EE448/528 Vesion.0 John Stensby Chapte 9 Eigenvalues, Eigenvectos and Canonical Foms Unde Similaity Eigenvectos and Eigenvectos play a pominent ole in many applications of numeical linea algeba and matix theoy. In this chapte, we povide basic esults on this subject. Then, we use these esults to establish necessay and sufficient conditions fo the diagonalization of a squae matix unde a similaity tansfomation. Finally, we develop the Jodan canonical fom of a matix, a canonical fom the has many applications. et T : U U be a linea opeato on a vecto space U ove the scala field F. We ae inteested in non-zeo vectos X which map unde T into scala multiples of themselves. That is, we ae inteested in non-zeo vectos X U that satisfy T[X ] λx (9-) fo some scala λ F. Such a vecto X is said to be an eigenvecto coesponding to the eigenvalue λ. Example et I : U U be the identity opeato. Fo evey X U, I[X ] X. Hee, λ is an eigenvalue of I, and evey non-zeo vecto in U is an eigenvecto. Example et T : R 2 R 2 otate in a counte clock wise diection T( X) X evey vecto by π/2 adians. The scala field is R, the set of eal numbes. ote that no non-zeo vecto is a scala π/2 multiple of itself. Hence T : R 2 R 2 has no eigenvalues o eignevectos. This lack of eigenvalues and eignevectos will not occu if we use F C, the field of complex numbes. Hence, in applications whee eignevalues CH9.DC age 9-
EE448/528 Vesion.0 John Stensby play a ole, we use the complex numbe field. et X and λ be an eigenvecto and eigenvalue, espectively, so that T[X ] λx. et c F C be any non-zeo scala. Then we have T[cX ] λ(cx ) (9-2) so that cx is an eignevecto. Hence, eigenvectos ae defined up to an abitay, non-zeo, scala. Two o moe linealy independent eigenvectos can be associated with a given eigenvalue. In fact, fo a given eigenvalue λ, the set S λ {X U : T(X ) λx } (9-3) is a subspace known as the eigenspace associated with λ (note that 0 is in the eigenspace, but 0 is not an eigenvecto). Finally, the dimension of eigenspace S λ is known as the geometic multiplicity of λ. In what follows, we use γ to denote the geometic multiplicity of an eigenvalue. Fo a given basis, the tansfomation T : U U can be epesented by an n n matix A. In tems of this basis, a epesentation fo the eigenvectos can be given. Also, the eigenvalues and eigenvectos satisfy (A - λi)x 0. (9-4) Hence, the eigenspace associated with eigenvalue λ is just the kenel of (A - λi). While the matix epesenting T is basis dependent, the eigenvalues and eigenvectos ae not. The eigenvalues of T : U U can be found by computing the eigenvalues of any matix that epesents T. et n n matix A epesent T : U U with espect to some fixed basis. Then the eigenvalues ae the oots of the n th -ode chaacteistic polynomial CH9.DC age 9-2
EE448/528 Vesion.0 John Stensby A - λi det(a - λi) 0. (9-5) (note the notation intoduced hee: A means the deteminant of matix A). The eigenvalues can be complex o eal-valued. They can occu as simple oots o as multiple oots of the chaacteistic polynomial. The numbe of times (i.e., the multiplicity) that λ appeas as a oot of det(a - λi) is called the algebaic multiplicity of λ. We use α k to denote the algebaic multiplicity of eigenvalue λ k. A basis fo the eigenspace associated with λ can be found by computing a basis fo the kenel of (A - λi). Example A F H G I 3 3 3 5 3 so that J λ I - A 6 6 4 K λ 3 3 3 λ + 5 3 6 6 λ 4 2 ( λ + 2) ( λ 4) 0. The distinct eigenvalues ae λ -2 and λ 2 4. Eigenvalue λ -2 has algebaic multiplicity α 2, and eigenvalue λ 2 4 has algebaic multiplicity α 2. ow we find the eigenvectos. Conside fist the eigenvalue λ -2. The matix Y [ A λi] Y λ 2 F H G 3 3 3 3 3 3 6 6 6 I K J has a nullity of two, and X [ 0] T and X 2 [- 0 ] T ae two linealy independent eigenvectos that span the two dimensional eigenspace associated with λ -2. Hence λ -2 has geometic and algebaic multiplicities of γ α 2. ow, conside λ 2 4. The matix Y [ A λi] Y λ2 4 F HG I KJ 3 3 3 3 9 3 6 6 0 CH9.DC age 9-3
EE448/528 Vesion.0 John Stensby has a nullity of, and X 2 [ 2] T spans the one-dimensional eigenspace associated with λ 2 4. Eigenvecto Indexing Fom time to time, subscipts and supescipts need to placed on eigenvectos (and the genealized eigenvectos that ae intoduced below). In the liteatue, thee is not one indexing scheme that is pedominant (thee ae faults with all eigenvecto indexing schemes). otice the indexing scheme that was intoduced by the pevious example. n some eigenvectos, we placed two subscipts; we wote X jk. The fist subscipt (the "j" subscipt) associates the eigenvecto with one of the numeically distinct eigenvalues (each of which can have an algebaic multiplicity geate than one); we have j d, whee d is the numbe of distinct eigenvalues. The second subscipt (the "k" subscipt) odes the eigenvecto in the set of independent eigenvectos associated with the "j" eigenvalue; we have k γ j, whee γ j is the geometic multiplicity of the "j" eigenvalue. Sometimes, we place only one subscipt on an eigenvecto. This one subscipt may associate the eigenvecto with a distinct eigenvalue, o it may ode the eigenvecto in a set of independent eigenvectos (o it may do both). When one subscipt appeas, its meaning can be infeed fom context (o its meaning will be stated explicitly). Finally, on an eigenvecto, subscipts ae used only when necessay; we will dop all subscipts when they ae not needed to claify notation. Eigenvalues of Simila atices Recall that n n matices A and B ae said to be simila if thee exists a nonsingula n n matix such that A - B. The matix epesenting a linea tansfomation depends on the undelying basis; howeve, all matices that epesent a linea tansfom ae simila to one anothe. Futhemoe, they have the same eigenvalues and eigenvectos. Theoem 9- Simila matices have the same eigenvalues and eigenvectos. oof: This follows diectly fom the basic definitions since eigenvalues and eigenvectos ae associated with an undelying linea tansfomation and not with any paticula matix o vecto CH9.DC age 9-4
EE448/528 Vesion.0 John Stensby epesentation. et β, β 2,..., β n and β, β 2,..., β n denote the old and new bases, espectively, fo the vecto space; in tems of a non-singula tansfomation matix, the old and new bases ae elated as shown by (3-27). X and A denote old epesentations fo the eigenvecto and matix, espectively. X - X and A - A denotes new epesentations fo the eigenvecto and matix, espectively (see (3-29) and (3-4)). With espect to the new basis, the old eigen poblem AX λx becomes (A - )X λ(x ). (9-6) Afte multiplication on the left by -, (9-6) becomes the new eigen poblem A X λx. (9-7) So, while a similaity tansfomation changes the matix and vecto epesentations, it does not change the undelying linea tansfomation o its eigenvalues/eigenvectos. Theoem 9-2 Fo eigenvalue λ, the geometic multiplicity γ does not exceed the algebaic multiplicity α. oof: The geometic multiplicity γ of eigenvalue λ is defined independently of any matix epesenting linea tansfomation T : U U. The chaacteistic equation, eigenvalues and eigenvectos ae the same fo all matices that epesent T. Hence, to epesent tansfomation T, we can choose the matix that makes obvious the poof of this theoem. et γ be the dimension of eigenspace S λ (γ is the geometic multiplicity of λ). et eigenvectos X,..., X γ be a basis fo eigenspace S λ (eigenvecto subscipts ae used hee as an index into the set of basis vectos). This linealy independent set of eigenvectos can be extended to a basis CH9.DC age 9-5
EE448/528 Vesion.0 John Stensby X, X2,, Xγ, Xγ +, Xγ + 2,, Xn 442 443 444 4443 (9-8) eigenvectos that span eigenspace S any othe independent vectos λ of n-dimensional U. The vectos X γ+,..., X n can be abitay as long as they ae independent of each othe and independent of the fist γ eigenvectos. ow, T(X i) λx i fo i γ. With espect to (9-8), the matix A epesenting T has the fom A λ γ cols λ λ A n-γ cols A γ, n γ n γ, n γ γ ows n-γ ows (9-9) Sub-matix A γ,n-γ is γ (n-γ) and A n-γ,n-γ is (n-γ) (n-γ). These sub-matices ae non-zeo, in geneal; the values they contain ae of no concen to us. Fom inspection of (9-9), it is evident that the algebaic multiplicity of λ is at least equal to γ. Hence, fo any eigenvalue, the algebaic multiplicity geometic multiplicity (α γ). Theoem 9-3 et λ, λ 2,..., λ s be any s distinct eigenvalues, and let X, X 2,..., X s (subscipts ae used hee to associate an eigenvecto with a distinct eigenvalue) be the associated eigenvectos. These s eigenvectos ae linealy independent. oof (by contadiction) Suppose the set of s vectos is dependent. Re-ode the eigenvectos so that the fist k ae linealy independent and the emaining s-k vectos ae dependent on the fist k vectos. Then, we can wite the unique epesentation CH9.DC age 9-6
EE448/528 Vesion.0 John Stensby k Xs cixi, c i F, (9-0) i Since X s 0, thee ae non-zeo c i in (9-0). Apply the linea tansfomation T to (9-0) and obtain k λsxs ciλixi i. (9-) Thee ae two possibilities. Fist, if λ s 0, then λ i 0, i k, since λ,..., λ s ae distinct. λ s 0 implies that X,..., X k ae dependent, a contadiction. The second possibility is that λ s 0, so that we can wite Xs k F λ i i Xi i H G I c s K J λ. (9-2) Since thee ae non-zeo c i, and λ i /λ s due to distinct eigenvalues, (9-2) is diffeent than (9-0), a contadiction (since epesentation (9-0) is unique). Hence, fo eithe possibility, we have a contadiction, and the s eigenvectos X, X 2,..., X s ae independent. ote that the convese of this theoem is not tue (independent eigenvectos ae not always associated with distinct eigenvalues). Theoem 9-3 tells us a lot about matices with distinct eigenvalues (distinct eigenvalues ae a common occuence in pactical applications). atices with distinct eigenvalues have linea independent eigenvectos. When this occus, it is possible to use the n independent eignevectos to fom a basis of n-dimensional U, a useful thing to do when poving theoems. et n n matix A epesent T : U U with espect to some fixed basis. Suppose T has n linealy independent eigenvectos X, X 2,..., X n, (subscipts ae used to index the eigenvectos in CH9.DC age 9-7
EE448/528 Vesion.0 John Stensby this set of n independent eigenvectos), and we use them as a basis of n-dimensional space U. We want to find the matix D that epesents linea tansfomation T with espect to this basis of eignevectos. Use these independent eigenvectos to define the n n tansfomation matix X X X 2 n. (9-3) Then, with espect to the eigenvecto basis, the matix D that epesents T is D - A. But this implies that A D, a esult that can be witten as A X X X n X X nx 2 λ λ2 2 λ n X X X 2 n λ λ2 λn. (9-4) But, Equation (9-4) leads to the obsevation that λ D λ2 X X Xn A X X Xn 2 2 λn (9-5) Hence, when a basis of eigenvectos is used, the n n matix epesenting T: U U is diagonal with the eigenvalues appeaing on the diagonal. If n n matix A has distinct eigenvalues, then thee is a basis of eigenvectos that can be used as columns of n n matix. And, with a similaity tansfomation, matix can be used to diagonalize matix A. oe geneally, if each eigenvalue of A has equal geometic and algebaic CH9.DC age 9-8
EE448/528 Vesion.0 John Stensby multiplicities, then thee ae n linealy independent eigenvectos, and A can be diagonalized as descibed above. The convese is tue as well. That is, if an n n nonsingula matix exists such that - A is diagonal, then we can conclude. The eigenvalues of A appea on the diagonal of - A, and 2. The columns of ae n linealy independent eigenvectos of matix A. We have agued the following theoem. Theoem 9-4 An n n matix A is simila to a diagonal matix D if and only if thee ae n linealy independent eigenvectos of A. Futhemoe, the eigenvalues of A must appea on the diagonal of D. Example A 2 det(a - λi) λ 2 + so that λ ± j ae the eigenvalues. λ +j has the eigenvecto X [ -j] T. λ 2 -j has the eigenvecto X 2 [ +j] T. The eigenvalues ae distinct, so X and X 2 ae independent and A j + j - Example A eigenvectos ae 0 0 0 0 0 2 j 0 0 j has eigenvalues λ (α 2), and λ 2 2 (α 2 ). The λ X [ 0 0] T and X 2 [0 0] T λ 2 2 X 2 [- 0 ] T ote that λ has equal algebaic and geometic multiplicities of two. Hence, X, X2 and X2 compise a basis of eigenvectos, and we have CH9.DC age 9-9
EE448/528 Vesion.0 John Stensby 0 0 0 0 0 Example A - A 2 0 3 0 0 2 0 0 0 0 0 0 2 has eigenvalues λ (α 2), and λ 2 2 (α 2 ). Since nullity(a - λ I), we know that λ l has a geometic multiplicity of γ but an algebaic multiplicity of α 2. Hence, thee is no basis of eigenvectos, and matix A cannot be diagonalized unde similaity. When thee is not a basis of eigenvectos, n n matix A cannot be diagonalized. Howeve, we show that a nonsingula n n matix exists such that - A is almost diagonal; ou - A has eigenvalues on its diagonal and s immediately above some of the diagonal eigenvalues. This new almost diagonal matix is called the Jodan Canonical Fom fo A, and it has many applications in engineeing and the applied sciences. Fist, we must intoduce the subject of genealized eigenvectos. Genealized Eigenvectos et A be an n n matix. Fo an eigenvalue λ, vecto X is said to be a genealized eigenvecto of ank k > 0 if k ( A λi) X 0. (9-6) k ( A λi) X 0 An odinay eigenvecto X is a genealized eigenvecto of ank k since (A - λi)x 0 and (A - λi) 0 X X 0. We develop a chain of genealized eigenvectos. Fo a given eigenvalue λ, let X be a genealized eigenvecto of ank k. Define the chain of k genealized eigenvectos as CH9.DC age 9-0
EE448/528 Vesion.0 John Stensby k X X k X k (A - λi) X (A - λi) X k 2 X 2 k (A - λi) X (A - λi) X. (9-7) k- 2 X (A - λi) X (A - λi) X A supescipt on a vecto is not a powe; it is used to indicate ank, and it is used as an index! n X k, the k is used as a ank indicato and index; k is not a powe (aising a vecto to a powe is an opeation that has not been defined!). ow, settle down, get ove it! n a vecto, the only time we will use a supescipt is when we ae woking with vectos in a chain of genealized eigenvectos (we have aleady descibed how we want to use the subscipt position(s)). n genealized eigenvectos, supescipts ae standad in the liteatue. Fo each i, i k, X i is a genealized eigenvecto of ank i since (A - λi) i X i (A - λi) i (A - λi) k-i X (A - λi) k X 0, (9-8) (A - λi) i- X i (A - λi) i- (A - λi) k-i X (A - λi) k- X 0. (9-9) ote that X is an "odinay" eigenvecto since (A - λi)x (A - λi)(a - λi) k- X (A - λi) k X 0. (9-20) As mentioned above, we call X, X 2,..., X k a chain of genealized eigenvectos. ow, we examine some popeties that chains have. CH9.DC age 9-
EE448/528 Vesion.0 John Stensby Theoem 9-5 A chain X, X 2,..., X k of genealized eigenvectos is linealy independent. oof (by contadiction) Fo the moment, assume that the vectos in the chain ae dependent. Then thee exists constants c, c 2,..., c k, not all zeo, such that c X + c 2 X 2 +... + c k X k 0. (9-2) Fist, note that fo i, 2,..., k- we can wite (A - λi) k- X i (A - λi) k- (A - λi) k-i X (A - λi) 2k-(i+) X 0, (9-22) a esult we will use vey soon. ow, apply (A - λi) k- to both sides of (9-2) to obtain (A - λi) k- { c X + c 2 X 2 +... + c k X k } 0. (9-23) Use (9-22) in (9-23) to obtain c k (A - λi) k- X k 0. (9-24) But, we know that (A - λi) k- X k 0. Hence, we must have c k 0 so that (9-2) becomes c X + c 2 X 2 +... + c k- X k- 0. (9-25) n this equation, epeat the pocedue that stats with (9-22). That is, multiply (9-25) by (A - λi) k-2, and epeat the above agument (that poduced c k 0) to each the conclusion that c k- 0. CH9.DC age 9-2
EE448/528 Vesion.0 John Stensby bviously, this same agument can be epeated a sufficient numbe of time to conclude that c i 0, i k. This contadiction leads to the conclusion that the chain X, X 2,..., X k is compised of linea independent genealized eigenvectos. Theoem 9-6 et λ λ 2 be two eigenvalues of n n matix A. Suppose X is a genealized eigenvecto of ank k associated with λ and Y is a genealized eigenvecto of ank m associated with λ 2. Define the two chains X k X, and X i (A - λ I)X i+ (A - λ I) k-i X fo i k-, k-2,..., (9-26) Y m Y, and Y j (A - λ 2 I)Y j+ (A - λ 2 I) m-j Y fo j m-, m-2,..., (9-27) The set of k+m vectos descibed by (9-26) and (9-27) ae linealy independent. Equivalently, any genealized eigenvecto fom one chain is independent of the vectos in the othe chain. oof (by contadiction) Suppose thee is an i, i k, fo which X i is linealy dependent on the chain Y, Y 2,..., Y m. Then, thee exists constants c,..., c m, not all zeo, such that m i j X c j Y j (9-28) ultiply (9-28) by (A - λ I) i, and use the fact that (A - λ I) i X i (A - λ I) i (A - λ I) k-i X 0 (9-29) to obtain CH9.DC age 9-3
EE448/528 Vesion.0 John Stensby m i (A - λ I) c j Y j 0 (9-30) j ow, multiply (9-30) by (A - λ 2 I) m-, and use the facts i) (A - λ 2 I) m- (A - λ I) i (A - λ I) i (A - λ 2 I) m- ii) (A - λ 2 I) m- Y j 0 fo j m-, m-2,..., to obtain (A - λ I) i (A - λ 2 I) m- c m Y m c m (A - λ I) i Y 0 (9-3) ow, Y is an "odinay" eigenvecto: AY λ 2 Y, so (9-3) becomes c m (λ 2 - λ )Y 0. (9-32) Since λ 2 λ we must have c m 0 so that (9-30) becomes m i (A - λ I) c j Y j 0. (9-33) j ow epeat the agument that stated with (9-30) and poduced c m 0. That is, multiply (9-33) by (A - λ 2 I) m-2, follow the agument, and conclude that c m- 0. Continue this pocess to the conclusion that c i 0 fo i m, m-, m-2,...,. This contadiction (the c i 's ae not all zeo) leads to the conclusion that X i is independent of Y, Y 2,..., Y m. Hence, the two chains Y, Y 2,..., Y m and X, X 2,..., X k contain m+k linealy independent vectos. Theoem 9-7 et Y and X be genealized eigenvectos of ank m and k, espectively, associated with the same eigenvalue λ. Define the two chains CH9.DC age 9-4
EE448/528 Vesion.0 John Stensby X k X, and X i (A - λi)x i+ (A - λi) k-i X fo i k-, k-2,..., (9-34) Y m Y, and Y j (A - λi)y j+ (A - λi) m-j Y fo j m-, m-2,..., (9-35) If the "odinay" eigenvectos Y and X ae independent, then so ae the two chains (i.e., (9-34) and (9-35) descibe m+k independent vectos). oof Simila to the poof of Theoem 9-6. Theoems 9-5, 9-6 and 9-7 povide the basis of ou genealized eigenvecto theoy. ote that we have shown an impotant esult. Associated with eigenvalue λ ae γ distinct chains of genealized eigenvectos (γ is the geometic multiplicity of λ). Each chain is "anchoed" by an "odinay" eigenvecto (of ank one). In these γ chains, the total numbe of genealized eigenvectos is α, the algebaic multiplicity of λ. And, these α vectos ae linealy independent. ote that we have not discussed how many vectos ae in each chain. We have agued only that thee ae a total of α genealized eigenvectos divided into γ chains associated with λ. While n n matix A may, o may not, have n independent eigenvectos, it always has n independent genealized eigenvectos. Eigenvecto Indexing - Revisited It's time once moe to conside genealized eigenvecto indexing. A genealized eigenvecto can have two subscipts and one supescipt. The meaning of the two subscipts ae given above in the section on eigenvecto indexing (which is woth eading again). The supescipt is used as both a ank indicato and index into a chain. Fo example, conside the genealized eigenvecto l X jk. The "j" subscipt associates the genealized eigenvecto with eigenvalue λ j ( j d, whee d is the numbe of numeically distinct eigenvalues). The "k" CH9.DC age 9-5
EE448/528 Vesion.0 John Stensby subscipt associates the genealized eigenvecto with a paticula chain of independent genealized eigenvectos fo λ j ( k γ j, whee γ j is the geometic multiplicity of λ j ). As descibed above, supescipt l is a ank indicato, and it is an index into the k th chain of genealized eigenvectos associated with λ j. Finally, note that X jk is the k th "odinay" eigenvecto associated with λ j. isting of all Genealized Eigenvectos et λ, λ 2,..., λ d denote the numeically distinct eigenvalues of an n n matix A. Fo k d, eigenvalue λ k has an algebaic multiplicity of α k and a geometic multiplicity of γ k. Futhemoe, fo k d, eigenvalue λ k is associated with γ k sepaate chains of genealized eigenvectos containing a total (in all of the γ k chains) of α k independent genealized eigenvectos. Finally, taken all togethe, fo the d numeically distinct eigenvalues, a total of n genealized eigenvectos exist, consideing all of the vectos in all of the chains. we wite We can list these n genealized eigenvectos. Using the indexing scheme outline above, The α genealized eigenvectos fo λ ae divided into γ chains The α2 genealized eigenvectos fo λ2 ae divided into γ 2 chains R S T 2 X X X X X X 2 2 2 h h2 2 X X X R S T 2 h γ γ γ γ 2 X2 X2 X X X X 2 22 22 X X X h2 2 h22 22 2 h γ γ γ 2 2 2γ 2 2 2 R 2 h Xd Xd X The αd genealized eigenvectos fo λd ae d 2 h Xd2 Xd2 Xd S divided into γ d chains X X X 2 h T 2 2 d d2 2 dγ d dγ d dγ d dγ d. (9-36) CH9.DC age 9-6
EE448/528 Vesion.0 John Stensby Hee, h kj, k d, j γ k, denotes the numbe of genealized eigenvectos in the j th chain associated with the numeically distinct eigenvalue λ k. Intege h kj has to be computed as outlined in the example given below. As stated in the list given above, we have αk γ k hkj. (9-37) j Also, we denote the total numbe of chains as d ν γ k. (9-38) k Finally, fo an n n matix A, we have d d γ k n αk hkj. (9-39) k k j An n n matix A may, o may not, have n linealy independent eigenvectos. Howeve, it always has n linealy independent genealized eigenvectos. Example Reconside the pevious example whee A 2 0 3 0 0 2 Eigenvalue λ has an algebaic multiplicity of α 2 and a geometic multiplicity of γ ; X [ 0 0] T is an "odinay" eigenvecto fo λ. Eigenvalue λ 2 2 has geometic and algebaic multiplicities of ; X 2 [5 3 ] T is an "odinay" eigenvecto fo λ 2. We ae one CH9.DC age 9-7
EE448/528 Vesion.0 John Stensby eigenvecto shot; the matix A cannot be diagonalized by a similaity tansfomation. Howeve, we can find two genealized eigenvectos associated with λ. et's find a chain of length two associated with λ. These two genealized eigenvectos, when combined with X 2, will poduce a basis of genealized eigenvectos. Fist, find a non-zeo X such that 0 2 (A λι) X X 0 0 3 0 0 0 (A λι) 2 0 2 0 2 X 0 0 5 X 0 0 3 0 0 3 0 0 3X 0 0 0 0 0 0 0 Clealy, X [0 0] T is a genealized eigenvecto of ank 2, and we use this vecto to wite 2 X 0 0 X ( A λ) X 0 2 0 0 3 0 0 0 0 0 0 { X, 2 X} is a chain of length two associated with λ. The vectos X [ 0 0] T, X 2 [0 0] T, X 2 [5 3 ] T fom a basis of genealized eigenvectos. With espect to this basis, let's find the matix A that epesents the undelying tansfomation. Define the 3 3 non-singula matix X X X 2 2, and compute A - A. We compute A by consideing the equivalent equation A A X X 2 X2 A A X X X 2 2 so that CH9.DC age 9-8
EE448/528 Vesion.0 John Stensby AX X X X 2 0 0 0 2 0 + + AX 2 X X 2 0 X2 0 + + AX 0 X X X 2 2 6 0 0 2 2 2 + + As a esult, we see that A A 0 0 0 0 0 2. ote that A has two blocks on its diagonal; we wite A as J A, J, J 2 0 J 2 2 atix A is known as the Jodan Canonical Fom fo matix A. Jodan Canonical Fom This pocedue can be applied to tansfom any n n matix into its block-diagonal Jodan canonical fom. et λ, λ 2,..., λ d be the numeically distinct eigenvalues of n n matix A. Fo k d, let λ k have algebaic multiplicity α k and geometic multiplicity of γ k. As outlined above, eigenvalue λ k is associated with γ k chains containing a total of α k genealized eigenvectos, and CH9.DC age 9-9
EE448/528 Vesion.0 John Stensby each chain is "anchoed" by an "odinay" eigenvecto. As listed by (9-36), thee ae a total of n linealy-independent genealized eigenvectos split up into ν chains. We use these n genealized eigenvectos to define the n n tansfomation matix h h h [ X X X X X X 442 443 44244 3 2 γ 2 2 γ γ 444 24443 chain # fo λ chain #2 fo λ chain # γ fo λ h h h X X X X X X 444 2444 3 4 42 444 3 2 22 2γ 2 2 2 22 22 2γ 2 444 24443 2 γ2 chain # fo λ chain #2 fo λ chain # γ fo λ 2 2 2 2 (9-40) X h d h 444 2444 3 4 42 444 3 h d d2 dγ d Xd Xd2 Xd2 X dγ X dγ 444 d d 24443 chain # fo λd chain #2 fo λd chain # γ fo λ d d ]. By using the similaity tansfomation A - A, matix, given by (9-40), can be used to tansfom n n matix A into its Jodan Canonical Fom. This canonical fom is a block diagonal matix A A J J2 Jν (9-4) made fom ν blocks J k, k ν, one block fo each chain of genealized eigenvectos. ote that (9-4) is equivalent to A A, a matix equation that can be witten as 442 443 h dγ dγ d d 444 24443 st chain th ν chain A A[ X X h X X dγ d ] CH9.DC age 9-20
EE448/528 Vesion.0 John Stensby 442 443 h dγ dγ 444 d d 24443 st chain th ν chain [ X X h X X d γ d ] J J2 A Jν (9-42) et's examine the stuctue of a typical block. Conside 2 X X X h jk jk, jk,, jk, the k th chain associated with λ j, the j th distinct eigenvalue. The Jodan block fo this chain is J p, whee j p k + γ i i (9-43) Fom the basic definition of this chain, we have h X jk h jk h jk h jk h jk h jk X ( A λ ji) X AX λ jx + X (9-44) h jk 2 h jk h jk h jk h jk 2 X ( A λ ji) X AX λ jx + X X ( A λ ji) X 2 AX 2 λ jx 2 + X, whee we have omitted the common subscipts jk on all genealized eigenvectos. Fom A A (see (9-42)), we have the equiement 2 h X X X J A X X X jk 2 h jk jk jk p jk jk jk jk, (9-45) whee p is given by (9-43). Howeve, fom (9-44) and the equiement A X jk λ j X jk, it is easy to see that CH9.DC age 9-2
EE448/528 Vesion.0 John Stensby h jk columns J p λ j 0 0 λ j 0 0 λ j λ j 0 λ j h jk ows (9-46) That is, J p is an h jk h jk matix with λ j on its diagonal, "s" on its fist "supe diagonal", and zeos eveywhee else. Computational ocedue fo Jodan Fom Fo many low-dimensional poblems of pactical inteest, the Jodan fom can be computed "by hand" without too much effot. A computational pocedue fo computing the Jodan fom is outlined below.. Compute the eigenvalues and "odinay" eigenvectos of n n matix A; detemine the algebaic and geometic multiplicities of the eigenvalues. The distinct eigenvalues ae λ, λ 2,..., λ d ; fo k d, eigenvalue λ k has algebaic multiplicity α k and geometic multiplicity γ k. 2. In γ distinct chains, compute a total of α independent, genealized eigenvectos fo λ. To accomplish this, compute (A - λ I) i fo i, 2,... until the ank of (A - λ I) k is equal to the ank of (A - λ I) k+. Then, compute a ank k genealized eigenvecto and its k-long chain. If k α, go to step #3. thewise, look fo a second ank-k vecto and its chain. If a second ank k vecto does not exist, look fo one of ank k-, and so on, until we have γ distinct chains of α genealized eigenvectos. 3. Repeat step #2 fo the emaining eigenvalues λ 2,..., λ d. 4. Wite down the Jodan fom. Fo eigenvalue λ j, the k th chain is of length h jk (detemined in step #2), and thee is an h jk h jk Jodan block with λ j on its diagonal. CH9.DC age 9-22
EE448/528 Vesion.0 John Stensby In the Jodan fom, the odeing of the blocks is not citical. Howeve, it is common to keep sequential all blocks associated with the same eigenvalue. Example A 3 0 0 0 0 0 0 2 0 0 0 0 2 0 0 0 0 0 0 0 0 Compute the eigenvalues and algebaic multiplicities. ote that det(a - λi) λ(λ - 2) 5, and this implies that λ 2 with α 5 and λ 2 0 with α 2. Futhemoe, eigenvalue λ 2 has the two independent eigenvectos T X 0 0 0 0, T X2 0 0 0 0 so γ 2. Also, λ 2 0 has the single eigenvecto T X2 0 0 0 0, so γ 2. ow, compute (A - λ I) i, fo inceasing i until the ank no longe changes. ( A 2I) 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 has ank equal to 4. CH9.DC age 9-23
EE448/528 Vesion.0 John Stensby 2 ( A 2I) 3 ( A 2I) 4 ( A 2I) 0 0 2 2 0 0 0 0 2 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 2 0 0 0 0 2 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 4 4 0 0 0 0 4 4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 8 8 0 0 0 0 8 8 has ank equal to 2. has ank equal to. has ank equal to. The ank of (A - 2I) 3 is equal to the ank of (A - 2I) 4 ; hence, thee is a ank 3 genealized eigenvecto that is in K((A - 2I) 3 ) but not in K((A - 2I) 2 ). It is easily computed as X 3 [0 0 0 0 0] T since (A - 2I) 3 3 X 0 but (A - 2I) 2 3 X 0. ow, we compute the fist 3-long chain associated with λ. 2 2 0 X A I 2 3 0 2 X X 2 A 2I X 3 0 X 3 ( ), ( ),. 0 0 0 0 0 0 0 0 0 0 CH9.DC age 9-24
EE448/528 Vesion.0 John Stensby Since α 5, thee ae two moe genealized eigenvectos associated with λ ; inspection of (A - 2I) 2 and (A - 2I) eveals whee they ae. Thee is a genealized eigenvecto of ank 2 that is in K((A - 2I) 2 ) but not in K(A - 2I). This ank 2 vecto is X 2 2 [0 0 0 0 ] T ; note that (A - 2I) 2 2 X 2 0 but (A - 2I) 2 X2 0. Hence, ou second chain associated with λ 2 is 0 0 X2 A 2I X2 2 2 X2 2 0 ( ),. 2 0 0 0 0 0 We have 5 genealized eigenvectos associated with λ 2; thee ae no moe. With the eigenvecto X T 2 0 0 0 0, we have a basis of 6 genealized eigenvectos that we can use to wite the tansfomation matix X X 2 X 3 X2 X2 2 X2 2 0 0 0 0 2 0 0 0 0 0 0 2 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0. We wite down (no computation is necessay) the Jodan canonical fom A 2 0 0 0 0 0 2 0 0 0 0 0 2 0 0 0 0 0 0 2 0 0 0 0 0 2 0 0 0 0 0 0 0. CH9.DC age 9-25
EE448/528 Vesion.0 John Stensby ote that A contains the thee Jodan blocks 2 0 J 0 2 0 0 2 2 0, J 2 0 2, J3. It is easy to see that A satisfies A A (so that A - A). As a atab execise, ente and A as descibed above, and type inv()*a* at the command pompt. atab will etun the Jodan canonical fom A given above. Jodan Fom - Sensitivity Issues Computation of the Jodan fom is laboious and time consuming. Also, the Jodan fom in computationally unstable ; in some cases, a vey small petubation of A can put back all of the missing eigenvectos and emove the supediagonal of ones. Because of the possible stability poblems, many numeical analysis compute pogams do not include the Jodan fom (the Jodan fom is not in atab pope; it is in atab s symbolic algeba toolbox). The Jodan fom has seveal applications in state space contol theoy. Geneally speaking, contol enginees will not design a system having a stuctue that is extemely sensitive to small petubations. In Chapte 0, we use the Jodan fom to compute functions of matices. CH9.DC age 9-26