖U − (^1) ‖ (^) ∞6.. #30. Let U = (uij ) be a nonsingular upper triangular matrix. Consider Cond(U ) = Since ‖U ‖∞ = 1 max≤i≤n∑ j=1^ n |uij | w impliese have ‖U ‖∞ ≥ max{uii} and if we denote uij the elements of U −^1 we know that uii = 1 /uii and this ‖U −^1 ‖∞ ≥ max{ 1 /uii} = (^) min^1 {uii} and so we have Cond(U ) = ‖U ‖∞‖U −^1 ‖∞ ≥ max{uii} · (^) min^1 {uii}.
are Fitsurthermore, elements u^ thisii and,^ result as ais classical^ valid^ for result^ any^ consisten ‖O‖ ≥ |λt^ |norm, for ev^ eryas^ w eigene^ willv^ aluesshow. λ^ Sinceof U ,^ thewe^ geteigenvalues^ λi^ of^ U ‖U ‖ ≥ uii ∀ i ⇒ ‖U ‖ ≥ max{uii} and since the eigenvalues μi of U −^1 are μi = 1 /uii, we have ‖U −^1 ‖ ≥ | (^) u^1 ii | ∀ i ⇒ ‖U ‖ ≥ (^) min^1 {uii} and this two last results imply Cond(U ) = ‖U ‖‖U −^1 ‖ ≥ max{uii} · (^) min^1 {uii}. For example, define the matrix Un = LnLt n, where
Ln =
...^1 ...^2 ...^3 ...
1 2 3 ... n
and so we have Cond(Un) = Cond(LnLt n) = Cond(Ln)^2 ≥ (n/1)^2 = n^2
- Chapter 6. #34 (a) Computing the inverse of A 2
C >> h A a = p [ t 0 e. r 0 0 6 0. 0 P 1 , r 1 o , 1 b ; 3 le , 1 m , 1 9 ; 1. , 2 , 3 ];
b fo=r[ 2 i=. (^010) : 301 ; 3 ; 3 ]; end^ p(i)=i; B x 1 = = f asoctlovrelluu(pA(B);,p,b); [ xB 2 , p=] s =o (^) lfvaeclutopr(lBup,p(A,b));;
x 1 > =x 1 03 .. (^3333333509111111019974784609) ^ - >^1 .x^323349444495980 x 20 =. 33330111100370
-^31 ..^3333335499141414149978013478
1 >. 0 aeb-s 010 (A* x* 1 - b) =
- 789039094449320
- 04614086890342
abs(A*x 2 - b)= 1. 0 e- 015 *
- 444089020985006
- 88817841970013 The residual is 5 orders of magnitude smaller with partial pivoting.
C >> h A a p =[ t 1 e , r 1 / 6 2. , 1 P / 3 r ; o 1 / b 2 l , e 1 m / 3 , 1 2 / 7 4 ;. 1/3,1/4,1/5]
A = 01 .. 5000000000000000000000000000 00 .. 3530303030303030303030303030 00 .. 2353030303030303030303030303
b=[ 10 ;. 13 ;^313 ]^33333333333 0.^25000000000000 0.^20000000000000 b = (^11) 1
The residual is 5 orders of magnitude smaller with partial pivoting.
C >> h A a p =[ t 1 e , r 1 / 6 2. , 1 P / 3 r ; o 1 / b 2 l , e 1 m / 3 , 1 2 / 4 7 ;. 1/3,1/4,1/5]
A = 01 .. 5000000000000000000000000000 00 .. 3530303030303030303030303030 00 .. 2353030303030303030303030303
b=[ 10 ;. 13 ;^313 ]^33333333333 0.^25000000000000 0.^20000000000000 b = (^11) x=(A^1 \b) x = (^3) - 2. (^040). 00000000000000000000000211
- 00000000000010
deltaA =[ 1 , 1 / 2 , 1 / 3 + 0. 00003 ; 1 / 2 , 1 / 3 , 1 / 4 ; 1 / 3 , 1 / 4 , 1 / 5 ] deltaA = 10 .. 0500000000000000000000000000 00 .. 5303030303030303030303030303 00 .. 3235033060030303030303303003
- 33333333333333 0. 25000000000000 0. 20000000000000
de>l (^) tdaexl t=a x=delta 2 A. 9 \b 9190728344492
- 2293 .. 9976370622492173831747997627
an>s c=o nd 0 (.A 01 ) 0 *n 04 or 8 m 79 (A 66 - deltaA 1481 )/norm(A)-norm(dx)/norm(deltax)
Which is positive as expected. Inequality verified.
