Exam, Solutions

Transcription

1 Exam, Solutons Note that many of the answers are more detaled than s requred for full cre Queston he robenus norm he robenus norm of a matrx M, M whch s an overall measure of the sze of the matrx, s defned as the square root of the sum of the squares of the absolute values of ts entres, e, M,, m A Show that M tr MM, where or any matrx M wth elements, M element n oston (, k ) of the roduct M s the adont of M m, the elements of the adont are M m So the, MM are MM, k, M M k, m mk So the MM, whch s the sum of the elements on the dagonal, s gven by trace of tr MM MM m m,, m herefore,, M m, tr MM, B If A s self-adont, show that A, where the are the egenvalues of A If A s self-adont, then A traa tr A he trace s the sum of the egenvalues, and the egenvalues of the square of a matrx are the squares of ts egenvalues (snce f Av v, then Av AAv ( ) A( v) Av v) C Show that f U s untary, then untary transformaton M UMU, e, the robenus norm s nvarant under UMU tr UMU UMU We work out the adont of UMU by takng the adont of UMU U M U the terms, n reverse order: UMU U M U UM U, and UMU tr UMU UMU tr UMU UM U tr UMM U tr UMU UM U Snce U s untary, U U So

2 nally, snce the trace s ndeendent of coordnates (or, snce tr( XY ) tr( YX ) for any two matrces X and Y ), UMU tr UMM U tr MM M Queston Reresentatons one grou nsde of another grou Say G s a (fnte) grou, U s a reresentaton (wth U : g Ug a homomorhsm from G to untary transformatons n Hom( V, V ), and H s a roer subgrou G (e, H s a subgrou and H G) A Show that U s a reresentaton of H Snce U : g Ug s a mang from G nto untary transformatons for whch Ugg UgUg, t necessarly has these roertes for a subset of G B Show that f U s an rreducble reresentaton of H, then t s also an rreducble reresentaton of G If U s rreducble on H, then (accordng to the defnton of rreducblty) there s no drect-sum decomoston V V V n whch all of the transformatonsu h act searately n V and V (e, there s no bass n whch all of the matrces U h have the same block-dagonal form), for all h H Snce G ncludes all of H, these conons necessarly hold for U consdered as a reresentaton of G C If U s an rreducble reresentaton of G, then s t necessarly an rreducble reresentaton of H? If yes, rovde a roof; f no, rovde an examle n whch U becomes reducble when restrcted to a subgrou ycally, even f U s rreducble on G, t s reducble on H As an examle, take G S3, the ermutaton grou on three obects S 3, or, equvalently, the rotatons and reflectons of an equlateral trangle It has an rreducble reresentaton of dmenson, corresondng to these matrx transformatons It also has a subgrou H consstng of ust the rotatons hs s a commutatve grou, so all of ts reresentatons are necessarly one-dmensonal D Gven the above setu, wth U an rreducble reresentaton of G, and further assume that the character of U s zero for all elements of G that are not n H (ths behavor s not unusual, for examle see art E of Q3 below) hen show that U s always reducble on H If U (e, U restrcted to H ), s rreducble, then du (, U ) = By the trace formula, H H H

3 du ( H, UH) = cu( g) H gîh G æ ö G æ é ù ö = cu( g) cu( g) cu( g) ú H = + G gîh H G ê, èç ø çè ú ëgîh gïh ûø G æ ö G G = cu ( g) d( U, U) H ç = = çèg gîg ø H H he last two equaltes follow from the trace formula and the hyothess that U an rreducble reresentaton of G Snce H G (because H s a roer subgrou G ), the last quantty s larger than Queston 3 Another way to construct grou reresentatons Here we wll develo another way to construct grou reresentatons Let G be a (fnte) grou, H s a subgrou, and recall that, for any b G a rght coset of H, denoted Hb, s the set of all elements of g G that can be wrtten as g hb We showed n class that G s the dsont unon of all of ts dstnct rght cosets (ncludng the rght coset corresondng to the dentty H He) hs means that the number of rght cosets s G H Any element c Gcan be vewed as actng as a ermutaton on the cosets, snce ( Hb) c s also a coset, namely, ( Hb) c H ( bc) A Show that the acton on cosets yelds a ermutaton reresentaton of G, that we wll denote here as Q We need to show that the mang from grou elements to ermutatons of cosets s a homomorhsm from the grou G to the ermutatons on cosets; f so, then Q: g Qg wll be the mang from G nto the corresondng ermutaton matrces o show homomorhsm, we need to show that the ermutaton corresondng to the grou element c, followed by the ermutaton corresondng to the grou element c, s the same as the ermutaton corresondng to the grou element cc hs follows from assocatvty: (( Hb) c ) c ( Hbc ) c H ( bc c ) Hb( c c ) B What does the above constructon reduce to f H G? What does t reduce to f H s ust the dentty element? If H G, then there s only one coset, H tself, whch s the entre grou So the constructon yelds the trval one-dmensonal reresentaton that mas every grou element to If H e, then every element of G s ts own coset, so the constructon yelds the regular reresentaton C Assume that H s s not all of G, e, that there s at least some element of G that s not n H Show that the reresentaton we constructed s NO rreducble

4 rst, note that snce every Q g s a ermutaton matrx, ts trace Q ( g ) must be a non-negatve nteger, G and that Q () e, the number of cosets We aly the trace formula, to show that Q contans at H least one coy of the trval reresentaton I, for whch I g : dqi (, ) = cq( g) ci( g) = cq( g) (snce for the trval reresentaton, I ( g) ) hs G gîg G gîg exresson must be an nteger Snce ( e Q ) 0 and all of the other character terms cannot be negatve, the fnal rght hand sde cannot be zero So dqi (, ) ³, whch means that that Q contans at least one coy of the trval reresentaton I, It remans to be shown that Q mas at least one grou element to somethng that s not the dentty We show ths by choosng some g H If t were the case that Q g s the dentty ermutaton, then Q g would ma the coset H to tself, whch means that Hg H hs means that hg H for every h, whch mles hgh hg h g h h g H, a contradcton D Now assume that H s a normal subgrou of G (A normal subgrou s a subgrou for whch every rght coset s also a left coset, e, that bh Hb Equvalently, for a normal subgrou H and any h H and any g G, then ghg H ) Determne the character of ( h) for h H We show that the ermutaton corresondng to rght multlcaton of cosets by h leaves all cosets unchanged hat s, for any coset Hb, Hbh Hb o do ths, note that for any hbh Hbh, hbh h ( ) bhb b h bhb b he mddle term bhb s n H because H s assumed normal So the rght hand sde s a member of Hb Snce rght multlcaton of cosets by h leaves all cosets unchanged, Q h s the dentty ermutaton Its dmenson s the number of cosets, whch s G H So ts trace, whch s the character ( h ), s G H E As n D, but now determne the character () c for c H Q Q We show that the ermutaton corresondng to rght multlcaton of cosets by c H moves every coset to a dfferent one hat s we show that for any coset Hb, Hbc Hb o do ths, assume the contrary and note that f any hbc Hbc and any hb Hb were equal, then hbc hb bch hb c b h hb he last quantty s n H because H s assumed normal So ths s a contradcton, and hence Hbc Hb Snce rght multlcaton of cosets by c moves every coset to a dstnct one, Q c s a ermutaton wth no s on ts dagonal So ts trace, whch s the character ( c ), s 0 Q Q Queston 4 Coherence and network dentfcaton

5 Say S () t and S () t are ndeendent nose sources wth ower sectra P ( ) S and P ( ) S, whch are connected to two observable oututs R () t and R () t by the followng network,where L are lnear flters wth transfer functons L ( ) A nd the ower sectra P ( ) R and P ( ) R of the two oututs Let s ( ) be ourer transforms of the nut sgnals over some tme nterval, and let r k ( ) be ourer transforms of the outut sgnals over ths nterval Because of the defnton of the ower sectrum, lm s s = PS Because the two nuts are ndeendent, lm s s = 0 for Because of the network, rk( ) L k( ) s ( ) herefore, æ ö æ ö PR = lm ( ) ( ) lm ( ) ( ) ( ) ( ) k rk w r k w = Lk w s w Lk w s w ç è ç øè ç ø = lm L s L s = lm L L s s k k k k,, = lm L k L k s s( w) = L k L k lm s s = L P k S B nd the cross-sectrum of R () t and R () t

6 æ ö æ ö PRR = lm r r = lm L s L s ç è ç øèç ø = lm L s L s = lm L L s s,, = lm L L s s = L L lm s s = L L P = L L P + L L P S S S C nd the coherence of R () t and R () t C RR PRR L L PS + L L PS = = P P R R ( L PS + L PS )( L PS + L PS ) D When s the magntude of the coherence equal to? he magntude-squared of the numerator of the coherence (art D), obtaned by multlyng the numerator by ts comlex conugate, s L L P + L L P L L P + L L P ( S S )( S S ) = L L P + L L P + S S S S ( + ) P P L L L L L L L L If the coherence s, ths must equal the magntude-squared of the denomnator he dfference between the above quantty and the magntude-squared of the denomnator s D= P P L L L L + L L L L - S S S S ( ) ( + ) P P L L L L ( = P P L L L L + L L L L - S S L L ( w) L L - L L L L ) S S S S ( )( ) = P P L L -L L - L L + L L =-P P L L -L L he coherence s when the above quantty s zero E Provde an nterretaton for the answer n D or the cross-sectrum to be zero, at least one of the three terms n the above exresson must be zero hat s, ether there s only one nut, or, L L - L L = 0 In the latter case,

7 L L = Say ths common quotent s q ( ) It follows that L L rk( ) L k( ) s( ) L k( ) s( ) L k( ) q( ) s( ), L ( )( q( )) s ( ) k e, that the system s ndstngushable from one n whch R () t and R () t can be vewed as beng derved from a sngle common sgnal Queston 5 Couled neurons (or oulatons): lnear systems vew Consder the network below, n whch one neural oulaton (the + at the to) s self-exctng wth lnear dynamcs secfed by the mulse resonse A() t, a second neural oulaton (the + at the bottom) s self-exctng wth lnear dynamcs secfed by Dt (), and the two neurons are couled to each other va the lnear dynamcs secfed by the mulse resonses Ct () and B() t Determne the transfer functon between the nut st () and the outut x() t, n terms of the transfer functons A ( ), B ( ), C ( ), and D ( ) Soluton: Lookng at the outut of A: a( ) A ( ) x ( ) Lookng at the outut of C: c( ) C ( ) x ( ) Lookng at the outut of D: d ( ) D ( ) y ( ) D ( ) c ( ) d ( ) D ( ) C ( ) x ( ) d ( ) whch mles

8 C ( ) D ( ) d ( ) x ( ) D ( ) Lookng at the outut of B: b ( ) B ( ) y ( ) B ( ) c ( ) d ( ) C ( ) D ( ) B ( ) C ( ) x( ) x( ) D ( ) C ( ) D ( ) B ( ) C ( ) x ( ) D ( ) B ( ) C ( ) x ( ) D ( ) Lookng at the three sgnals that sum to form x() t : x( ) b ( ) s( ) A ( ) x( ) B ( ) C ( ), x( ) s( ) A ( ) x( ) D ( ) whch mles that B ( ) C ( ) x( ) A ( ) s( ) D ( ) So the transfer functon that relates x( t) to st ( ) s: x ( ) s ( ) B ( ) C ( ) A ( ) D ( ) Queston 6 Couled neurons (or neural oulatons), dynamcal systems vew Let x() t and yt () reresent the fluctuatons n actvty of two neural oulatons; for convenence, we set the mean level of both to zero Consder the followng dynamcs: dx 3 ax by Ax, where a 0, d 0 (the oulatons are self-exctng), b 0, c 0 (so the dy cx dy oulatons nhbt each other), and A 0 (so there s no runaway actvty n ether drecton of x() t ) A What are the ossble knds of behavor near ( xy, ) (0,0), and for whch arameter values do they occur? Soluton:

9 a (0,0) s a fxed ont, and the Jacoban of the lnearzed system s J c roots of the ts equaton, trace( J) det( J) 0, e, ( a d) ( ad bc) 0 hese roots are gven by ( ad) ( ad) 4( ad bc) b Is egenvalues are d ( ad) ( ad) 4bc Snce b and c are both ostve, the exresson under the radcal must be real So both egenvalues of J are real Snce they sum to a d, a ostve number, they must ether be both ostve, or one ostve and one negatve he transton occurs when the smaller root s zero, namely, ( ) ( ) 4 ad ad bc 0 ( ad) ( ad) 4bc (hs could also have been seen drectly, snce a zero egenvalue mles ad bc that det( J) ad bc 0 ) So one knd of behavor s that (0,0) s an unstable node; another s that t s a saddle ont, wth the frst occurrng f ad bc -- where the self-exctaton terms domnate the mutual nhbton terms dx B Sketch the nullclnes (the loc n the ( x, y) lane n whch 0 and dy 0 ) for the regmes dentfed n A dx he nullclne for 0 s 3 a A 3 0 ax by Ax, e, y x x b b hs s a cubc that ascends near the orgn wth (ostve) sloe a, and then eventually descends for large 0 b x he nullclne for dy 0 s 0 cx dy, e, c y x, a lne through the orgn wth (ostve) sloe c If the d d determnant det J ad bc s ostve, the sloe of the cubc s larger: ad bc 0 whch mles a c hs s shown n the left anel below Conversely, f det J 0, the sloe of the lne s larger, b d as a c hs s n the rght anel below b d fgure;set(gcf,'poston',[ ]); xv=[-5::5]; % sublot(,,); a=7;b=;c=0;d=5;a=05; nullx=(/b)[axv-axv^3]; nully=(/d)cxv; lot(xv,nullx,'k');hold on; lot(xv,nully,'r');hold on; ttle('determnant > 0');

10 legend('x nullclne','y nullclne'); set(gca,'xlm',[-5 5]); set(gca,'ylm',[-0 0]); xlabel(srntf('a=%4f b=%4f c=%4f d=%4f A=%4f',a,b,c,d,A')); % sublot(,,); a=7;b=4;c=;d=5;a=05; nullx=(/b)[axv-axv^3]; nully=(/d)cxv; lot(xv,nullx,'k');hold on; lot(xv,nully,'r');hold on; ttle('determnant < 0'); legend('x nullclne','y nullclne'); set(gca,'xlm',[-5 5]); set(gca,'ylm',[-0 0]); xlabel(srntf('a=%4f b=%4f c=%4f d=%4f A=%4f',a,b,c,d,A')); C Are there any other fxed onts for the regmes dentfed n A? Under what crcumstances?

11 dx dy he full set of fxed onts are the solutons of 0, e, solutons of the two equatons 3 0 ax by Ax rom the second equaton, y cx / d Substtutng nto the frst equaton: 0 cx dy bcx ax Ax, whch mles 0 adx bcx dax, e, xad ( bcdax) 0 hs has roots at d ad bc x 0 (the fxed ont at the orgn that we already knew about) and at x where x, da rovded thatthe exresson under the radcal s ostve hat s, f the determnant det J ad bc s ostve, there are two fxed onts other than the orgn, at ( x, cx / d) D Lnearze the system near those fxed onts We need to aroxmate the rght hand sdes of dx 3 ax by Ax dy cx dy near ( x, y) ( x, y) ( x, cx / d) (he analyss at ( x, y) ( x, cx / d) or dx, we do ths by aylor seres exanson: ax by Ax ( x x) ( ax by Ax ) ( y y ) ( ) (, ) (, ) ax by Ax x x y x y y ( x, y) ( x, y) ( xx)( a3 Ax ) ( y y )( ) ( xy, ) ( x, ) b ( xy, ) ( x, y) y ( xx)( a3 Ax) b( y y) ad bc ( a3 )( xx) b( y y) d or dy, ths s a lnear functon of x and y, so the aylor exanson s trval and exact: dy cx dy c( x x) d( y y ), snce cx dy 0 So, near ( x, y ) ( x, cx / d), the system can be aroxmated by dx ad bc ( a3 )( xx) b( y y) d dy cx ( x) d( yy) E What knd of behavor does the system have near those fxed onts?

12 We need to fnd the egenvalues of the matrx ad bc a 3 b J d, e, the roots of trace( J ) det( J ) 0 c d ad bc It has trace trace( J ) ad 3, and determnant d ad bc det( J ) a3 d bcad 3( ad bc) bcdet J d trace( J ) trace( J) 4det( J) Snce det J 0, det( J ) 0, and the exresson under the radcal s necessarly ostve, and larger than trace( J ) herefore the roots are both real, and one s ostve and one s negatve herefore the fxed ont ( x, y ) s a saddle ont he same argument holds for the other fxed ont, ( x, y )