September 20 Homework Solutions


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1 College of Engineering nd Compuer Science Mechnicl Engineering Deprmen Mechnicl Engineering A Seminr in Engineering Anlysis Fll 7 Number 66 Insrucor: Lrry Creo Sepember Homework Soluions Find he specrum ll he eigenvlues nd eigenvecors for he mri, A, he righ. We use he bsic equion h he eigenvlues, of mri, A, cn be found from he equion DeA I =. Wriing A I nd using he epression for he by deerminn gives A De A I Here we see cse of wo generl resuls: he deerminn of ringulr mri is he produc of he erms on he digonl; nd he eigenvlues of ringulr mri re equl o he elemens on he digonl. In his cse we hve hree unique eigenvlues, =, =, nd =. Noe h he ordering of he eigenvlues is rbirry; ofen hey re numbered in rnk order. Regrdless of how he eigenvlues re numbered, he eigenvecors mus be numbered in consisen mnner. In solving for he eigenvecors, we recognize h i will no be possible o obin unique soluion. The eigenvecors re deermined only o wihin muliplicive consn. If we denoe he componens of ech eigenvecor s,, nd, he soluion for ech eigenvecor is found by subsiuing he corresponding eigenvlues ino he equion A I =, nd solving for he corresponding i s he componens of he eigenvecor. This is shown for ech eigenvecor in he ble below. + = + 6 =  = = 6 =  = + + = + 6 = = For =, we see h he soluion o he hird equion is =. Wih his soluion, he second equion gives = 6 =. We re lef wih no equion for, so we conclude h his cn be ny vlue nd we denoe i is. Thus gives he firs eigenvecor s = [ ] T. For =, he second nd hird equion boh give =. The firs equion gives = + =. If we pick o be n rbirry quniy, sy b, hen = b nd he second eigenvecor becomes = [b b ] T. For =, ere he hird equion gives us no informion, bu he second equion ells us h = 6. If we pick = c n rbirry quniy, hen = c nd he firs equion gives =  / = c c/ = 7c/. So we hve he hird eigenvecor s = [7c/ c c] T. 6 Find he principl es nd he corresponding epnsion or compression fcors for he elsic deformion mri, A, shown he righ. As shown in he e, his mri describes sreching where every old coordine,, moves o new coordine, y., such h y = A. A
2 Sepember homework soluions MEA, L. S. Creo, Fll 7 Pge The principl direcions for his deformion re he direcions for which he posiion vecor chnges only in mgniude. Thus, he new posiion vecor, y, hs he sme direcion s he old posiion vecor,. This mens h he wo vecors re reled by muliplicive consn. E. g., y =. This mens h we hve wo equions: he generl equion y = A, for ny coordine direcion nd y =, in he principl coordine direcion. We cn only sisfy hese wo equions if A = for he principl coordines. This gives n eigenvlues/eigenvecor problem. In ddiion, he eigenvlue defines he fcor by which he coordine is epnded or conrced when he coordines re muliplied by he eigenvlue, Thus, he eigenvlue is he epnsion or compression fcor. We firs solve DeA I =, s shown below, o ge he eigenvlues. De A I The eigenvlues re found by he usul formul for he soluion of qudric equion. 6, If we denoe he componens of ech eigenvecor s, nd, he soluion for ech eigenvecor is found by subsiuing he corresponding eigenvlues ino he equion A I =, nd solving for he corresponding i s he componens of he eigenvecor. Here we hve only wo equions o solve, nd one of he unknowns will be rbirry. For =, he wo equions re / / nd /. Boh hese equions give us. Thus, our firs eigenvecor is = [ ] T, where is rbirry. For = ½, he wo equions re / nd / /. Boh hese equions give us so h we cn wrie our second eigenvecor s = [b  b] T. If we regrd he coordine s lying long horizonl is nd he coordine s lying long vericl is, we cn find he principl coordine direcions corresponding o he wo eigenvlues s follows. o n n.6.6 for = b o n n.7.7 for = / b Here we find h he firs principl direcion is.6 o wih n epnsion by fcor of he vlue of he eigenvlue; he second principl direcion is.7 o wih conrcion by fcor of ½. Noe h he numbering of firs nd second re rbirry. Also, we see h when we re ineresed in only he direcion of vecor, hving common fcor in ll componens does no ffec he resul.
3 Sepember homework soluions MEA, L. S. Creo, Fll 7 Pge Is he mri he righ symmeric, skewsymmeric or orhogonl? Find he eigenvlues of he mri nd show how his illusres Theorems or Theorem in secion.. A cos sin sin cos The mri is no skewsymmeric since i hs nonzero componens on is principl digonl. For skewsymmeric mri A = A T, ll erms on he principl digonl mus be zero. The mri is lmos symmeric, bu = sin = sin. Thus he mri is no symmeric. If he mri is orhogonl, he inner produc of ech pir of rows mus vnish. There re hree such producs ignoring order. Since he firs row hs no elemens in he second or hird columns nd he second nd hird rows hve no elemens in he firs column, he inner produc of he firs row, wih he second nd hird row is zero. The inner produc of he second nd hird row is + [cos][sin] + [sin][cos] =. Since ll possible inner producs re zero we conclude h his mri is orhogonl. We find he eigenvlues in he usul wy by solving he equion DeA I =. De A I cos sin sin cos [cos ] The nonzero erms in he deerminn resul in he following equion. De A I [cos ] [cos cos sin ] [ cos ] [sin ] [ sin ] [cos ] [cos ] [sin ][ sin ] [sin ] We see h one roo of his equion gives he eigenvlue =. We find he oher wo eigenvlues by seing he remining fcor in he finl row equl o zero. This requires he soluion of he following equion for he roos of he qudric equion. cos cos cos [ sin ] cos i sin Here i =  indicing h we hve pir of comple eigenvlues. This illusres he generl fc h rel mrices cn hve comple eigenvlues, bu such eigenvlues occur s pirs of comple numbers h re conjuge o ech oher. The mgniude of ech comple eigenvlue is equl o cos sin. We see h he eigenvlues of his orhogonl mri sisfy Theorem on pge 7, which ses h he eigenvlues of n orhogonl mri re rel or comple conjuge pirs nd hve mgniude of one. Give geomericl inerpreion of he coordine rnsformion y = A, wih he mri A s shown on he righ. A cos sin The individul componens of he rnsformion y = A, wih his mri re shown below. sin cos
4 Sepember homework soluions MEA, L. S. Creo, Fll 7 Pge cos sin sin cos cos sin sin cos y y y This rnsformion from he old coordine sysem o he new y coordine sysem represens roion in he plne perpendiculr o he is. This coordine remins he sme in he new sysem s indiced by he firs equion y =. The remining wo equions, y = cos  sin nd y = cos + sin give he locion of he sme poin in he new coordine sysem. These equions represen counerclockwise roion of he coordine sysem hrough n ngle. Find bsis for eigenvecors nd digonlize he mri, A, shown he righ. A We sr by solving DeA I =, s shown below, o ge he eigenvlues. I A De We see h he roos of his equion re nd so h he wo eigenvlues giving he higher eigenvlue he lower inde re = nd =. If nd re he wo componens of ech eigenvecor, we find hese componens by solving he equion A I =. For =, we hve o solve he equions + = nd =. Boh of hese equions give =. We cn pick = = for simpliciy nd wrie our firs eigenvecor s = [ ] T. For =, he equion A I =.gives + = for ech equion. Thus, =  for he second eigenvecor. If we choose = so h = , our second eigenvecor is = [ ] T. We hve he generl resul h he mri, X, whose columns re eigenvecors of he mri A cn be used o cree digonl mri, = X  AX, provided h X hs n inverse. The elemens of he digonl mri re he eigenvlues. Thus in his cse we epec h he mri will hve [ ] in is firs row nd [ ] in is second row. The X mri in his cse will be, nd X  = from he equion for such n inverse in he lecure noes nd in he Kreyszig e. You cn redily verify h XX  = I. Since our eigenvecors could be muliplied by ny nonzero fcor nd sill be eigenvecors, we cn wrie he mos generl form of he X mri in erms of rbirry numbers nd b s X = b b. In his cse, X  = b b b. We sill hve XX  = I for ny choice of nd b h re no zero. The following operions show h X  AX is digonl mri of eigenvlues in his cse. AX X
5 Sepember homework soluions MEA, L. S. Creo, Fll 7 Pge 6 Find ou wh kind of conic secion or pir of srigh lines is represened by he equion 6 + =. Trnsform i o principl es. Epress T = [ ] in erms of he new coordine vecor y T = [y y ]. We cn wrie his equion s follows in mri form. T A To deermine he kind of figure h his equion represens, we find he principl es by finding he eigenvlues of he A mri. De A I We see h he roos of his equion re nd so h he wo eigenvlues wih he higher eigenvlue hving he lower inde re = nd =. This mens h he sndrd equion h is wrien in erms of he eigenvlues is y + y = y =. This is pir of srigh lines y =. To find he principl es we hve o find he eigenvecors. If nd re he wo componens of ech eigenvecor, we find hese componens by solving he equion A I =. For =, we hve o solve he equions = nd =. Boh of hese equions give =. We cn pick = nd = for simpliciy nd wrie our firs eigenvecor s = [ ] T. For =, he equion A I =.gives = nd + = for ech equion. Thus, = for he second eigenvecor. If we choose = so h =, our second eigenvecor is = [ ] T. Following emple 6, we use hese wo eigenvecors o form he X mri h reles he new y nd old coordine sysems by he equion = Xy. Before doing his, we normlize he eigenvecors so h he wonorm of ech eigenvecor is one. The mgniude of ech eigenvecor is + =. Thus, if we divide ech componen by, he norm of ech eigenvecor will be one. Doing his nd consrucing he X mri gives. Xy y y We cn lso obin X  from he equion in he course noes nd he Kreyszig e for he inverse of woby wo mri. Doing his gives he following relion beween he originl nd rnsformed coordine sysem. y y y X Here we see h X = X . This is rue becuse X is symmeric, orhogonl mri. For n orhogonl mri, X  = X T. For symmeric mri, X T =X. So, for symmeric, orhogonl mri, X = X .
6 Sepember homework soluions MEA, L. S. Creo, Fll 7 Pge 6 7 Deermine nd skech disks h conin he eigenvlues of he mri A he righ. A Apply Gerschgorin s heorem which ses h eigenvlues re loced in disk on he comple plne wih cener given by he digonl elemens nd rdius equl o he sum of he bsolue vlues in he row, ecluding he digonl elemens. For he firs row, he cener is nd he rdius is  + =. For he second row, he cener is nd he rdius is + = 6. For he hird row, he cener is 7 nd he rdius is + = 6. The resuling Gerschgorin disks re shown on he ne pge. We see h he disks overlp on he rel is so h we cnno sepre ou ny eigenvlues. In ddiion, A is no symmeric so we cnno be sure h he eigenvlues line on he rel is. 7 Find he similriy rnsformion, T  AT such h he Gerschgorin disk wih cener for he mri A he righ is reduced o / of is originl vlue. A The required similriy rnsformion uses digonl mri, T, where ll erms on he digonl re one, ecep he one in he row whose disk rdius we seek o reduce. To see h his is he cse, consider he bsic form of T s digonl mri wih componens jij. The ypicl erm in he inverse mri, T  will be ij/j. The componens of he T  AT produc re found from he usul rules for mri muliplicion wih he recogniion h ij = unless i = j. Thus he only nonzero erms in summion in which ij is fcor is he erm for which he i = j. [ T AT] ij n k ik n k m km In he rnsformion mri we seek here, ll he digonl erms will be one, ecep for he chosen row, sy row k, in which k will be lrge fcor. Thus here re hree possible resuls from his rnsformion wih he single digonl fcor, F, no equl o one. m mj n k ik k kj j j i ij
7 Sepember homework soluions MEA, L. S. Creo, Fll 7 Pge 7 [ T AT ] ij F ij F ij ij i i i j nd fcor F loced in row j j nd fcor F loced in row i j or fcor F no loced in row i or row j This leds o he bsic rule for using T  AT rnsformion o reduce he size of Gerschorgin disk rdii. To reduce he rdius round he cener from row k by fcor of /F, use digonl mri wih ll ones on he digonl ecep for vlue of F in row k. Be creful hough. This will increse he vlues of he elemens in column k by fcor of F in he resul. In his cse we wn o decrese he disk rdius by fcor of / for he disk wih cener given by he firs row. To do his we use digonl T mri wih in he firs row nd s for ll oher digonl elemens. This mri nd he resul of he rnsformion is shown below.. T AT. We see h his rnsformion hs decresed he disk rdius for he firs row by fcor of s desired. However, i hs lso incresed he disk rdius for he second nd hird rows from. in he originl mri o. in he rnsformed mri. This pproch o reducing he disk rdius is bsed on he fc h he eigenvlues of T  AT, for ny inverible mri, T, re he sme s he eigenvlues of A. We cn esily show his since he eigenvlues of T  AT re defined by he equion T  AT =. If we premuliply his equion by T, we ge he resul h TT  AT = IAT = AT =T. This finl equion shows h he eigenvlues of A nd T  AT re he sme. The eigenvecors re differen, however. The eigenvecors of T  AT, re, he eigenvecors of A re T.
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