Jordan Canonical Form of a Nilpotent Matrix. Math 422

Transcription

1 Jordan Canonical Form of a Nilpotent Matrix Math Schur s Triangularization Theorem tells us that every matrix A is unitarily similar to an upper triangular matrix T However, the only thing certain at this point is that the the diagonal entries of T are the eigenvalues of A The off-diagonal entries of T seem unpredictable and out of control Recall that the Core-Nilpotent Decomposition of a singular matrix A of index k produces a block diagonal matrix C L similar to A in which C is non-singular, rank (C) =rank A k, and L is nilpotent of index k Isitpossible to simplify C and L via similarity transformations and obtain triangular matrices whose off-diagonal entries are predictable? The goal of this lecture is to do exactly this for nilpotent matrices Let L be an n n nilpotent matrix of index k Then L k =and L k = Let s compute the eigenvalues of L Suppose x = satisfies Lx = λx then =L k x = L k (Lx) =L k (λx) =λl k x = λl k (Lx) = λ L k x = = λ k x thus λ =is the only eigenvalue of L Now if L is diagonalizable, there is an invertible matrix P and a diagonal matrix D such that P LP = D Since the diagonal entries of D are the eigenvalues of L, and λ =is the only eigenvalue of L, wehave D = Solving P LP =for L gives L = Thus a diagonalizable nilpotent matrix is the zero matrix, or equivalently, a non-zero nilpotent matrix L is not diagonalizable And indeed, some off-diagonal entries in the simplified form of L will be non-zero Let L be a non-zero nilpotent matrix Since L is triangularizable, there exists an invertible P such that P LP = The simplification procedure given in this lecture produces a matrix similar to L whose non-zero entries lie exclusively on the superdiagonal of P LP An example of this procedure follows below Definition Let L be nilpotent of index k and define L = I For p =,,,k, let x C n such that y = L p x = The Jordan chain on y of length p is the set L p x,,lx,x ª Exercise Let L be nilpotent of index k If x C n satisfies L k x =, prove that L k x,, linearly independent Example Let L = then L = and L = Let x = a b then c Lx = a b = b +c c and L x = a b = c c c Lx,x} is Note that L x, Lx, x ª is linearly independent iff c = Thus if x =[] T, then y = L x =[9] T and the Jordan chain on y is 9, 8 9, Form the matrix P = L x Lx x = 9 8 9

2 then P LP = is the Jordan form of L = Since λ =is the only eigenvalue of an n n non-zero nilpotent L, the eigenvectors of L are exactly the non-zero vectors in N (L) Butdim N (L) <nsince L is not diagonalizable and possesses an incomplete set of linearly independent eigenvectors So the process by which one constructs the desired similarity transformation P LP involves appropriately extending a deficient basis for N (L) to a basis for C n This process has essentially two steps: Construct a somewhat special basis B for N (L) Extend B toabasisforc n by building Jordan chains on the elements of B Definition A nilpotent Jordan block is a matrix of the form A nilpotent Jordan matrix is a block diagonal matrix of the form J J J m, () where each J i is a nilpotent Jordan block When the context is clear we refer to a nilpotent Jordan matrix as a Jordan matrix Theorem 5 Every n n nilpotent matrix L of index k is similar to an n n Jordan matrix J in which the number of Jordan blocks is dim N (L) the size of the largest Jordan block is k k for j k, the number of j j Jordan blocks is rank L j rank L j + rank L j+ the ordering of the J i s is arbitrary Whereas the essentials of the proof appear in the algorithm below, we omit the details Definition If L is a nilpotent matrix, a Jordan form of L is a Jordan matrix J = P LP The Jordan structure of L is the number and size of the Jordan blocks in every Jordan form J of L Theorem 5 tells us that Jordan form is unique up to ordering of the blocks J i Indeed, given any prescribed ordering, there is a Jordan form whose Jordan blocks appear in that prescribed order

3 Definition 7 The Jordan Canonical Form (JCF) of a nilpotent matrix L is the Jordan form of L in which the Jordan blocks are distributed along the diagonal in order of decreasing size Example 8 Let us determine the Jordan structure and JCF of the nilpotent matrix 5 5 L = 5 the number of Jordan blocks is dim N (L) = Then L = has rank and L =therefore the index (L) =and the size of the largest Jordan block is Let then the number n i of i i Jordan blocks is r = rank L = r = rank L = r = rank L = r = rank L = r = rank L = n = r r + r = n = r r + r = n = r r + r = Obtain the JCF by distributing these blocks along the diagonal in order of decreasing size Then the JCF of L is This leaves us with the task of determining a non-singular matrix P such that P LP is the JCF of L As mentioned above, this process has essentially two steps A detailed algorithm follows our next definition and a key exercise Definition 9 Let E be any row-echelon form of a matrix A Let c,,c q index the columns of E containing leading s and let y i denote the c th i column of A The columns y,,y q are called the basic columns of A Exercise Let B = [b b p ] be an n p matrix with linearly independent columns Prove that multiplication by B preserves linear independence, ie, if {v,,v s } is linearly independent in C p,then {Bv,,Bv s } is linearly independent in C n

4 Nilpotent Reduction to JCF Given a nilpotent matrix L of index k, set i = k and let S k = {y,,y q } be the basic columns of L k Extend S k S i to a basis for R L i N (L) in the following way: (a) Let {b,,b p } be the basic columns of L i and let B =[b b p ] (b) Solve LBx =for x and obtain a basis {v,,v s } for N (LB) then {Bv,,Bv s } is a basis for R L i N (L) by Lemma below o (c) Form the matrix [y y q Bv Bv s ] its basic columns ny,,y q,bv β,,bv βj form a basis for R L i N (L) containing S k S i Let S i = nbv β,,bv βj o (d) Decrement i, let {y,,y q } be the basic columns of L i, and repeat step until i = Then S k S = {b,,b t } is a basis for N (L) For each j, if b j S i, find a particular solution x j of L i x = b j and build a Jordan chain ª L i x j,,lx j,x j Set p j = L i x j Lx j x j the desired similarity transformation is defined by the matrix P =[p p t ] Lemma In the notation of the algorithm above, {Bv,,Bv s } is a basis for R L i N (L) Proof Note that Bv j R (B) and LBv j =implies Bv j N (L) But R (B) =R L i implies that Bv j R (B) N (L) for all j The set {Bv,,Bv s } is linearly independent by Exercise To show that {Bv,,Bv s } spans R (B) N (L), lety R (B) N (L) Since y R (B), there is some u C n such that y = Bu since y N (L), =Ly = LBu Thus u N (LB) and we may express u in the basis {v,,v s } as u = c v + +c s v s Therefore y = Bu = B (c v + + c s v s )=c Bv + +c s Bv s and {Bv,,Bv s } spans Example Let s construct the matrix P that produces the Jordan form of L in Example 8 Since L has index, we L and find one basic column y =[,,,,, ] T Then S = {y } contains the basic column of R L set i = Extend S to a basis for R (L) N (L): (a) Row-reducing, we find that the basic columns of L are its first three columns Thus we let 5 B = 5

5 (b) Compute LB and solve LBx =to obtain a basis for N (LB): 5 5 LB = = 5 5 v =, v = Then n Bv =[ ] T,Bv =[ 5 7 ] T o is a basis for R (L) N (L) (c) Form the matrix [y Bv Bv ]= 5 7 columns and are its basic columns, hence {y,y = Bv } is a basis for R (L) N (L) containing S Let S = n[ 5 o 7 ] T and decrement i then i = Extend S S = {y,y } to a basis for R L N (L) =N (L): (a) Since L = I, we let B = I (b) Since LB = L, we findabasisforn (LB) =N (L): v =, v =, v = Then a basis for N (L) is {Bv,Bv,Bv } = {v,v,v } 5 5

6 (c) Form the matrix [y y v v v ]= columns,, and are its basic columns, hence {y,y,v } is a a basis for N (L) containing S S Let S = n[ o ] T and decrement i then i =and the process terminates having produced the basis S S S for N (L) Let S = {b }, S = {b }, and S = {b } For j =,,, build a Jordan chain on b j S i of length i + by finding a particular solution x j of L i x = b j When j =we see by inspection that L e = b Thus p = L e Le e = 5 When j =we solve Lx = b : A particular solution is x =[ ] T hence 5 7 p =[Lx x ]= 5 When j =we have L = I, and the unique solution of L x = b is x = b S Thus the Jordan chain on b consists only of b and we have p =[ ] T Finally, we form the matrix 5 7 P =[p p p ]= 5 Then as expected, the JCF of L is J = P LP =

7 Exercise A Hessenberg matrix H and a Jordan matrix J appear below Find an invertible matrix P such that J = P HP (Note: Some texts define the JCF with s below the main diagonal as in H) H = Exercise Prove that the Jordan matrices J = J =, J = are not similar invertible) (Hint: Show that if P =(p ij ) is a matrix such that PJ = J P, then P is not Exercise 5 A nilpotent matrix L is given below Find matrices P and J such that P LP = J has Jordan form: L = 5 Exercise Consider the 5 5 matrix: L = a Show that L is nilpotent and determine its index of nilpotency b Find the Jordan Form J of L c Find an invertible matrix P such that J = P LP Exercise 7 Determine the Jordan structure of the following 8 8 nilpotent matrix: L =