Chapter 1. Linear Algebra

Transcription

1 Liner lger - hpter. Liner lger.. Introduction The solution of lgeric equtions pervdes science nd engineering. Therodynics is n re rife ith eples ecuse of the uiquitous presence of equiliriu constrints. Therodynic constrints re typiclly lgeric equtions. For eple, in the to-phse equiliriu of inry syste, there re three lgeric constrints defining the equiliriu stte: I II therl equiliriu, or equlity of tepertures in the to phses, T T ; echnicl I II equiliriu, or equlity of pressures in the to phses, p p ; nd cheicl equiliriu, or I II equlity of cheicl potentils of coponent i, in the to phses, i i. Under ost circustnces, these constrints re lgeric equtions ecuse there re no differentil or integrl opertors in the equtions. In the discussion of lgeric equtions, it is nturl to divide the topic into the solution of liner nd nonliner equtions. The theticl freor for the ethodicl solution of liner lgeric equtions is ell-estlished. There re rigorous techniques for the deterintion of the eistence nd uniqueness of solutions. There re estlished procedures for Eigennlysis. The discussion of the solution of non-liner lgeric equtions is postponed ecuse it is not s stright-forrd nd requires ost of the tools tht e develop in our solution of liner lgeric equtions... Linerity We hve lredy hinted t the iportnce of eing le to distinguish hether n eqution is liner or nonliner since the solution technique tht e dopt is different for liner nd nonliner equtions. We egin ith discussion of liner opertors. In thetics, n opertor is syol or function representing theticl opertion. pertors tht re filir to undergrdutes include eponents, logriths, differentition nd integrtion. We cn investigte the linerity of ech of these opertors y pplying the folloing test of linerity L y L Ly (.)

2 Liner lger - here L is liner opertor, operting on the vrile, y is nother vrile, nd nd re constnts. We cn directly chec the four opertors listed ove for linerity. The differentil opertor, d L t t, cn e sustituted into eqution (.) to yield dt d dt d d (.) dt dt t y t t y t The differentil opertor is indeed liner ecuse e no fro differentil clculus tht constnt cn e pulled out of the differentil nd tht the differentil of su is the su of the differentils. Siilrly, the integrl opertor, (.) to yield t to t yt dt t dt yt t t L t t t t o dt, cn e sustituted into eqution dt (.) to to The integrl opertor is indeed liner ecuse e no fro integrl clculus tht constnt cn e pulled out of the integrl nd tht the integrl of su is the su of the integrls. n The eponentil opertor, L, cn e sustituted into eqution (.) to yield n n n y y (.4) Eqution (.4) is not generlly true. It is true for n. Hoever, it is not true for ny other integer (positive, negtive or zero). We cn deonstrte directly tht eqution (.4) is not true for n through lgeric nipultion. y y y y (.5) Eqution (.4) is lso not true for frctionl eponents, such s n. y y y (.6) Siilrly, the logrith opertor, the inverse opertor of the eponentil opertor is not liner. log n y log log y (.7) n n

3 Liner lger - Without deonstrtion, e lso stte tht ll trigonoetric functions re nonliner. In the solution of lgeric equtions, the first step is therefore to deterine if the eqution is liner. n eqution is liner if it does not contin ny nonliner opertions. If e consider only one eqution ith one vrile, the liner eqution hs the for, (.8.) hich e ill choose to rerite in the ore generl for of function s f (.8.) Eples of nonliner lgeric equtions include f (.9.) f f sin (.9.) ep (.9.c) If there is single nonliner ter in the eqution, the entire eqution is nonliner. In the considertion of systes of equtions, if single eqution in the syste is nonliner, then the entire syste ust e treted s nonliner... Mtri Nottion syste of n lgeric equtions contining unnon vriles hs the for generl for f i j i j j for i = to n (.), i For eple, syste ith n = lgeric eqution nd = vriles hs the for f f,,,,,, (.) It is conventionl to dopt short-hnd nottion, non s tri nottion, nd epress eqution (.) s (.)

4 Liner lger - 4 here the tri of constnt coefficients,, is n,,,,, n,,, n, (.) The to lines underline in the nottion indicte tht the tri is to-diensionl tri. We refer to s n n tri ecuse it contins n ros (equtions) nd coluns (vriles). The vectors nd re nd n (.4) The single underline in the nottion nd indictes tht nd re vectors, or onediensionl trices. We refer to nd s colun vectors of size nd n respectively. The crucil thing to reeer out tri nottion is tht it is convention to siplify the nottion. It does not introduce ny ne theticl rules. It does not dd or chnge the rules of lger. The solution to the set of eqution cn e otined folloing the filir rules of lger, lthough such nipultions ecoe cuersoe hen the nuer of equtions is lrge. Therefore e ill shortly dopt set of nottions for tri opertions, hich consist of sequence of lgeric rules. In eqution (.), e present the first tri opertion, tri ultipliction,. Mtrices cn only e ultiplied if the inner indices tch. In this cse is of size n nd is of size. Since the lst inde of tches the first inde of, they cn e ultiplied. If the indices do not tch, there is not tri ultipliction. For eple, the tri ultipliction cnnot e perfored ecuse is of size nd is of size n nd the inner indices, nd n re not the se. The tri resulting fro vlid tri ultipliction is of size defined y the to outer indices of the fctor trices. Thus yields tri of size n, colun vector of length n. The i th eleent of n n tri,, nd n colun vector,, is defined s

5 Liner lger - 5 j i for i = to n (.5), j j Siilrly, the i,j th eleent of n n tri,, nd n p colun vector, B, is defined s i,, j for i = to n nd for j = to p (.6).4. The Deterinnt nd Inverse We no consider the solution of eqution (.) or lterntively eqution (.). When n= eqution nd = vrile, e hve (.5), This of course hs the generl solution,,. This siple prole illustrtes the issue of eistence of solution. The solution only eists if,. We cn net consider set of liner lgeric equtions ith n= equtions nd = vriles, e hve,,,, (.6) Through series of lgeric nipultions, e cn rrive t the solution nd (.7) Note tht oth nd hve the se denointor. This denointor is given specil ne, the deterinnt, det. det (.8) It is cler tht solution only eists if det. This is ectly prllel to the single eqution cse given ove, here the deterinnt of the one eqution cse is siply det.

6 Liner lger - 6 We cn net consider set of liner lgeric equtions ith n= equtions nd = vriles, e hve,,,,,,,,, (.9) Through series of lgeric nipultions, e cn rrive t the solution det det det (.) here det (.) Note tht oth, nd hve the se denointor. It is cler tht solution only eists if det. There is no theoreticl reson tht e could not continue to solve systes of n liner lgeric equtions ith n vriles for ritrry n. Hoever, prcticlly speing it ecoes very tie consuing. Note in ll these cses tht the deterinnt is strictly function of. The deterinnt is not function of nd. t this point, e siply etrpolte the theticl oservtion tht unique solution to eists only if the deterinnt of the tri eists. It turns out tht the solution to the prole given in eqution (.7) nd the solution to the prole given in eqution (.) nd the solution for the generl nn prole cn e epressed in tri nottion s Through series of lgeric nipultions, e cn rrive t the solution (.) here is clled the inverse tri of. In ddition to providing the solution to s given in eqution (.), the inverse ll hs the dditionl property,

7 Liner lger - 7 I (.) here I is the identity tri, defined s I (.4) ne cn lso derive eqution (.) s follos I We cn oserve directly fro the eples given ove tht for the sll, nd systes, (.5.) det det (.5.) det (.5.c) lerly fro eqution (.5), the inverse does not eist if the deterinnt is zero. If the inverse of tri eists, is sid to e non-singulr. If the inverse of the tri does not eist, is sid to e singulr.

8 Liner lger Eleentry Ro pertions There eists ethodicl procedure for generting inverses nlyticlly. With the uiquitous presence of coputers, it is unliely tht ny student ill ever hve ny need to perfor such procedure. It is not even perfectly cler tht it is essentil to include such procedure in odern tetoo. Nevertheless, since students y e clled upon to generte inverses of sll systes in n eintion in hich coputers re not ville, e present the procedure here. The procedure uses three eleentry ro opertions. The first eleentry ro opertion is the ultipliction of ro y constnt. ro c ro (.6) The second eleentry ro opertion is sitching the order of ros. lerly, the order tht the equtions re ritten should not influence the vlidity of the equtions. ro ro (.7) The third eleentry ro opertion is the replceent of n eqution y the liner cointion of tht eqution ith other equtions. In other ords, either eqution in (.8) cn e replced y ro ro ro (.8) nd the resulting syste of equtions ill still yield the se result. These three eleentry ro opertions provide the necessry tools to (i) deterine the eistence nd unique of solutions, (ii) deterine the inverse of if it eists nd (iii) provide the solution to. For tri, the procedure for finding the inverse is given elo. First, e crete n ugented I tri. If is n nn tri nd I is n nn identity tri, then I is n n(n) tri. defined s,, n I (.9) n, n, n In generl, one perfors eleentry ro opertions tht convert the side of the ugented tri to I. t the se tie, one perfors the se eleentry ro opertions to the I side of the ugented tri, hich converts it to the inverse of. We cn illustrte the process for tri.

9 Liner lger - 9 I () Put one in the digonl eleent of RW. RW RW () Put zeroes in ll the entries of LUMN ecept RW. RW RW RW () Put one in the digonl eleent of RW. RW RW (4) Put zeroes in ll the entries of LUMN ecept RW. RW RW RW

10 Liner lger - hich cn e siplified s: Here e hve converted the tri on the left hnd side to the identity tri. The tri on the right hnd side is no the inverse s cn e seen through coprison of eqution (.5.). We cn lern severl things out the inverse fro this deonstrtion. The ost iportnt thing is tht if the deterinnt is zero, the inverse does not eist (ecuse e divide y the deterinnt to otin the inverse.) Never clculte n inverse until you hve first shon tht the deterinnt is not zero..6. Rn nd Ro Echelon For To deterine the eistence nd uniqueness of the solution to ugented, e ust crete n tri. If is n nn tri nd is n n colun vector, then n(n+) tri. defined s is n n,, n, n, n n (.)

11 Liner lger - In order to deterine the eistence nd uniqueness of solution to, e need to put the tri nd the ugented tri into ro echelon for (ref) (or reduced ro echelon for (rref)) using sequence of eleentry ro opertions. Ro echelon for is lso clled upper tringulr for, in hich ll eleents elo the digonl re zero. For n ritrry tri, e hve We cn put this tri into ro echelon for ith one eleentry ro opertion, nely RW RW RW hich yields ref Eqution (.) is the ro echelon for of. Reduced ro echelon for siply requires dividing ech ro of the ro echelon for y the digonl eleent of tht ro, RW RW RW RW hich yields rref Eqution (.) is the reduced ro echelon for of ecuse it is in ro echelon for nd it hs ones in the digonl eleents.

12 Siilrly, the ugented Liner lger - tri cn e put into ro echelon for or reduced ro echelon for using precisely the se set of eleentry ro opertions. For eple, e hve for the ro echelon for,,,, RW RW RW,, ref, For eple, e hve for the reduced ro echelon for RW RW RW RW rref,,,,,,, det In order to evlute the eistence nd uniqueness of solution, e lso require the rn of tri. The rn of tri is the nuer of non-zero ros in tri hen it is put in ro echelon for. The rn of tri in ro echelon for is the se s the rn of tri in reduced ro echelon for. onsider the folloing upper tringulr trices.

13 Liner lger - u u u U u u (.) u The rn of this tri is. The deterinnt of this tri is non-zero. If the deterinnt of n nn tri is zero, then the rn ( ) is less thn n. u u u U u u (.) Non-squre trices cn lso e put in ro echelon for. onsider the ugented n(n+) tri of the for: u u u v U u u v (.) u v The rn of this tri is still defined s the nuer of non-zero ros in the ro echelon for of the tri. The rn of the tri shon ove is. For ugented trices, the non-zero eleent cn pper on either tri. The rn of the folloing tri is still three. u u u v U u u v (.4) v In n ugented tri, oth sides of ro ust e zero for the ro to e considered zero. The rn of the folloing tri is to. u u u v U u u v (.5) The rn provides the nuer of independent equtions in the syste. For eple, consider the eple given elo. The third eqution is liner cointion of the first to n

14 Liner lger - 4 equtions. This tri cn e put in ro echelon for using the folloing eleentry ro opertions, (.6) c c c This tri cn e put in ro echelon for using the folloing eleentry ro opertions, RW RW crw RW RW RW hich yields RW (.7).7. Eistence nd Uniqueness of Solution The eistence nd uniqueness of solution to cn e deterined ith either the ro echelon for or the reduced ro echelon for of nd s follos. In deling ith liner equtions, e only hve three choices for the nuer of solutions. We either hve,, or n infinite nuer of solutions. No Solutions: rn( ) n nd rn( ) rn( ) ne Solution: rn( ) rn( ) n Infinite Solutions: rn( ) rn( ) n

15 Liner lger - 5 When rn( ) rn( ), your syste is over-specified. t lest one of the equtions is linerly dependent in the tri ut is ssigned to n inconsistent vlue of in the vector. There re no solutions to your prole. When rn( ) rn( ) n, you hve properly specified syste ith n equtions nd n unnon vriles nd you hve one, unique solution. When rn( ) rn( ) n, then you hve less equtions thn unnons. You cn pic n rn() unnons ritrrily then solve for the rest. Therefore you hve n infinite nuer of solutions. We ill or one eple of ech cse elo. Eple.. ne Solution to here Let s find () the deterinnt of () the inverse of (c) the solution of (d) the solution of Solution: () The deterinnt of is (y eqution.) det. () Becuse the deterinnt is non-zero, e no there ill e n inverse. Let s find it. STEP NE. Write don the initil tri ugented y the identity tri. I STEP TW. Using eleentry ro opertions, convert into n identity tri. () Put one in the digonl eleent of RW. RW RW RW

16 Liner lger - 6 / / / () Put zeroes in ll the entries of LUMN ecept RW. RW RW RW RW RW RW RW RW RW RW / / / / / / / / / () Put in the digonl eleent of RW. RW RW RW / / / / / / / / / / (4) Put zeroes in ll the entries of LUMN ecept RW. RW RW RW RW RW RW/ * RW RW RW / * RW 5 / / / / / / / / / (5) Put in the digonl eleent of RW. RW RW RW / 5/ / / / / /

17 Liner lger - 7 (6) Put zeroes in ll the entries of LUMN ecept RW. / * 5/ * RW RW RW RW RW RW RW RW RW RW 5 We hve the inverse. 5 (c) The solution to is. 5 (d) The solution to is We see tht e only need to clculte the inverse once to solve oth nd. Tht s nice ecuse finding the inverse is lot hrder thn solving the eqution once the inverse is non. Eple.. No Solutions to

18 Liner lger In this cse, hen e copute the deterinnt, e find tht det. The deterinnt is zero. No inverse eists. To deterine if e hve no solution or infinite solutions find the rns of nd. In ro echelon for, ecoes: ref By inspection of the ro echelon for, rn. In ro echelon for, ecoes ref By inspection of the ro echelon for, rn. Since rn rn, there re no solutions to. Eple.. Infinite Solutions to onsider the se tri,, s s used in the previous eple.. The deterinnt is zero nd the rn is. No consider different vector. In ro echelon for, ecoes:

19 Liner lger - 9 ref By inspection of the ro echelon for, rn. Since ) ( ) ( n rn rn, there re infinite solutions. We cn find one eple of the infinite solutions y folloing stndrd procedure. First, e ritrrily select ) rn( n vriles. In this cse e cn select one vrile. Let s e. Then sustitute tht vlue into the ro echelon for of nd solve the resulting syste of ) ( rn equtions. ref When ref No solve ne prole here nd coe fro the non-zero prts of ref. This prole ill lys hve n inverse. The solution is given y So one eple of the infinite solutions is

20 Liner lger -.8. Eigennlysis Eigennlysis involves the deterintion of eigenvlues nd eigenvectors. It is prt of liner lger tht is etreely iportnt to scientists nd engineers in rod vriety of pplictions. Here e first provide the theticl freor for otining eigenvlues nd eigenvectors. Then e provide n eple. For n nn squre tri, there re n eigenvlues, though they need not ll e different. If the deterinnt of the tri is non-zero, ll of the eigenvlues re non-zero. If the deterinnt of the tri is zero, t lest one of the eigenvlues is zero. To clculte the eigenvlues,, for n nn tri, one egins y sutrcting the eigenvlue fro ll digonl eleents. I,, n,,, n,, n, n n, n (.8) Second, the deterinnt of I is set to zero, det I (.9) Third you ust solve this eqution for. Eqution (.9) is clled the chrcteristic eqution nd is polynoil in of order n. Thus, this eqution hs n roots. s ith ny polynoil eqution, the roots y e cople. The n roots of eqution (.9) re the n eigenvlues. Ech eigenvlue hs ssocited ith it n eigenvector. Thus, if there re n-eigenvectors, there re lso n-eigenvlues. The i th eigenvector,, for the tri, is otined y solving I i i i (.4) for i. This eqution defines the eigenvectors nd cn e solved n ties for ll n eigenvlues to yield n eigenvectors. For tri, e hve

21 Liner lger - I,,,, (.4) The deterinnt of I is set to zero, det I ( ) det (.4),,, For tri, the chrcteristic eqution is qudrtic polynoil. The to roots of the eqution re given y the qudrtic forul,, (,, ) (,, ) 4det (.4) The eigenvectors re given y, i, I i i i (.44),, i This set of liner equtions ust e solved. The first step in solving syste of liner eqution is finding the deterinnt. Becuse the eigenvlues re solutions to det I (fro eqution (.4)), the deterinnt of the tri in eqution (.44) is lys zero. Since the vector in eqution (.44) is the zero vector, the rn of I is equl to the rn of the ugented tri I, hich is less thn n, rn I rn I n (.45) onsequently, there re lys infinite solutions for ech eigenvector. nother y to thin of this is tht eigenvectors provide directions only, ut not gnitude. For tri, e cn rndoly set the second eleent of the eigenvector to n ritrry vrile,. Solving the first eqution in eqution (.44) yields the eigenvectors,,, i,, i for i to n, for itrry (.46) Typiclly, eigenvectors re reported s norlized vectors, here the gnitude of the vector is one. The gnitude of n ritrry vector,, of length n is defined s

22 Liner lger - n i i (.47) Therefore norlized vector,, is given y (.48) By construction, the gnitude of this norlized vector is one. The norlized eigenvectors for the eple is then,,,,, for itrry to n for i i i i i i (.49) Even the norlized eigenvectors still hve to equivlent epressions, hich involves ultipliction y -. norlized eigenvector ultiplied y - is still norlized eigenvector. For coon list of eigenvectors, e dopt the convention tht the rel coponent of the first eleent of ech eigenvector should e positive. Eple.4. onsider the tri,. The chrcteristic eqution is given y 4 det I The eigenvlues re given y nd. Fro eqution (.46), the unnorlized eigenvectors re for =, The norlized eigenvectors re

23 Liner lger - If e follo the convention tht the first eleent should e positive in our eigenvectors, then our norlized eigenvectors re Eple.5. Norl Mode nlysis Students frequently s for n eple of the physicl ening of eigenvlues nd eigenvectors. Such ening cn e directly oserved in the ppliction of norl ode nlysis. Belo e provide norl ode nlysis of siple one-diensionl odel of cron dioide in the idel gs stte. onsider cron dioide olecule odeled s three prticles connected y to springs, ith the cron to in the iddle, s shon in Figure.. The positions in the lortory fre of reference re suscripted ith n. The positions s devitions fro their equiliriu positions reltive to the olecule center-of-ss contin only the essentil infortion for this prole nd do not hve the suscript. We odel the interction eteen olecules s Hooin springs. For Hooin spring, the potentil energy, U, is spring () spring () U ( nd the force, F, is ) F ( ),,, Figure.. n idelized odel of the cron dioide olecule. here is the spring constnt (units of g/s ). We cn rite Neton s equtions of otion for the three olecules: F F F

24 Liner lger - 4 Knoing tht the ccelertion is the second derivtive of the position, e cn rerite the ove equtions in tri for s (first divide oth side of ll of the equtions y the sses) dt d here Tody, e re not interested in solving this systes of ordinry differentil equtions. We re content to perfor n eigennlysis on the tri. First e hve I The chrcteristic eqution for this tri is det I Rerrnging this cuic polynoil for yields det I In this for, y inspection the roots to the chrcteristic eqution re

25 Liner lger - 5 The eigenvectors for ech of these eigenvlues re given y i i I I Since the equtions re not linerly independent, e cn reove, s vrile nd set it equl to. Then our syste ecoes:,, This tri hs non-zero deterinnt. We cn solve it uniquely to yield,, nd the eigenvector tht corresponds to is To find the second eigenvector, the eigenvector tht corresponds to I

26 Liner lger - 6 s efore, e reove the third eqution nd reove, s vrile nd set it equl to. Then our syste ecoes:,, With this ne tri, e cn clculte tht the deterinnt is non-zero nd the solution is,, nd the eigenvector tht corresponds to is To find the third eigenvector, the eigenvector tht corresponds to I s efore, e reove the third eqution nd reove, s vrile nd set it equl to. Then our syste ecoes:,, With this ne tri, e cn clculte tht the deterinnt is non-zero nd the solution is

27 Liner lger - 7,, nd the eigenvector tht corresponds to is So e hve the three eigenvlues nd the three eigenvectors. So ht? Wht good do they do us? For virting olecule, the squre root of the solute vlue of the eigenvlues fro doing n eigennlysis of Neton s equtions of otion, s e hve done, re the norl frequencies. You see tht the units of the eigenvlues re /sec, so the squre root hs units of frequency (or inverse tie). For cron dioide, the three norl frequencies re: The frequency of zero is no frequency t ll. It is not virtionl ode. In fct, it is trnsltion of the olecule. We cn see this y eining the eigenvectors. The eigenvector tht corresponds to or is This is description of the norl virtion ssocited ith frequency of zero. It sys tht ll tos ove the se ount in the -direction. See Figure..

28 Liner lger - 8 The eigenvector tht corresponds to or is This eigenvector descries virtion here oth the oygen ove y fro the cron eqully nd the cron does not ove. The eigenvector tht corresponds to or is This eigenvector descries virtion here oth the oygen ove to the right nd the cron ove ore to the left, in such y tht there is no center of ss otion. The norl odes of otion provide coplete, independent set of virtions fro hich ny other virtion is liner cointion. Figure.. Norl odes of one-diensionl odel of cron dioide. The top ode is trnsltionl ode ith. The iddle ode is virtionl ode ith ode is virtionl ode ith.. The otto

29 Liner lger Sury of Logiclly Equivlent Stteents t this point, e hve identified the ost coon tss required in the solution of syste of liner lgeric equtions. n eple of the nlyticl ethod y hich nuericl vlues cn e otined y hnd hs een presented. The vlue in presenting these hnd clcultions lies in developing n understnding of the generl ehvior of systes of liner equtions. It is unliely tht e ill ever e clled upon (outside of es) to clculte eigenvlues or inverses y hnd. Nevertheless, noing ht to epect fro n nlyticl understnding etter prepres us to e sense of the nuericl tools nd etter understnd hy nuericl tools fil. For eple, e cn s progr to copute the inverse of tri ith deterinnt of zero. Depending on the softre, vriety of cryptic essges y e provided hen the code crshes. Shoing tht the deterinnt is zero first, llos us to understnd tht not ll of our equtions ere independent. lterntively, soe softre ill siply return soe tri ithout ever notifying the user tht the inverse does not eist. gin, the sic understnding provided ove cn go long y in interpreting the results of theticl softre. To this end, e cn identify surize the logiclly equivlent stteents out n nn tri,. If ny one of these stteents is true, ll the others re true. If nd only if det( ) If nd only if det( ) then inverse eists then inverse does not eist then is non-singulr then is singulr then rn( ) n then rn( ) n then there re no zero ros in the ro echelon for of then there is t lest one zero ro in the ro echelon for of then hs one, unique solution then hs either no solution or infinite solutions ll eigenvlues of re non-zero t lest one eigenvlue of is zero

30 Liner lger -.. Sury of MTLB onds In the tle elo, sury of iportnt liner lger conds in MTLB is given. Entering tri =[,;,] (cos seprte eleents in ro, seicolons seprte ros) (esiest for direct dt entry) =[ ] (ts seprte eleents in ro, returns seprte ros) (useful for copying dt fro tle in Word or Ecel) Entering colun vector =[;;] (n n vector) deterinnt of tri det() (sclr) inverse of n nn tri inv() (nn tri) solution of = =\ or =inv()* (n vector) Entering ro vector =[,,] ( n vector) rn of tri rn() (sclr) trnspose of n n tri or n n vector = (n tri or n vector) reduced ro echelon for of n nn tri rref() (nn tri) eigenvlues nd eigenvector of n nn tri [,ld]=eig() ( is n nn tri here ech colun is n eigenvector, ld is nn tri here ech digonl eleent is n eigenvlue, off-digonls re zero).

31 Liner lger -.. Proles Proles re locted on course esite.