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Sections 4.4-4.8 of a linear algebra text or lecture notes (apparently Matthews, going by the folder name). It proves that a matrix is non-derogatory exactly when its characteristic and minimal polynomials agree, computes powers of Jordan blocks by the binomial theorem, and shows A^m tends to 0 when all eigenvalues have modulus below 1. It defines e^A, proves its properties, and solves X'=AX with a worked 3x3 example.

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We now present some interesting applications of the Jordan canonical form. 4.4 Non{derogatory matrices and transformations If chA=mA, we say that the matrix Aisnon-derogatory . THEOREM 4.5 Suppose that chTsplits completely in F[x]. Then chT=mT,9 a basisforVsuch that [T] =Jb1(c1):::Jbt(ct); wherec1;:::;ctare distinct elements of F. PROOF. ( chT=tY i=1chJbi(ci)=tY i=1(xci)bi; mT= lcm ((xc1)b1;:::; (xct)bt) = (xc1)b1:::(xct)bt= chT: )Suppose that ch T=mT= (xc1)a1(xct)at. We deduce that the dot diagram for each pi= (xci) consists of a single column of bidots, where pbi ijjmT; that is, dimFNh;pi= 1 for h= 1;2;:::;bi: Then, for each i= 1;2;:::;t we have the following sequence of positive integers: 1(pi(T))<(p2 i(T))<<(pbi i(T)) =ai: Butai=bihere, as we are assuming that ch T=mT. In particular, it follows that (ph i(T)) =hforh= 1;2;:::;biandh= 1 gives (pi(T)) = 1 = i: So the bottom row of the i-th dot diagram has only one element; it looks like this: bi8 >< >: ... 78 and we get the secondary decomposition Kerpbi i(T) =CT;vi1: Further, if=11[[t1, wherei1is the elementary Jordan basis forCT;vi1, then [T] =tM i=1 iM j=1Jeij(ci)) =tM i=1Jbi(ci); as required. 4.5 Calculating Am, whereA2Mnn(C). THEOREM 4.6 Letc2F. (a) Jm n(c) =2 6666666666664cm0 0m 1 cm1cm 0m 2 cm2m 1 cm1 0 ..................m m 0 0m m 0 .................. 0m m m 1 cm1cm3 7777777777775 if1mn1; (b) Jm n(c) =2 666664cm0 0 0m 1 cm1cm 0 0m 2 cm2m 1 cm1 0 0 ............m n1 cmn+1m n2 cmn+2m 1 cm1cm3 777775 79 ifn1m, wherem k is the binomial coecient m k =m! k!(mk)!=m(m1)(mn+ 1) k!: PROOF.Jn(c) =cIn+N, whereNhas the special property that Nkhas 1 on thek{th sub{diagonal and 0 elsewhere, for 0 kn1. Then because cInandNcommute, we can use the binomial theorem: Jm n(c) = (cIn+N)m =mX k=0m k (cIn)mkNk =mX k=0m k cmkNk: (a). Let 1mn1. Then in the above summation, the variable kmust satisfy 0kn1. HenceJm n(c) is annnmatrix havingm k cmkon thek{th sub{diagonal, 0 kmand 0 elsewhere. (b). Letn1m. Then Jm n(c) =mX k=0m k cmkNk=n1X k=0m k cmkNk; asNk= 0 ifnk. HenceJm n(c) is annnmatrix havingm k cmkon the k{th sub{diagonal, 0 kn1 and 0 elsewhere. COROLLARY 4.1 LetF=C. Then lim m!1Jm n(c) = 0 ifjcj<1: PROOF. Suppose that jcj<1. Letn1m. Then Jm n(c) =n1X k=0m k cmkNk: But for xed k;0kn1; cmk!0 asm!1 . For m k =m(m1)(mk+ 1) k! 80 is a polynomial in mof degreekand jmjcmkj=jmje(mk) logcj=mje(mk) logjcj!0 asm!1; as logc= logjcj+iarg c and logjcj<0. The last corollary gives a more general result: COROLLARY 4.2 LetA2Mnn(C)and suppose that all the eigenvalues of Aare less than 1in absolute value. Then lim m!1Am= 0: PROOF. Suppose chA= (xc1)a1(xct)at, wherec1;:::;ctare the distinct eigenvalues of Aandjc1j<1;:::;jctj<1. Then ifJis the Jordan canonical form of A, there exists a non{singular matrixP2Mnn(C), such that P1AP=J=tM i=1 iM j=1Jeij(ci): Hence P1AmP= (P1AP)m=Jm=tM i=1 iM j=1Jm eij(ci): HenceP1AmP!0 asm!1 , becauseJm eij(ci)!0. 4.6 Calculating eA, whereA2Mnn(C). We rst show that the matrix limit lim M!1 In+A+1 2!A2++1 M!AM exists. We denote this limit by eAand write eA=In+A+1 2!A2++1 m!Am+=1X m=01 m!Am: To justify this denition, we let Am= [a(m) ij]. We have to show that In+A+1 2!A2++1 M!AM ij=a(0) ij+1 1!a(1) ij++1 M!a(M) ij 81 tends to a limit as M!1 ; in other words, we have to show that the series 1X m=01 m!a(m) ij converges. To do this, suppose that jaijj;8i; j: Then it is an easy induction to prove that ja(m) ijjnm1mifm1: Then the above series converges by comparison with the series 1X m=01 m!nm1m: 4.7 Properties of the exponential of a complex matrix THEOREM 4.7 (i)e0=In; (ii)ediag (1;:::;n)=diag(e1;:::;en); (iii)eP1AP=P1eAP; (iv)eLt i=1Ai=Lt i=1eAi; (v) ifAis diagonable and has principal idempotent (spectral) decomposi- tion: A=c1E1++ctEt; then eA=ec1E1++ectEt; (vi) d dtetA=AetA; ifAis a constant matrix; (vii)eA=p(A), wherep2C[x]; 82 (viii)eAis non{singular and (eA)1=eA; (ix)eAeB=eA+BifAB=BA; (x) eJn(c)=2 666666664ec0 0 0 ec=1! ec0 0 ec=2!ec=1!ec 0 ............... ec=(n2)!...ec=1!ec0 ec=(n1)!ec=(n2)!ec=2!ec=1!ec3 777777775: (xi) etJn(c)=2 666666664etc0 0 0 tetc=1! etc0 0 t2etc=2! etc=1!etc 0 ............... tn2etc=(n2)!...tetc=1!etc0 tn1etc=(n1)!tn2etc=(n2)!t2etc=2!tetc=1!ec3 777777775: (xii) If P1AP=J=tM i=1 iM j=1Jeij(ci): then P1eAP=J=tM i=1 iM j=1eJeij(ci): PROOF. (i) e0=1X m=01 m!0k=In; 83 (ii) LetA= diag (1;:::;n). Then Am= diag (m 1;:::;m n) 1X m=01 m!Am= diag 1X m=0m 1 m!;:::;1X m=0m n m!! = diag (e1;:::;en): (iii) eP1AP=1X m=01 m!(P1AP)m =1X m=01 m!(P1AmP) =P1 1X m=01 m!Am! P =P1eAP: (iv) and (v) are left as exercises. (vi) Using the earlier notation, Am= [a(m) ij], we have etA=1X m=01 m!(tA)m =1X m=0tm m!Am ="1X m=0tma(m) ij m!# d dtetA=" d dt1X m=0tma(m) ij m!# ="1X m=1tm1a(m) ij (m1)!# ="1X m=0tma(m+1) ij (m)!# 84 =1X m=0tm m!Am+1 =AetA: (vii) Let deg mA=r. Then the matrices In; A;:::;Ar1are linearly inde- pendent over C, as if mA=xrar1xr1a0; then mA(A) = 0)Ar=a0In+a1A++ar1Ar1: Consequently for each m1, we can express Amas a linear combina- tion overCofIn; A; :::; Ar1: Am=a(m) 0In+a(m) 1A++a(m) r1Ar1 and hence MX m=01 m!Am=MX m=0a(m) 0 m!In+MX m=0a(m) 1 m!A++MX m=0a(m) r1 m!Ar1; or [t(M) ij] =s0MIn+s1MA++sr1MAr1; say. Now [t(M) ij]!eAasM!1 . Also the above matrix equation can be regarded as n2equations in s0M; s1M;:::;sr1;M: Also the linear independence of In; A;:::;Ar1implies that this sytem has a unique solution. Consequently we can express s0M; s1M;:::;sr1;M as linear combinations with coecients independent of Mof the se- quencest(M) ij. Hence, because each of the latter sequences converges, it follows that each of the sequences s0M; s1M;:::;sr1;Mconverges tos0; s1;:::;sr1, respectively. Consequently r1X k=0skMAk!r1X k=0skAk and eA=s0In+s1A+sr1Ar1; a polynomial in A. 85 (viii) { (ix) Suppose that AB=BA. ThenetBis a polynomial in Band henceAcommutes with etB. Similarly, AandBcommute with eA+B. Now let C(t) =et(A+B)etBetA; t2R: ThenC(0) =In. Also C0(t) = (A+B)et(A+B)etBetA +et(A+B)(B)etBetA +et(A+B)etB(A)etA = 0: HenceC(t) is a constant matrix and C(0) =C(1). That is In=eA+BeBeA; (12) for any matrices AandBwhich commute. The special case B=Athen gives In=e0eAeA=eAeA; thereby proving that eAis non{singular and ( eA)1=eA. Then multiplying both sides of equation (12) on the left by eAeBgives the equation eAeB=eA+B. Inx4.8 we give an application to the solution of a system of dierential equations. (x) LetJn(c) =cIn+N, whereN=Jn(0). Then eJn(c)=ecIn+N=ecIneN = (ecIn)1X m=01 m!Nm =n1X m=0ec m!Nm: (xi) Similar to above. 86 4.8 Systems of dierential equations THEOREM 4.8 IfX=X(t)satises the system of dierential equations _X=AX; fortt0, whereAis a constant matrix, then X=e(tt0)AX(t0): PROOF. Suppose _X=AXfortt0. Then d dt(etAX) = (AetA)X+etA_X = (AetA)X+etA(AX) = (AetA)X+ (AetA)X = (AetA+AetA)X = 0X= 0: Hence the vector etAXis constant for tt0. Thus etAX=et0AX(t0) and X=etAet0AX(t0) =e(tt0)AX(t0): EXAMPLE 4.3 Solve _X=AX, where A=2 40 42 15 3 14 23 5: Solution:9Pwith P1AP =J2(1)J1(1) =2 41 0 0 11 0 0 013 5 and P1(tA)P=2 4t0 0 tt0 0 0t3 5: 87 Thus P1etAP=etJ2(1)J1(1) =etJ2(1)etJ1(1) =2 4et0 0 tetet0 0 0et3 5=K(t), say. SoetA=PK(t)P1. Now X=etAX0=etP2 41 0 0 t1 0 0 0 13 52 4a b c3 5 =etP2 4a at+b c3 5; where for brevity we have set2 4a b c3 5=P1X0. 88