Errors for Linear Systems


 Gervais Spencer
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1 Errors for Lnear Systems When we solve a lnear system Ax b we often do not know A and b exactly, but have only approxmatons Â and ˆb avalable. Then the best thng we can do s to solve Âˆx ˆb exactly whch gves a dfferent soluton vector ˆx. We would lke to know how the errors of Â and ˆb nfluence the error n ˆx. Example: Consder the lnear system Ax b wth ( ( We can easly see that the soluton s x ( x x 2 and solve the lnear system Aˆx ˆb. Ths gves the soluton vector ˆx sde vector has caused a large change n the soluton vector. ( Now let us use the slghtly dfferent rght hand sde vector ˆb ( 2 0 ( In ths case a small change n the rght hand Vector norms In order to measure errors n vectors by a sngle number we use a socalled vector norm. A vector norm measures the sze of a vector x R n by a nonnegatve number and has the followng propertes 0 x 0 αx α x + y + y for any x, y R n, α R. There are many possble vector norms. We wll use the three norms, 2, defned by x + + x n 2 ( x x n 2 /2 max{ x,..., x d } If we wrte n an equaton wthout any subscrpt, then the equaton s vald for all three norms (usng the same norm everywhere. If the exact vector s x and the approxmaton s ˆx we can defne the relatve error wth respect to a vector norm as ˆx x. Example: Note that n the above example we have ˆb b 0.0, but ˆx x. That means that the relatve error of the soluton s 00 tmes as large as the relatve error n the gven data,.e., the condton number of the problem s at least 00. Matrx norms A matrx norm measures the sze of a matrx A R n n by a nonnegatve number. We would lke to have the property Ax for all x R n ( where s one of the above vector norms, 2,. We defne as the smallest number satsfyng (: Ax : sup x R n x 0 max Ax x R n By usng the, 2, vector norm n ths defnton we obtan the matrx norms, 2, (whch are n general dfferent numbers. It turns out that and are easy to compute:
2 Theorem: max,...,n,...,n max,...,n,...,n Proof: For the nfnty norm we have a (maxmum of row sums of absolute values a (maxmum of column sums of absolute values Ax max a x max a x max a mplyng max a. Let be the ndex where the maxmum occurs and defne x sgn a, then and Ax max a. For the norm we have ( Ax a x ( a x max a mplyng max a. Let be the ndex where the maxmum occurs and defne x and x 0 for, then and Ax max a. We wll not use 2 snce t s more complcated to compute (t nvolves egenvalues. Note that for A, B R n n we have AB B snce The followng results about matrx norms wll be useful later: ABx Bx B. Lemma : Let A R n n be nonsngular and E R n n. Then E < mples that A + E s nonsngular. A Proof: Assume that A + E s sngular. Then there exsts a nonzero x R n such that (A + Ex 0 and hence A Ax Ex E The left nequalty for b : Ax follows from A b A. As > 0 we obtan A E. Lemma 2: For gven vectors x, y R n wth x 0 there exsts a matrx E R n n wth Ex y and E y. Proof: For the nfntynorm we have x for some. Let a R n be the vector wth a, a k 0 for k and let E ya, then ( a x mples Ex y and ( ya v a v y wth a v v mples E y. For the norm we use a R n wth a sgn(x snce a x and a v v. For the 2norm we use a x/ 2 snce a x 2 and a v a 2 v 2 v 2. Condton numbers Let x denote the soluton vector of the lnear sytem Ax b. If we choose a slghtly dfferent rght hand sde vector ˆb then we obtan a dfferent soluton vector ˆx satsfyng Aˆx ˆb. We want to know how the relatve error ˆb b / nfluences the relatve error / ( error propagaton. We have A(ˆx x ˆb b and therefore A (ˆb b A ˆb b.
3 On the other hand we have Ax. Combnng ths we obtan A ˆb b. The number cond(a : A s called condton number of the matrx A. It determnes how much the relatve error of the rght hand sde vector can be amplfed. The condton number depends on the choce of the matrx norm: In general cond (A : A and cond (A : A are dfferent numbers. Example: In the above example we have cond (A A ( and therefore ˆb b 00 ( whch s consstent wth our results above (b and ˆb were chosen so that the worst possble error magnfcaton occurs. The fact that( the matrx A n our example has a large condton number s related to the fact that A s close to the sngular matrx B. The followng result shows that cond(a ndcates how close A s to a sngular matrx: Theorem: A B mn B R n n, B sngular cond(a Proof: ( Lemma shows: B sngular mples A B A. (2 By the defnton of A there exst x, y R n such that x A y and A y matrx E R n n such that Ex y and E y sngular and A B E A.. By Lemma 2 there exsts a. Then B : A E satsfes Bx Ax Ex y y 0, hence B s ( (.0.99 Example: The matrx A s close to the sngular matrx B so that A B By the theorem we have that 0.0 cond or cond (A (A 00. As we say above we have cond (A 00,.e., the matrx B s really the closest sngular matrx to the matrx A. When we solve a lnear system Ax b we have to store the entres of A and b n the computer, yeldng a matrx Â wth rounded entres â fl(a and a rounded rght hand sde vector ˆb. If the orgnal matrx A s sngular then the lnear system has no soluton or nfntely many solutons, so that any computed soluton s meanngless. How can we recognze ths on a computer? Note that the matrx Â whch the computer uses may no longer be sngular. Answer: We should compute (or at least estmate cond(â. If cond(â < ε M then we can guarantee that any matrx A whch s rounded to Â must be nonsngular: â a ε M a mples Â A εm for the nfnty or norm. Therefore Â A Â ε M ε M nonsngular by the theorem. ε M and cond(â < ε M ε M ε M mply Â A Now we assume that we perturb both the rght hand sde vector b and the matrx A: Â <. Hence the matrx A must be cond(â
4 Theorem: Assume Ax b and Âˆx ˆb. If A s nonsngular and Â A / A there holds cond(a ˆb b Â cond(a Â A A + Proof: Let E Â A, hence Aˆx ˆb Eˆx. Subtractng Ax b gves A(ˆx x (ˆb b Eˆx and therefore A ( ˆb b + E ˆx. Dvdng by and usng / gves A ˆb b + E ˆx Now we have ˆx + ˆx x + ˆx x. By puttng ˆx x on the left hand sde and solvng for t we obtan the asserton. If cond(a Â A we have that both the relatve error n the rght hand sde vector and n the matrx are magnfed by cond(a. If cond(a Â A A Â A then by the theorem for the condton number the matrx Â may actually be sngular, so that the soluton ˆx s no longer well defned. Computng the condton number We have seen that the condton number s very useful: It tells us what accuracy we can expect for the soluton, and how close our matrx s to a sngular matrx. In order to compute the condton number we have to fnd A. Ths takes n 3 +O(n 2 operatons, compared wth n3 3 +O(n2 operatons for the LUdecomposton. Therefore the computaton of the condton number would make the soluton of a lnear system 3 tmes as expensve. For large problems ths s not reasonable. However, we do not need to compute the condton number wth full machne accuracy. Just knowng the order of magntude s suffcent. Assume that we pck a vector c and solve the lnear sytem Az c. Then z A c and z A c or A z c. Ths gves us a lower bound for A, and the cost of ths operaton s only n 2 + O(n. The trck s to pck c such that z c becomes as large as possble, so that the lower bound s close A. There are a number of heurstc methods avalable whch acheve farly good lower bounds: ( Pck c (±,..., ± and pck the sgns so that the forward susbsttuton gves a large vector, ( pckng c : z and solve A z c often mproves the lower bound. The Matlab functons condest(a and /rcond(a use smlar deas to gve lower bounds for cond (A. Typcally they gve an estmated condton number c wth c cond (A 3c and requre the soluton of 2 or 3 lnear systems whch costs O(n 2 operatons f the LU decomposton s known. (However, the Matlab commands condest and rcond only use the matrx A as an nput value, so they have to compute the LU decomposton of A frst and need n3 3 + O(n2 operatons. Computaton n machne arthmetc and resduals When we run Gaussan elmnaton on a computer each sngle operaton causes some roundoff error, and nstead of the exact soluton x of a lnear system we only get an approxmaton ˆx. As explaned above we should select the pvot canddate wth the largest absolute value to avod unnecessary subtractve cancellaton, and ths usually s a numercally stable algorthm. However, there s no theorem whch guarantees ths for partal pvotng (row nterchanges. (For full pvotng wth row and column nterchanges some theoretcal results exst. However, ths algorthm s more expensve, and for all practcal examples partal pvotng seems to work fne.
5 Queston : How much error do we have to accept for ˆx x? Ths s the unavodable error whch occurs even for an deal algorthm where we only round the nput values and the output value to machne accuracy, and use nfnte accuracy for all computatons. When we want to solve Ax b we have to store the entres of A, b n the computer, yeldng a matrx Â and a rght hand sde vector ˆb of machne numbers so that Â A ε M and ˆb b ε M. An deal algorthm would then try to solve ths lnear system exactly,.e., compute a vector ˆx such that Âˆx ˆb. Then we have cond(a cond(aε M (ε M + ε M 2 cond(aε M f cond(a /ε M. Therefore the unavodable error s 2 cond(aε M. Queston 2: After we computed ˆx how can we check how good our computaton was? The obvous thng to check s ˆb : Aˆx and to compare t wth b. The dfference r ˆb b s called the resdual. As Ax b and Aˆx ˆb we have ˆb b cond(a where ˆb b / s called the relatve resdual. We can compute (or at least estmate cond(a, and therefore can obtan an upper bound for the error /. Actually, we can obtan a slghtly better estmate by usng (ˆb A b A ˆb b ˆb b cond(a ˆx ˆx ˆb b wth the weghted resdual ρ :. Note that / ˆx δ mples for δ < ˆx δ x + (ˆx x δ ( + δ δ whch s the same as δ up to hgher order terms O(δ 2. If ˆb b / s not much larger than ε M then the computaton was numercally stable: Just perturbng the nput slghtly from b to ˆb and then dong everythng else exactly would gve the same result ˆx. But t can happen that the relatve resdual s much larger than ε M, and yet the computaton s numercally stable. We obtan a better way to measure numercal stablty by consderng perturbatons of the matrx A: Assume we have a computed soluton ˆx. If we can fnd a slghtly perturbed matrx Ã such that Ã A ε, Ãˆx b (2 where ε not much larger than ε M, then the computaton s numercally stable: Just perturbng the matrx wthn the roundoff error and then dong everythng exactly gves the same result as our computaton. How can we check whether such a matrx Ã exsts? We agan use the weghted resdual ˆb b ρ : ˆx. Then:. If ˆx s the soluton of a slghtly perturbed problem (2 we have ρ ε.
6 2. If ρ ε then ˆx s the soluton of a slghtly perturbed problem (2. Proof:. Let E Ã A. Then (A + Eˆx b or ˆb b Eˆx yeldng ˆb b ˆb b E ˆx, ˆx E ε. 2. Let y : b ˆb. Usng Lemma 2 we get a matrx E wth Eˆx y and E y ˆx. Then Ã : A + E satsfes Ãˆx (A + Eˆx ˆb + (b ˆb b and E b ˆb ˆx ε. Summary Recommended method for solvng lnear systems on a computer:. Gven A fnd L, U, p usng Gaussan elmnaton wth pvotng, choosng the pvot canddate wth the largest absolute value. 2. Solve Lu b (where b b p by forward substtuton and Ux y by back substtuton. Do not compute the nverse matrx A. Ths takes about 3 tmes as long as computng the LU decomposton. The condton number cond(a A characterzes the senstvty of the lnear system: If Ax b and Aˆx ˆb we have ˆb b cond(a. The unavodable error due to the roundng of A and b s approxmatvely 2 cond(aε M. If cond(â /ε M then the matrx Â could be the machne representaton of a sngular matrx A, and the computed soluton s usually meanngless. You should compute an approxmaton to the the condton number cond(a A. Here A can be approxmated by solvng a few lnear systems wth the exstng LU decomposton (condest n Matlab. In order to check the accuracy of a computed soluton ˆx compute the resdual r : Aˆx b and the weghted resdual ρ : r ˆx. we get an error bound ˆx x ˆx cond(aρ we should have that ρ s not much larger than ε M, otherwse the computaton was not numercally stable: The error s much larger than errors resultng from roundng A and b. Gaussan elmnaton wth the pvotng strategy of choosng the largest absolute value s n almost all cases numercally stable. We can check ths by computng the weghted resdual ρ. If ρ s much larger than ε M we can obtan a better result by teratve mprovement: Let r : Aˆx b and solve Ae r usng the exstng LU decomposton. Then let ˆx : ˆx e.
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