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# LINEAR ALGEBRA I MTH 341

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7 Linear Algebra Mth 341 Winter 1998 Example Population Movement Bent Petersen This example deals with population movements The example involves only as much linear algebra as we studied in the first couple of lectures Much more can be said about population models but more linear algebra is required in particular matrix multiplication eigenvalues and eigenvectors We suppose that we have a population P distributed among 3 locales call them L1 L2 and L3 Each year 5 percent of the population of L1 moves to L2 and 8 percent moves to L3 Also each year 10 percent of the population of L2 moves to L1 and 6 percent moves to L3 Finally each year 7 percent of the population of L3 moves to L2 and 2 percent moves to L1 The percentages are relative to the populations at the beginning of the year and we assume the process goes on year after year with the same percentages In addition we assume that the total population is conserved gt withlinalg Warning new definition for norm Warning new definition for trace When we specify numbers with a decimal point Maple treats them as oating point numbers of limited precision Since our percentages have only two digits we could get in trouble with roundoff Since the progression of events in row reduction depends on whether or not certain computed quantities are exactly 0 even a small amount of roundoff can cause big trouble Fortunately Maple does exact arithmetic with rational numbers so we avoid the necessity of studying roundoff in our calculations by converting our data to rational numbers In reality of course the data are known only approximately and so we should study but we will not how the solutions vary with small pertubations in the data That almost amounts to studying the propogation of roundoff in the calculations Here s a Maple procedure to convert a oating point matrix to a rational matrix There is no error checking or other hand holding so use it properly gt ratmatrix proc Amatrix gt local jk gt for j from 1 to rowdimA do gt for k from 1 to coldimA do gt Ajk convert Ajk rational gt od od A end Let x0 y0 and 20 be the start of the year populations of L1 L2 and L3 respectively and let xl yl 21 be the end of year populationsThen from the given data we clearly have Page 1 87x0 10y0 0220 x1 05 x0 84y0 0720 yl 08 x0 06y0 9120 21 The coef cient matrix A in these equations is called the transition matrix or migration matrix gt A matrix33 8710020584O708069l 87 10 02 A 05 84 07 08 06 91 Unfortunately we have not yet developed the tools to ask some really interesting questions such as what is the longterm behavior However we can ask if there is a steadystate situation That is is there an initial distribution of the population which remains invariant under the population movements described above Setting xx0xly y0yl and 22021 above we obtain a homogeneous system of linear equations for x y and z with coef cient matrix B given by gt B evalmAdiaglll 13 10 02 B 05 16 07 08 06 09 It is not really necessary for a proper discussion but let us recall that the total population is P Thus we have the extra equation xyzP We can include this equation in our system by using Maple s aptly named stack command Then we obtain an inhomogeneous system of 4 linear equations in 3 unknowns with coef cient matrix C 39 gt C stackBmatrixl3 lll 13 10 02 C 05 16 07 08 06 09 l l l and inhomogenous term Tgt b matrix4l 000P Page 2 quot0000 gt solnl linsolveCb 2825484765 P solnl 2797783934P 4376731302P We can ask Maple to solve this system for us If convert the entries in C to rational numbers first we get the solution gt soan linsolveratmatrixCb 102 P 361 101 P 361 158 P 361 301112 Let us check by how much these solutions differ We will use 20 digits precision to compute the difference since if we use the default precision used to compute the solutions Maple will probably just return 0 Z gt evalf evalmsolnlsoln2 20 4570637119 104013 4819944598 104013 609418283 104113 Since Maple defaults to 10 digits precision these errors show that roundoff was not a problem Before we get too cocky though let us note gt rrefB Ot IO D IOO f gt rref ratmatrixB Page 3 l 0 79 101 0 158 0 0 0 The first row reduction of B is not correct If we had elected to work directly with the homogenous system of linear equations solved for x y and z with the thought that we would substitute into x y z P later we would have obtained x0 y0 and 20 and we would have been pretty confused If we convert B to a rational matrix first this method does work Since the matrices A and B are the ones that one usually works with in problems of this type it is important to be careful about roundoff Converting to rational numbers may seem silly since it implies absolute precision in the data However the entries in B are not arbitrary Indeed the entries in each column must sum to exactly 0 why Converting to rational numbers enables us to preserve exactly the consequences that follow from this property Of course we should still study how the solution varies with small pertubations in the data UDDEEUDURI We have not studied matrix multiplication yet but we are about to do so The rest of this note will then make sense Let s start with the entire population in L1 and then let them do their thing for 50 years gt popOmatrix3lP00 P popO 0 0 Z gt evalmA 50amppopO 2825778886P 2797657846 P 4376563276 P Let us see also what happens if we start with the population evenly distributed fgt poplmatrix3l P3P3P3 Page 4 l P o 1 P P P 3 P gt evalm AA50amppopl 2825601809 P 27977463 84 P 4376651813 P Recall the equilibrium solution that we found above gt evalmsolnl 2825484765 P 2797783934 P 4376731302 P Do you suspect a theorem here The fact is that our matrix A here is an example of a regular stochastic matrix The key property which you guessed from the above is theorem 18 on page 288 of our text Page 5 Introduction to Maple and Linear Algebra Mth 341 Summer 1995 Oct 26 1996 These notes are a compilation of seven sets of notes distributed in Mth 3417 Linear Algebra7 during the OSU 1995 Summer Session The original notes were created in Maple V37 exported as LATEX7 edited and converted to AMSB IEX This compilation was lightly edited7 converted to ETEXZg and compiled with the Maple V4 ETEX style le It should be realized that some of the more idiosyncratic behavior of Maple V3 referred to below may have changed in Maple V47 or may change in future versions 1 Handout a Before doing any linear algebra with maple its a good idea to load the limng package gt with1ina1g Warning new definition for norm Warning new definition for trace The colon which terminates this command supresses outputi Without it weld get a blitz of messages The rede nition error messages can be ignored Matrices can be de ned as follows gt A matrix36 03664537858939129615 0 3 76 6 4 75 A I 3 77 8 75 8 9 3 79 12 79 6 15 Linear Algebra Summer 1995 Introduction to Maple The rst two parameters specify the size of the matrix The third parameter is a list indicated by square brackets of the entries in the matrix7 in rowmajor order that is7 rowbyrow Maple has a builtin command to compute the reduced row echelon canonical form gt rref A 1 0 72 3 0 724 0 1 72 2 0 77 0 0 0 0 1 4 A synonym for rref reduced row echelon form is gaussjord GaussJordan form gt gaussjordA 1 0 72 3 0 724 0 1 72 2 0 77 0 0 0 0 1 4 We can also do the row reduction by hand gt swaprowA12 gt mulrowquot 113 77 8 75 8 1 E E 3 0 3 76 6 4 75 3 79 12 79 6 15 Here the double quote is the ditto operator 7 Maple interprets it to mean the previous expression in our case the matrix 1 gt addrowquot13 3 758 1 3 03766475 072474726 Mth 5 41 2 Bent E Petersen Linear Algebra Summer 1995 Introduction to Maple gt addrowquot321 77 8 75 8 1 3 3 0 1 72 2 2 1 0 72 4 74 72 6 gt addrowquot2173 22 16 1 0 72 3 7 7 3 3 0 1 72 2 1 0 72 4 74 72 6 gt addrowquot232 22 16 1 0 72 3 7 7 3 3 0 1 72 2 2 1 70 0 0 0 2 8 7 gt mulrowquot312 22 16 1 0 72 3 7 7 3 3 0 1 72 2 2 1 70 0 0 0 1 4 7 gt addrowquot31 223 1 0 72 3 0 724 0 1 72 2 2 1 0 0 0 0 1 4 gt addrowquot32 2 1 0 72 3 0 724 0 1 72 2 0 77 0 0 0 0 1 4 Sure beats doing the arithmetic by hand Mth 5 41 3 Bent E Petersen Linear Algebra Summer 1995 Introduction to Maple 2 Handout b Here s some more Maple examplesl gt with1ina1g Warning new definition for norm Warning new definition for trace gt Amatrix03 64 1213 2303 1459 0 73 76 4 71 72 71 3 72 73 0 3 1 4 5 79 Note here we de ned the matrix A by feeding a list of its rows to the matrix function We didnlt have to specify the size of the matrix Sometimes though it s more convenient to specify the size7 as for example gt bmatrix41 9117 To solve the system of equations AI b we can form the augmented matrix gt Caugment Lb 0 73 76 4 9 71 72 71 3 1 72 73 0 3 71 1 4 5 79 77 and row reduce gt rref C OOOH OOHO O H O Mth 5 41 4 Bent E Petersen Linear Algebra Summer 1995 Introduction to Maple At this point we can readoff the solutions As you might expect though7 Maple does provide a facility for solving linear equations directly gt 1insolveAb Here tl is a parameter let s denote it by t since the underscore is unaesthetic and we donlt really need the subscript Then Maplels response means 11 3t 5 12 72t 7 3 13 t 14 i 0 Lets see What happens When therels no solution gt Amatrix22 1111 bmatrix21 12 gt 1insolveAb Maple appears to be silent7 but has in fact returned the empty setl We can generate random matrices useful for examples gt Arandmatrix45 brandmatrix41 Mth 5 41 5 Bent E Petersen Summer 1995 Introduction to Maple Linear Algebra 54 75 b 39 99 761 gt 1insolveAb 13620907 78454729 7 13630726 27261452 itl 13466201 19319399 t 7 6815363 13630726 7 1 18178923 19875703 t 13630726 27261452 17576855 38671600 t r 1 6815363 6815363 t1 1 The fact that we have one parameter here tells us that there is only one free variable The remaining 4 variables are bound pivotali Thus the rank of A must be 41 Lets see if Maple agrees gt rankA 4 Sure enough7 we were right about the rank Note that Maple gave us the exact solution abovel Some times this is inconvenienti We can round off the answer and express it in decimal notation as follows gt eva1fquotquot 719992796422 2877863182 7751 71975859686 1l417341894t1 1333672396 7 7290771966 tl 2579004963 7 5674180524 t1 t1 The double quote here refers to the expression prior to the previous one 3 Handout c In this note we will see by example how to determine if a given vector is a linear combination of other given vectorsi gt with1ina1g Warning new definition for Warning new definition for norm trace Mth 5 41 Bent E Petersen Linear Algebra Summer 1995 Introduction to Maple Consider the following vectors gt umatrix41 21 31 vmatrix41 1350 wmatrix41 2 113 gt bmatrix411617373 Weld like to know if we can write I as a linear combination of u7 v and w7 that is can we nd scalars 11712713 such that b 11 u 12 v 13 wl If we let A be the matrix with columns u7 v7 w gt A augmentuvw 2 71 2 l 3 71 A2 73 5 l l 0 73 then we see that we are looking for a solution to the system Az 12 gt 1insolveAb Thusb73u6v72wl 4 Handout d In this handout we look at the problem of nding a onesided inversel First let s look at the theory Theorem 1 Let A be an m X n matrix Then A has a left inverse if and only if the n X n matrix ATA is invertible fA has a left inverse then m 2 n and ATA71 AT is a left inverse for A Proof If ATA is invertible then ATA71ATA I implies D ATA71AT is a left inverse for Al Conversely suppose D is any left inverse for A in general not unique Let E E R and suppose ATAE 6 Then 6 ETATAE AET A5 implies A5 6 But then 5 DAE D6 6 Thus ATAE39 6 implies E 6 and therefore the square matrix ATA is invertible If A has a left inverse D then DAE39 5 implies A5 6 has only the trivial solutionl Hence each column in A must be pivotal and so m 2 nl Mth 5 41 7 Bent E Petersen Linear Algebra Summer 1995 Introduction to Maple Theorem 2 Let A be an m X n matrix Then A has a right inverse if and only the m X m matrix AAT is invertible fA has a right inverse then n 2 m and AT AAT7 is a right inverse for A Proof If D is a right inverse for A then DT is a left inverse for ATi D We can use the explicit formulae above to nd left or right inversesl Keep in mind thought that these are usually not unique 7 we are selecting a particular one gt with1ina1g Warning new definition for norm Warning new definition for trace gt Amatrix23 121312 1 2 71 3 1 2 A Since A is 2 X 3 we have m lt n and so A cannot have a left inversel To check for a right inverse we compute AAT gt Beva1mA amp transposeA gt rref B Since the reduced row echelon form of B is I we know B is invertible Thus A has a right inverse given by C ATB II gt Ceva1mtransposeA amp inverseB 1 1 E C2 1 0 3 74 1 E Mth 5 41 8 Bent E Petersen Linear Algebra Summer 1995 Introduction to Maple We can check our proposed right inverse by multiplying gt eva1mA amp C It worksl Note Maple uses amp for matrix multiplication7 not l The reason is that matrix multiplication is not commutative Adding the ampersand prevents Maple from attempting simpli cations Which assume commutativityl Exercise 1 We commented above that A cannot have a left inverse It follows that the 3 X 3 matrix ATA is not inveitible Verify this fact by computing ATA and rowreducing it Use Maple of course 5 Handout e Letls look again at the problem of nding right inversesl We can handle left inverses by considering the transpose gt with1ina1g Warning new definition for norm Warning new definition for trace Consider the same example as in handout d gt Amatrix23 121312 1 2 71 3 1 2 gt Iddiag11 1 Id 0 To nd a right inverse for A we solve the equation AB 1 Maple can solve this equation With the same function call as is used to solve A5 b gt B 1insolve A Id Mth 5 41 9 Bent E Petersen Linear Algebra Summer 1995 Introduction to Maple Well this is certainly correct it is a solution but Maple doesnlt seem to realize that the two columns can be selected independently of each other The solution really should contain 2 parameters We use the subs function substitution to replace Maple s one parameter with 2 parameters7 s in the rst column7 and t in the second column gt Cstgtaugmentsubst 1 scol B 1 subsLt 1 1 colB2 C I at A augment subs tl scolB 1 7subs tl tcolB2 What a mess Does it really look like its supposed to do gt Cst 1 2 7 7 7 s 7 7 t 5 5 1H 3 5 s t Letls check that it works gt eva1m A amp Cst 1 0 0 1 Sure enough Cs7 t is a right inverse for any 8 and any ti Taking s 7415 and t 15 we get the same solution as in handout d gt C41515 1 H 5 1 g 74 E i We could never have obtained this solution directly from Maplels solution by substituting for tli Even though Maple is quite clever we do have to coax it a bit sometimes 6 Handout f In this handout we simply illustrate Maplels functions for computing eigenvalues and eigenvectorsi gt with1ina1g Warning new definition for norm Warning new definition for trace Mth 5 41 10 Bent E Petersen Linear Algebra Summer 1995 Introduction to Maple gt Amatrix33 134 2 340 313 71 4 72 A2 73 4 0 73 1 3 gt eigenvalsA 123 gt eigenvectsA 11111 211 H 31134 The format here is eigenvalue algebraic multiplicity set of linearly independent eigenvectorsHi gt Ptransposematrix33111132321341 1 1 1 3 P 1 E 3 3 1 7 4 2 gt eva1minverseP amp A amp P 1 0 gt Bmatrix33422242224 4 B2 2 2 gt eigenvalsB 822 gt eigenvectsB 871M111l7l2727110l7l7101ll Mth 5 41 11 Bent E Petersen Linear Algebra Summer 1995 Introduction to Maple Note we have 2 linearly independent eigenvectors for the eigenvalue 2 and so 3 linearly independent eigenvectors all together Thus B is diagonizab e gt Qtransposematrix33 111110 101 l 71 71 Q 1 1 0 l 0 l gt eva1minverseQ amp B amp Q 8 0 0 0 2 0 0 0 2 We can also use Maple to compute characteristic polynomials gt charpoly A1ambda A376A211A76 gt charpoly B1ambda A3712A236A732 Obviously Maple uses det AI 7 A for the chararcteristic polynomial rather than det A 7 AI as we did in c ass The question of diagonizability can be answered by computing the Jordan normal form this is a matrix similar to the given one and as close to being diagonal as possible Thus A is diagonizable if and only if jordanA is diagonal In any case7 the diagonal entries in jordanA are just the eigenvalues of A7 repeated according to algebraic multiplicityi gt jordanA l 0 0 0 2 0 0 0 3 gt jordanB 2 0 0 0 8 0 0 0 2 Mth 5 41 12 Bent E Petersen Linear Algebra Summer 1995 Introduction to Maple gt cmatrix33121333212 1 2 1 CI 3 3 3 2 1 2 gt jordanC 6 0 0 0 0 1 0 0 0 We see the eigenvalues of C are 600 and C is not diagonizablei Letls compute the eigenvectors of C gt eigenvectsC 67H12111710727l1011l Sure enough we have only 2 linearly independent eigenvectorsi Thus C is not diagonizablei 7 Handout g In this handout we continue to illustrate Maplels facilities for computing eigenvalues and eigenvectorsi gt with1ina1g Warning new definition for norm Warning new definition for trace gt Amatrix33124372569 4 A2 3 7 2 9 gt eveigenvalsA 166 1 17 ev11377137 9 1 3 1 13 83 1 17 1 13 166 1 771 if i 71 3 1 if 2 911332 f 9 1137 1 13 83 1 17 1 13 166 1 771 iii feel 3 1 777 2 9 1133 2 f 9 76113 2 76111088 7 Ix94182 27 9 Mthzuz 13 Bent E Petersen Linear Algebra Summer 1995 Introduction to Maple Here s one place where Maple s penchant for exact answers gets in the way We can convert the exact answer to an approximate oating point answer gt evalf ev 1374788906 7894602542 7 1210 8 14146713484 7 610 9 I The symbol I is the square root of 71 Thus we have complex answers with very small imaginary parts Letls increase the accuracy gt Digits18 gt eva1fev 137478890587268580 789460254283571530 7 1 10 17 I 414671348410885728 4110471 The imaginary parts are smaller In fact they are of the same order as the precision of our conversion This is pretty sure sign we are seeing a roundoff error If we really want a oating point answer we can force Maple to provide one from the beginning by putting at least one oating point number in the matrix gt Digits10 gt Bmatrix33 1024372569 10 2 4 B I 3 7 2 5 6 9 gt eigenvalsB 7894602543413747889014146713483 This answer less precise than the one we obtained above with the default precision of 10 digits however we no longer have the troublesome imaginary parts What about eigenvectors gt evt eigenvectsA 138 53 evt 11a17 a 7 1 Root0fz3 71722 41 Z 51 3 2 34 47 5 2 61 1 7 61 11221221 Here 1 runs over the roots of the characteristic polynomial that is the eigenvalues We can try a oating point conversion again gt evalf evt 78946025428 1 73078932392 9166765693 1 Mth 5 41 14 Bent E Petersen Linear Algebra Summer 1995 Introduction to Maple Sigh We only get one of the roots and the corresponding eigenvector More work is needed On the other hand if we force Maple to work in oating point from the beginning we get all the eigenvectors gt eigenvectsB 137478890171 3592673648 43466708379276415841 7 4146713483717 72987282688 7371720079 7 6035890728 7 7894602543471 946575123 7 2818195168 7 3074361518 Actually if we want to compute eigenvalues and eigenvectors in oating point the best way is to use the inert unevaluated function Eigenvals note the capitol 7E This function is designed for numeric matrices We force evaluation by using the evalf function gt eva1fEigenvalsB 7894602543413747889014146713483 We can also get the eigenvectors simply by providing a second variable gt eva1fEigenvalsBC 7894602543413747889014146713483 gt Reva1mC 946575123 3592673648 72987282688 R 2 72818195168 4346670837 7371720079 73074361518 9276415841 76035890728 If we compute R lBR we should get a diagonal matrix with the eigenvalues of B on the diagonal gt eva1m inverseR amp B amp R 78946025428 74510 8 71310 3 71710 7 1374788906 7710 3 72110 3 610 8 4146713484 Apart from a fairly large roundoff error we get what we expected Copyright 1995 1996 Bent E Petersen Permission is granted to duplicate this document for non7pro t educational purposes provided that no alterations are made and provided that this copyright notice is preserved on all copies Bent E Petersen 24 hour phone numbers Department of Mathematics of ce 541 7375163 Oregon State University home 541 7542315 Corvallis OR 973314605 fax 541 7370517 email petersenmath0rstedu web httpwwworstedufpeterseb Mth 5 41 15 Bent E Petersen 39An Example of Roundoff Mth 341 Linear Algebra Spring 2000 Bent Petersen gt restartwith linalg Warning new definition for norm Warning new definition for trace Let39s row reduce the following matrix quotby handquot so to speak gt Bmatrix33 13100205 16070806 09 13 10 02 B 05 16 07 08 06 09 gt addrow B1 2 1 addrow231 mulrow1 1 13 B1addrow 1 2 08 13 10 02 08 06 09 08 06 09 13 10 02 08 06 09 0 0 0 1000000000 7692307692 1538461538 08 06 09 0 0 0 1000000000 7692307692 1538461538 B 0 1215384615 07769230770 0 0 0 gt B2mulrowB1 21B122 B3addrow21 B2 1 2 gt 1000000000 7692307692 1538461538 BZ 0 1000000000 6392405066 0 0 0 1000000000 0 6455696204 B3 0 1000000000 6392405066 0 0 0 Page 1 I It is clear that the matrix B has rank 2 Now let39s let Maple do the row reduction directly 39 gt rref B l 0 0 0 0 l QOb O This time it looks as if B has rank 3 Which is it Let39s o the row reduction again by hand but this time we normalize our pivot in column 1 rst gt C1mulrowB11B11 1000000000 7692307692 l538461538 C 05 16 07 08 06 09 gt addrowC112 C121 C2addrow13 C131 1000000000 7692307692 l53846l538 0 1215384615 07769230769 08 06 09 1000000000 7692307692 1538461538 C2 0 1215384615 07769230769 0 1215384615 07769230770 gt C3mulrowC221C222 addrowC321 C312 C4addrow23 C332 1000000000 7692307692 1538461538 C3 0 1000000000 6392405065 0 1215384615 07769230770 1000000000 0 6455696203 0 1000000000 6392405065 0 1215384615 07769230770 1000000000 0 6455696203 C4 0 1000000000 6392405065 0 0 1 103910 Notice inthe third column third row we have a very small number which arises from roundolT in our calculations Since it is not zero it becomes a bogus pivot In a sense we are doing the calculations too accurately This bogus pivot is very much smaller than the presumed 2 place accuracy in our original data If we do the calculations with just 2 or 3 decimal places of accuracy we are less likely to be led astray In Maple we can set the accuracy for decimal calculations by setting the quantitiy quotDigitsquot The default is 10 Let39s try a few other values gt Digits2 rref B Page 2 gt Digits3 rref B l 0 a644 0 l a637 gt Digits4 rref B O O gt Digits19 rref B gt Digits20 rref B l 0 645569620253l6455697 0 l 639240506329ll392406 0 0 0 gt Digits160 rref B Ot IO D IOO l 0 0 gt Digits10 Reset to default We do have a problem here Let39s multiply B by 100 and convert the entries to integers If we row reduce the resulting matrix Maple will use exact rational arithmetic 39gt BBmatrix33 131025 167869 rrefBB 13 10 2 BB 5 l6 7 8 6 9 51 0 79 101 0 l 158 0 0 0 Thus if B is actually given exactly by the original expression then the rank is 2 But if the entries in B consist of experimental data which is not known exactly the situation is less clear Page 3 Here s digits gt gt gt gt gt gt gtr gt digits a little routine to compute the rank of B in oating point using a prescribed accuracy of n decimal rnk proc A n local rt tDigits Digitsn if rrefA33 1 then else r2 fi Digits t r3 end Remark This is not a good algorithm for computing the rank Let39s look at the sequence of results that we get by considering decimal precisions from 2 to 240 decimal 39gt seqrnkBn n2 240 2 2 3 3 3 3 2 2 3 3 3 3 3 3 3 3 3 3 2 2 3 3 3 3 2 2 3 3 3 3 3 2 3 3 3 3 2 2 3 3 2 3 2 3 2 3 3 2 2 3 3 3 2 2 2 3 3 3 3 3 2 3 3 3 2 2 3 3 3 3 3 2 2 3 3 3 3 2 2 3 3 3 3 2 3 3 3 3 2 2 3 3 3 3 3 3 3 3 3 3 2 2 3 3 3 3 2 2 3 3 3 3 3 2 3 3 3 3 2 2 3 3 3 2 2 3 3 3 2 2 2 3 3 3 3 3 2 3 3 3 2 2 3 3 3 3 3 2 2 3 3 3 3 2 2 3 3 3 3 2 3 3 3 3 2 2 3 3 3 3 3 3 3 3 3 3 2 2 3 3 3 3 2 2 3 3 3 3 3 2 3 3 3 3 2 2 3 3 3 2 2 3 3 3 2 2 2 3 3 3 3 3 2 3 3 3 2 2 3 3 3 3 3 2 2 3 3 3 3 2 2 3 3 3 3 2 3 3 3 One thing is abundantly clear extra precision does not cast any light on the problem Our sequence of rank estimates looks almost random The calculation of rrefB is inherently badly posed To see this let s perturb a bit the exact version of BB 100B and row reduce using exact arithmetic First recall BB gt evalmBB rref BB 101 158 0 Page 4 Now lets perturb one of the diagonal entries in BB by a very tiny little bit 39 gt CCevalmBBdiag1 00 10quot50 4 2 CC l00000000000000000000000000000000000000000000000000 5 l6 7 8 6 9 39 gt rref CC COD I Ob O D IOO Recall this is an w calculation At this point you may wonder is the rank of B two or three Ifwe view B as an exact matrix then the rank is 2 otherwise we should probably view it as having rank 3 if our interest is primarily in the rank any tiny pertubation of B is almost certain to have exact rank 3 On the other hand we have to admit that the rank 3 result arises from promoting a very small pivot to l and that in an applied problem it might make more sense to set it to 0 From a practical realworld point of view then we should regard the rank as 2 Solving a linear system throws some light on our dilema gt linsolveBvector 001 6455696203 10 6392405065 10 1 1012 gt linsolveBBvector 0 0 100 Note if we view B as exact ie the second case we have no solution for the given right hand side In the rst case the solution is found using oating point arithmetic and is much larger than seems reasonable Altering the precision has a big effect on the solution For example gt Digits 20 linsolve B vector 0 0 1 64556962025316455695 1021 63924050632911392405 1021 1 1022 These quotsolutionsquot which are unique clearly depend upon noise in the oating point calculations From the point of view of solving a linear system with coef cient matrix B we should think of B as having rank 2 and the example system above with inhomogeneous term 001 as being inconsistent that is as having no solution The matrix B was care llly chosen on the brink so to speak to yield a badly conditioned problem You might think that in real world problems an experimentally obtained matrix is likely to have maximal rank and lead to Page 5 better behaved problems Normally that would be almost true maximal rank does not suf ce but in the presence of symmetries exactly conserved quantities or relations between the elements of the matrix one may easily end up with very badly conditioned systems Solutions to real world problems should probably be accompanied by estimates of the error in the data and estimates of the in uence of such errors together with roundolT and other errors on the solution Even power ll systems such as Maple do not allow you to dispense with care ll thought Note in our example the matrix B has column sums 0 You can imagine that B describes some physical system with a conservation law that requires the column sums be My 0 Then all physically meaningful matrices B have rank at most 2 and we would feel no reservation about killing an exceptionally small pivot that threatens to yield rank 3 Page 6

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