Jordan Canonical Forms

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1 Chapter 8 Jordan Canonical Forms 8 Jordan canonical forms By Theorem 69, if A has an eigenvalue λ of multiplicity m such that dime λ (A) < m, then A can not be diagonalizable However, Theorem 8 below says that any matrix, even a non-diagonalizable matrix, is similar to a matrix very close to a diagonal matrix, called a Jordan canonical form The columns of Q consists of a maximal set of linearly independent eigenvectors and some more vectors, so called generalized eigenvectors Its proof may be beyond a beginning level, and can be found in advanced linear algebra books Theorem 8 Let A be a square matrix of order n, and let λ dime λ(a) = s Then it is similar to a matrix J of the following form, called the Jordan canonical form, or a Jordan canonical matrix: J = Q AQ = J 0 J 0 J s in which each J i, called a Jordan block, is a triangular matrix of the form λ 0 J i =, 0 λ corresponding to a basis vector of E λ (A) for a single eigenvalue λ of A 333,

2 334 Chapter 8 Jordan Canonical Forms We will just illustrate how to find the Jordan canonical form and the transition matrix by observing the statements in the theorem Remark: () Each eigenvalue λ appears in k = dime λ (A) Jordan blocks Thus the total number of Jordan blocks in a Jordan canonical form is the maximal number of linearly independent eigenvectors: ie, dim E λi (A) () Thus, A has a full set of n linearly independent eigenvectors, if and only if the number of Jordan blocks of A is equal to n, if and only if every Jordan block is a matrix That is, a diagonal matrix is a particular case of the Jordan canonical form (3) The deficient eigenvectors in the transition matrix Q are filled up by, the so called, generalized eigenvectors, which will be discussed in the next section Assuming Theorem 8, one can first determine the Jordan canonical form of A as shown in the following, and then try to find the generalized eigenvectors as discussed in the next section Example 8 Let A be a 5 5 matrix that has a single eigenvalue λ of multiplicity 5 Classify all possible Jordan canonical forms of A up to permutations of the Jordan blocks Solution: () If A has only one linearly independent eigenvector belonging to λ: ie, dime λ =, then the Jordan canonical form of A has only one block of the form: J () = Q AQ = λ λ λ λ λ () If it has two linearly independent eigenvectors belonging to λ: ie, dime λ =, then its Jordan canonical form has two blocks, either one of the forms: J () = λ 0 λ λ 0 0 λ 0 0 λ λ or J (3) = λ λ λ λ These two matrices J () and J (3) cannot be similar, because (J () λi) 3 = 0, but (J (3) λi) 3 0 (One can justify this by a direct computation)

3 8 Jordan canonical forms 335 (3) If it has three linearly independent eigenvectors belonging to λ: ie, dime λ = 3, then its Jordan canonical form has three blocks, either one of the forms: λ J (4) = λ 0 λ λ 0 λ λ or J (5) = λ λ 0 0 λ 0 0 λ Again, these two matrices J (4) and J (5) are not similar, because (J (4) λi) = 0, but (J (5) λi) 0 (4) If it has four linearly independent eigenvectors belonging to λ: ie, dime λ = 4, then its Jordan canonical form has four blocks of the form: λ J (6) λ = λ λ 0 λ (5) If it has five linearly independent eigenvectors belonging to λ: ie, dime λ = 5, then its Jordan canonical form has 5 blocks, each of which is just matrix: ie, a diagonal matrix, λ J (7) λ = λ λ λ Thus, seven Jordan canonical forms are possible Note that all of these seven possible Jordan canonical matrices have the same trace, determinant, characteristic polynomial, and eigenvalues as those of the matrix A, but no two of them are similar to each other, As shown in the case () (also in (3)) of Example 8, when dime λ =, two nonsimilar Jordan canonical matrices J () and J (3) were possible To determine the right Jordan canonical form of the given matrix A, one has to look at the ranks of (A λi) k for k =,,,n In general, if λ is an eigenvalue of multiplicity m λ of an n n matrix A and J is its Jordan canonical form, then rank(a λi) k = rank (J λi) k

4 336 Chapter 8 Jordan Canonical Forms for all positive integer k, since they have the same eigenvalues and the same number of eigenvectors Thus, to decide the order of each Jordan block in J, it is useful to examine the sequence {rank(j λi) k : k =,,m λ } Example 8 (The orders of Jordan blocks) Let A be a matrix with two distinct eigenvalues λ of multiplicity m λ = 7 and µ of multiplicity m µ =, and similar to the Jordan canonical matrix J with four Jordan blocks: J = J 0 J J 3 = 0 J 4 λ 0 0 λ 0 0 λ [ λ 0 λ ] [ λ 0 λ This shows that λ has 3 linear independent eigenvectors Let c λ = n m λ (= 8 7 = ) be the total multiplicities of the other eigenvalues of A Then () rank(j λi) c λ = 5 = 4 = the total number of s off the diagonal in the Jordan blocks belonging to λ The blocks of order 3 has two s and one in each block of order, so that there are one block of order 3 and two blocks of order () rank(j λi) c λ = = = the number of blocks of order 3 (J λi) = [ ] [ ] ] [ (µ λ) ] [µ] (3) rank(j λi) 3 c λ = = 0 Thus, the number of blocks of order for λ is 7 ( ) (3 ) = 0 One can easily show that the power (J i λi) k of the Jordan blocks J i belonging to λ becomes a zero matrix for some k m λ, and so rank(j λi) k decreases as k increases, and stops decreasing at c λ = n m λ for some

5 8 Jordan canonical forms 337 k m λ, while all the other blocks belonging to the other eigenvalues remain as upper triangular matrices with nonzero diagonal entries, whose ranks will be summed up to c λ = n m λ This justifies that the sequence {rank(a λi) k c λ : k =,,m λ } completely determines the orders of the blocks of J belonging to λ: () If k is the smallest positive integer such that rank(a λi) k c λ = l 0 = 0 and rank(a λi) k c λ = l 0, then k is the order of the largest blocks belonging to λ and l is the number of such blocks () rank(a λi) k c λ l = l is equal to the number of blocks of order k (3) rank(a λi) k 3 c λ 3l l = l 3 is equal to the number of blocks of order k, and so on (4) If l,,l i are determined, then l i =rank(a λi) k i c λ il (i )l l i is the number of blocks of order k (i ) with l 0 = 0, for i = 0,, k In summary, if λ,, λ t are all distinct eigenvalues of A with their multiplicities m λ,, m λt, respectively, so that m λ + +m λt = n, then one can determine the Jordan canonical form J of an n n matrix A by the above procedure However, for a matrix of large order, the evaluation of (A λi) k might not be simple at all, while, for matrices of lower order or relatively simple matrices, the computations may be simple Example 83 Find the Jordan canonical form of the matrix 4 A = Solution: Clearly, the eigenvalues of A are λ = and µ = 3 of multiplicities and, respectively Hence, there are two possibilities of the Jordan canonical form of A: J () = or J () = But, for c = 3 =, rank(a I) c = 0 and rank(a I) c = Thus, there is one Jordan block belonging to λ = of order, and so the Jordan canonical form of A must be J ()

6 338 Chapter 8 Jordan Canonical Forms Example 84 Determine the Jordan canonical form J of the matrix A = Solution: The characteristic polynomial of the matrix A is det(a λi) = λ 4 4λ 3 + 6λ 4λ + = (λ ) 4 Thus, the eigenvalue of A is λ = of multiplicity 4 Now the rank of the matrix 0 0 A I = is 3, since the first three columns are linearly independent and they span the last column Hence, the dim N(A I) = dim E =, ie, A has only one linearly independent eigenvector so that the Jordan canonical form J has only one block: J = Problem 8 Let A be a 5 5 matrix with two distinct eigenvalues λ of multiplicity 3 and µ of multiplicity Determine all possible Jordan canonical forms of A up to permutations of the Jordan blocks Problem 8 Find the Jordan canonical form for each of the following matrices: [ ] 0 4 i 0 (), () 0 4, (3) i

7 8 Generalized eigenvectors Generalized eigenvectors In the previous section, assuming Theorem 8, we have shown how to determine the Jordan canonical form of a matrix A by analyzing the sequence {rank(a λi) k : k =,,,m λ }, for each eigenvalue λ of A In this section, we discuss how to find a transition matrix Q, and provide a theoretical basis for the validity of the method Example 85 (Transition matrix for a Jordan block) Let A be a 3 3 matrix similar to a Jordan block of the form λ 0 Q AQ = J = 0 λ 0 0 λ Determine the transition matrix Q = [x x x 3 ] Solution: Clearly, λ is the only eigenvalue of the two similar matrices A and J of multiplicity 3 Since 0 0 rank(j λi) = rank 0 0 =, dim N(J λi) = dime λ = That is, J has only one linearly independent eigenvector To see what the columns of Q are, we expand AQ = QJ as Thus, we get [Ax Ax Ax 3 ] = [λx x + λx x + λx 3 ] Ax 3 = x + λx 3, or (A λi)x 3 = (A λi) x 3 = x Ax = x + λx, or (A λi)x = (A λi) x 3 = x Ax = λx, or (A λi)x = (A λi) 3 x 3 = 0 This shows that x is an eigenvector of A, and if we find a vector x 3 such that (A λi) 3 x 3 = 0 and (A λi) x 3 = x 0, then the other x i s are obtained from x 3 via (A λi) i x 3 = x 3 i for i =,, 3 with x 0 = 0 Such a vector x 3 is called a generalized eigenvector of rank 3, and the set {x, x, x 3 } is called a chain of the generalized eigenvectors belonging to λ, and the columns of Q consist of chains of generalized eigenvectors

8 340 Chapter 8 Jordan Canonical Forms Remark: Note that, in the Example 85, if we set Q = [x 3 x x ] = QP where P is an orthogonal permutation matrix that reverses the order of the columns of Q, then one can get Q AQ = P (Q AQ)P = P JP = λ 0 0 λ 0 0 λ = J T, Thus, J and J T are similar and represent essentially the same Jordan canonical form of A Thus we say that any (complex) square matrix A is similar to a unique Jordan canonical matrix up to a permutation matrix In this sense, it is called the Jordan canonical form of a matrix A In general, by expanding AQ = QJ, one can easily see that the columns of Q corresponding to the first columns of the Jordan blocks of J form a maximal set of linearly independent eigenvectors of A, and remaining columns of Q are generalized eigenvectors Definition 8 A nonzero vector x is said to be a generalized eigenvector of A of rank k belonging to an eigenvalue λ if (A λi) k x = 0 and (A λi) k x 0 Note that if k =, this is the usual definition of an eigenvector For a generalized eigenvector x of rank k belonging to an eigenvalue λ, define x k = x, x k = (A λi)x = (A λi)x k, x k = (A λi) x = (A λi)x k, x = (A λi) k x = (A λi)x Thus, for each l, < l k, (A λi) l x l = (A λi) k x = 0 and (A λi) l x l = (A λi) k x 0 Hence, the vector x l = (A λi) k l x is a generalized eigenvector of A of rank l Definition 8 The set of vectors {x, x,, x k } is called a chain of generalized eigenvectors belonging to the eigenvalue λ Note that, x = (A λi) k x is always an eigenvector belonging to λ, called the initial eigenvector of the chain Sometimes the length k of a

9 8 Generalized eigenvectors 34 chain is also called the rank of this initial eigenvector of the chain Note also that (A λi) l x i = 0 for l i The following series of three theorems shows that a transition matrix Q may be constructed from the chains of linearly independent generalized eigenvectors of A, and justifies the invertibility of Q Theorem 8 A chain of generalized eigenvectors S = {x, x,,x k } belonging to an eigenvalue λ is linearly independent Proof: Let us solve c x +c x + +c k x k = 0 for scalars c i, i =,, k If we multiply (on the left) both sides of this equation by (A λi) k, then for i =,, k, (A λi) k x i = (A λi) k (i+) (A λi) i x i = 0 Thus, c k (A λi) k x k = 0, and, hence, c k = 0 Do the same to the equation c x + +c k x k = 0 with (A λi) k and get c k = 0 Proceeding successively, we can show that c i = 0 for all i =,, k That is, the equation has only the trivial solution Hence, the set S is linearly independent Theorem 83 The union of chains of generalized eigenvectors of a square matrix A belonging to distinct eigenvalues is linearly independent Proof: Let {x, x,, x k } and {y, y,, y l } be the chains of generalized eigenvectors of A belonging to the eigenvalues λ and µ, respectively, and let λ µ We wish to show that the set of vectors {x,, x k, y,, y l } is linearly independent To solve the linear dependence of them, c x + + c k x k + d y + + d l y l = 0, for c i s and d j s, we multiply both sides of the equation by (A λi) k and note that (A λi) k x i = 0 for all i =,,k Thus we have (A λi) k (d y + d y + + d l y l ) = 0 Again, multiply this equation by (A µi) l and note that (A µi) l (A λi) k = (A λi) k (A µi) l, (A µi) l y l = y, (A µi) l y i = 0

10 34 Chapter 8 Jordan Canonical Forms for i =,,l Thus we obtain 0 = d l (A λi) k y Because (A µi)y = 0 (or Ay = µy ), this reduces to d l (µ λ) k y = 0, which implies that d l = 0 by the assumption λ µ and y 0 Proceeding successively, we can show that d i = 0, i = l, l,,,, so we are left with c x + + c k x k = 0 Since {x,, x k } is already linearly independent by Theorem 8, c i = 0 for all i =,,k Thus the set of generalized eigenvectors {x,, x k, y,, y l } is linearly independent The next step to produce Q such that AQ = QJ is to choose chains of linearly independent generalized eigenvectors for each eigenvalue so that the union of the chains is linearly independent Definition 83 Let λ be an eigenvalue of A The generalized eigenspace of A belonging to λ, denoted by K λ, is the set K λ = {x C n : (A λi) p x = 0 for some positive integer p} It turns out that dim K λ is the multiplicity of λ, and it contains the usual eigenspace N(A λi) The following theorem enables us to choose a basis for K λ, but we omit the proof even though it can be proved by induction on the number of vectors in S T Theorem 84 Let S = {x, x,, x k } and T = {y, y,, y l } be two chains of generalized eigenvectors of A belonging to the same eigenvalue λ If the initial vectors x and y are linearly independent, then the union S T is linearly independent Note that this theorem easily extends to a finite number of chains of generalized eigenvectors of A belonging to an eigenvalue λ, and the union of such chains will form a basis for K λ so that the matrix Q may be constructed from these bases for each eigenvalue as usual However, finding chains of generalized eigenvectors are not simple in general One may try to find them in two ways as follows: Method : () For each eigenvalue λ of multiplicity m λ, first find a maximal set of linearly independent eigenvectors x,y,,z which form a basis for the eigenspace E λ, and then solve (A λi)x = ax + by + + cz = x

11 8 Generalized eigenvectors 343 for suitable a, b,, c that make the systems (including those systems in step () below) consistent () Inductively, solve (A λi)x i = x i for i =,, until the equation becomes inconsistent When (A λi)x k+ = x k becomes inconsistent, k is the rank of the generalized eigenvector x k (3) If k < m λ, then, by the same method in step () above, find another suitable vector y in E λ which is linearly independent from x, and then repeat the above process (4) Do this until the ranks sum up to m λ (5) An inconvenience of this method is that one does not know what the initial eigenvectors of the chains and their ranks are at the outset Example 86 The characteristic polynomial of A = , is det(λi A) = λ 3 3λ + 3λ = (λ ) 3, and the eigenvalue of A is λ = of multiplicity 3 Its eigenvectors are the solutions of 4 3 x 0 (A I)x = y = z 0 Since the three equations are identical, one gets two linearly independent eigenvectors u = (, 0, ) and u = (0,, 3) A generalized eigenvector is the solution of (A I)x = c u + c u = x for some constants c i s In fact, the system x y z = c 0 + c 0 3 = c c c 3c has a solution if and only if c = c, and a solution is x = (0, 0, ) for x = (, 4, ) Since (A I)x 3 = x is inconsistent, the rank of the generalized eigenvector x is so that {x,x } is a chain Since u = (, 0, ) is linearly independent to x, we may take x 3 = u and so Q = [x x x 3 ] =

12 344 Chapter 8 Jordan Canonical Forms is a transition matrix so that Q AQ = J = Method : () Determine the Jordan canonical form together with the order of each block as explained in the previous section () For each eigenvalue λ and each block of order k belonging to λ, find the solution (A λi) k x = 0, but (A λi) k x 0, and then construct a chain of this generalized vector (3) An inconvenience of this method is that one has to compute (A λi) k for each eigenvalue λ and k, which is not simple if the order of A is large Example 87 For the same matrix A as in Example 86, we know that A has only two Jordan blocks for only one eigenvalue By direct computation we find that A I 0 and (A I) = 0 Thus, we can take any vector x such that (A I)x 0, which becomes a generalized eigenvector of rank Take x = (,, ), then x = (A I)x = (,, ) Now find another eigenvector of A belonging to λ = by solving (A I)x 3 = 0, and get x 3 = (, 0, ) Then for Q = [x x x 3 ] = 0, we get 0 Q AQ = Example 88 For a matrix A = so that Q AQ is the Jordan canonical matrix, find a transition matrix Q Solution: Write Q = [x x x 3 ], and we try to find the columns x i, i =,, 3 Method : In Example 83, the matrix J of A is determined as J = From AQ = QJ, by comparing the column vectors of both sides, one can get Ax = x, Ax = x + x, Ax 3 = 3x 3

13 8 Generalized eigenvectors 345 Thus x and x 3 are the eigenvectors of A belonging to λ = and λ = 3, respectively By a direct computation, they are found to be x = (, 0, 0) and x 3 = (3,, ) By solving the equation (A I)x = x, one gets x = (a,, 0) with any constant a, so that Q = a It is not hard to see that x, x, x 3 are linearly independent, so that Q AQ = J Method : From Example 83, the Jordan block belonging to is of order Thus we need to find a generalized eigenvector x of rank belonging to the eigenvalue, which is a solution of the following systems: (A I)x = (A I) x = x 0, x = 0 From the second equation, x has to be of the form (a, b, 0), and from the first equation we must have b 0 Let us take x = (0,, 0) for a generalized eigenvector of rank Thus we have (A I)x = x = (, 0, 0), (A I) x = (A I)x = 0 Thus x = (, 0, 0) is the only one linearly independent eigenvector belonging to An eigenvector belonging to λ 3 = 3 is easily found to be x 3 = (3,, ) Clearly, the set of vectors {x, x, x 3 } is linearly independent, and so Q = , so Q = Then Q AQ = [ J 0 = 0 J ],

14 346 Chapter 8 Jordan Canonical Forms where J = [ 0 ] and J = [3] Example 89 Find Q so that Q AQ = J is the Jordan canonical form of the matrix A = Solution: Method : In Example 84, we have found that 0 0 J = Write a transition matrix as Q = [x x x 3 x 4 ] From the expansion of AQ = QJ, one gets: Ax = x, Ax = x + x, Ax 3 = x 3 + x, Ax 4 = x 4 + x 3 There is only one linearly independent eigenvector belonging to λ =, which is x = (,,, ) Now a solution of (A I)x = x is x = (a, a+, a+, a + 3) for any a Take x = (0,,, 3) Inductively, the solutions of (A I)x 3 = x and (A I)x 4 = x 3 are x 3 = (b, b, b +, b + 3) for any b and set x 3 = (0, 0,, 3), and successively one can take x 4 = (0, 0, 0, ) Clearly, they are linearly independent and so Q = One may check Q AQ = J by a direct matrix multiplication Method : As we saw in Example 84, A has only one Jordan block Therefore, one has to find a generalized eigenvector of rank 4, which is a solution x of the following equations: (A I) 4 x = 0 (A I) 3 x = x 0

15 83 Applications of Jordan canonical form 347 But, a direct computation shows that the matrix (A I) 4 = 0 Hence, one can take any vector that satisfies the second equation as a generalized eigenvector of rank 4: Take x 4 = (, 0, 0, 0), and then x 3 = (A I)x 4 = x = (A I)x 3 = (, 0,, ), x = (A I)x = (,,, ) = Therefore, these vectors {x, x, x 3, x 4 } form the chain of linearly independent generalized eigenvectors Therefore, Q = , Q = 0 0 0, and Q AQ = = J Problem 83 Find a full set of generalized eigenvectors of the following matrices: (), () , 83 Applications of Jordan canonical form The Jordan canonical form of any square matrix A enables us to compute the power A k and the exponential matrix e A, and to solve many other problems related to the matrix A Let J be the Jordan canonical form of an arbitrary n n square matrix A such that Q AQ = J = J where Q is made of generalized eigenvectors of A and J i s are Jordan blocks J s,

16 348 Chapter 8 Jordan Canonical Forms 83 Computations of A n and e A I Note that we have A k = QJ k Q = Q J k J k s Q, for k =,, Hence, for A k, it is good enough to compute J k for each Jordan block J Now an m m Jordan block J belonging to an eigenvalue λ of A may be written as λ J = 0 λ = λ λ = λi + N Since I is the identity matrix, clearly IN = NI and k ( ) k J k = (λi + N) k = λ k j N j j But N k = 0 for k m Thus, by assuming ( k l) = 0 if k < l, m ( ) k J k = λ k j N j j j=0 ( ) ( ) k k = λ k I + λ k N + + λ k (m ) N m m λ k ( k ( ) λ k k ) ( λ k k ) m λ k m+ 0 λ k ( k ( ) λ k k ) m λ k m+ = λ k ( k ) λ k j=0 0 0 λ k Example 80 Compute A k, k =,,, for A =

17 83 Applications of Jordan canonical form 349 Solution: The characteristic polynomial of A is det(λi A) = λ 4 8λ 3 + 4λ 3λ+6 = (λ ) 4, and λ = is an eigenvalue of multiplicity 4 By a direct computation, one can see that rank(a I) =, rank(a I) = and rank(a I) 3 = 0 Thus, the Jordan canonical form J of A must be of the form J = Also one can easily find a transition matrix Q to be 0 0 Q = , and then Q = 0 0 such that Q AQ = J Therefore, where k 0 0 = 0 0 Hence, A k A k = QJ k Q = Q = = k k 0 + ( ) k 0 0 k k k ( k ( ) k k ) k 0 k ( k ) k 0 0 k k [ ] k k ( k Q,, ) 0 0 k k ( k ( ) k k ) k = QJ k Q = Q 0 k ( k ) k k Q 0 k

18 350 Chapter 8 Jordan Canonical Forms = k k k k k k(k ) k + k k k k 0 k k k k 0 k k k k k(k ) k k k + k Problem 84 Compute A k, k =,,, for 0 0 () A = , () B = For the computations of the exponential matrix e A, note that e A = e QJQ = Qe J Q e J 0 e J = Q Q, 0 e Js where J i s are the Jordan blocks Thus, it is enough to compute e J for a simple Jordan block J Let J = λi + N, where I and N are as in Section 83 Then, N k = 0 for k n, and! (n )! n e J = e λi e N = e λ N k 0 = e λ (n )! k! k=0 0 Example 8 Compute e A for A =

19 83 Application: Linear difference equations II 35 Solution: With the same notations as in Example 80, [ ] e A = e QJQ = Qe J Q e J 0 = Q 0 e J Q, where Hence, J = = I + e J = e I e N = e k=0 N k k! = I + N and J = [] = e! 0 It gives that e A = Qe J Q = e Q Q = e 3 e e 0 e e e 0 e e e e 83 Linear difference equations II Let A be the companion matrix of order k of a linear difference equation x n = Ax n, and let λ 0 be an eigenvalue of A with multiplicity m > Then, by Lemma 7, λ 0 has only one linearly independent eigenvector v of the form [λ0 k λ 0 ] T, which means that the size of the Jordan block corresponding to this eigenvector is the multiplicity m of λ 0 Therefore, if A has s distinct eigenvalues each of multiplicity m j, j =,,, s so that m +m + +m s = k, then the Jordan Canonical form of A has s Jordan blocks each of order m i : ie, Q AQ = J = J J s, where J j = λ j 0 λj The chain of generalized eigenvectors containing v j = [λj k λ j ] T consists of m j vectors: v ji, i =,,, m j, which, from AQ = QJ, satisfy (A λ j I)v j = 0, (A λ j I)v j = v j,, (A λ j I)v jmj = v j(mj )

20 35 Chapter 8 The Jordan canonical forms Beginning with v j = [λj k λ j ] T, one can easily compute inductively a chain of generalized eigenvectors: ie, if we denote a generalized eigenvector by v = [x k x x 0 ] T, then for v j, x n = λ n j for v j, x n = nλ n j for v j3, x n = n(n ) λ n j = ( n 0) λ n j, for n 0, = ( ) n λ n j, for n, = ( ) n λ n j, for n, for v jmj, x n = = ( ) n n (m m j λ j ) j, for n m j Thus the columns of the transition matrix Q corresponding to J j looks like: λ k j λ m j j ( k ) (k ) λ j ( mj ) m λ j j ( k ) (k ) λ j ( mj ( k ) (k ) (m m j λ j ) j ) m λ j 3 j 0 λ j λ j λ j Notice that the coefficients form the Pascal triangle Therefore, the general solution of a linear difference equation x n = Ax n is of the form, x n = s j= ( ( ) ( ) ( ) ) n n n c j λ n j + c j λ n j + + c jmj λ n m j+ j, 0 m j for n 0 Here we assume that ( n m) = 0 if n < m 833 Discrete dynamical systems II In Section 7, the discrete dynamical systems have been introduced, and it is shown that is always an eigenvalue of a stochastic matrix We now show that the magnitudes of all the eigenvalues of a stochastic matrix are bounded by, and moreover, if a stochastic matrix has positive entries, then its eigenvalues λ are all either λ < or λ =, and so no one is such that λ = but λ For this, we first try to estimate the bound for the magnitude of the eigenvalues of any square matrix A (including complex matrix) in terms of

21 83 Application: Discrete dynamical systems II 353 the absolute values of the entries of A For any square matrix A = [a j i ] of order k, let k R(A) = max{r i (A) = a j i : i k}, C(A) = max{c j (A) = s i = R i (A) a i i j= k a j i : j k}, Theorem 85 (Gerschgorin s Theorem) For any square matrix A of order k, every eigenvalue λ of A satisfies λ a ll s l for some l k i= Proof: Let λ be an eigenvalue with eigenvector x = [x x x k ] T Then k j= a i j x j = λx i, for i =,, k Take a coordinate x l of x with the largest absolute value Then clearly x l 0, and λ a l l x l = λx l a l l x l = a j l x j j l a j l x l = s l x l j l Since x l > 0, λ a l l s l Corollary 86 For any square matrix A of order k, every eigenvalue λ of A satisfies λ min{r(a), C(A)} Proof: Note that λ λ a l l + a l l s l + a l l = R l (A) R(A) Moreover, since λ is also an eigenvalue of A T, λ R(A T ) = C(A) Furthermore, if A is a stochastic matrix, then C(A) = Corollary 87 If λ is an eigenvalue of a stochastic matrix A, then λ We next give a quick explanation of the key properties of a positive matrix Theorem 88 (Perron-Frobenius Theorem) Let A be a matrix with positive entries If λ 0 is an eigenvalue of A such that λ 0 is the largest, then λ 0 is real and positive, and so are the components of its eigenvector x

22 354 Chapter 8 The Jordan canonical forms Proof: Let λ and z be such that Az = λz Then λ z = Az A z, where x = z is a nonnegative vector Thus the following number exists: λ 0 = max{ t R : Ax tx, for some x 0, x 0} Then clearly, λ 0 λ > 0 We claim that Ax = λ 0 x, which shows that λ 0 is the largest eigenvalue with a nonnegative eigenvector: Indeed, suppose not: Ax λ 0 x, ie, some components may be in equalities and some are in strict inequalities Then, since A is positive, one can easily see that we have strict inequality: A x > λ 0 Ax, Ay > λ 0 y, y = Ax, which contradicts the maximality of λ 0 Note that a stochastic matrix A has as an eigenvalue by () of Theorem 78 In the following, we will further show that, if a stochastic matrix A has all positive entries, then no eigenvalue of A other than λ = satisfies λ = Thus is the largest eigenvalue with a nonnegative eigenvector Theorem 89 Let A be a matrix with positive entries If λ is an eigenvalue of A such that λ = R(A), then λ = R(A) and dim E λ = with a basis u = [ ] T Proof: Let x = [x x x k ] T be an eigenvector belonging to λ Take a coordinate x l of x with the largest absolute value Then k λ x l = λx l = a j k l x j a j l x j j= j= k a j l x l = R l (A) x l j= R(A) x l Since λ = R(A), the three inequalities are actually equalities, ie, k a j k l x j = a j l x j j= k a j l x j = j= j= k a j l x l j= R l (A) = R(A)

23 83 Application: Discrete dynamical systems II 355 It is an easy exercise (see Problem 85 below) to show that the first equality means that all terms a j l x j, j =,, k, are nonnegative multiple of some complex number z with z = Thus a j l x j = c j z, or x j = c jz a j, for some l nonnegative real numbers c,, c k The second equality means that a j l = 0 or x j = x l for all j =,, k Since a j l > 0 by assumption, we have the second possibility: x j = x l = M for all j =,, k This means that, for each j =,, k, the above computation is valid: ie, R j (A) = R(A) for all j Thus one gets M = x j = c jz a j l = c j a j or x l j = Mz for all j =,, k, and so x = [x x k ] T = Mz[ ] T That is, u = [ ] T is a basis for E λ, so that dime λ = Moreover, A = k j= a j k j= a k j = R (A) R k (A) = R(A) = R(A) R(A) or Ax = R(A)x Thus if x is an eigenvector belonging to λ with λ = R(A), then it is an eigenvector belonging also to R(A) so that λ = R(A), Problem 85 () Prove that u+v u + v, for any complex numbers u and v, and the equality holds if and only if u = az and v = bz where a, b are nonnegative real numbers and z is a complex number such that z = () Use induction to show k i= z i k i= z i, and equality holds if and only if z i = c i z for some nonnegative real numbers c i and a complex number z such that z = Note that λ = C(A) also implies λ = C(A) since C(A) = R(A T ) If A is a stochastic matrix, then C(A) =, and so we have the following Corollary 80 Let A be a stochastic matrix with positive entries, and let λ be an eigenvalue of A other then Then λ < Moreover, dime = Problem 86 Prove that the dimensions of the eigenspaces of a common eigenvalue λ of A and A T are equal Therefore, if A is diagonalizable stochastic matrix with positive entries, then lim n An = L always exists by Theorem 76 and Corollary 80 For a general (nondiaginalizable) matrix, we have the following criteria Theorem 8 Let A be any square matrix Then lim n An exists if and only if the following conditions hold:

24 356 Chapter 8 The Jordan canonical forms () An eigenvalue λ of A satisfies λ <, or λ = () If is an eigenvalue of A, then dime is the multiplicity of Proof: Let J be the Jordan canonical form of A Then lim k Ak exists if and only if lim k Jk exists Let J i be a Jordan block of order m in J of the form as () in page 348 Then, for i < m, lim k ) k ( i λ k i = 0 if λ <, and lim k ) k ( i λ k i = if λ It follows that lim k Jk i exists if and only if either λ < or λ = and m = In the former case, lim k Jk i = 0, and in the latter case, lim k Jk i = [] and so each Jordan block of λ = is a matrix, which means dime = multiplicity of In general, it may happen that dime is less then the multiplicity of the eigenvalue λ =, as the following example [ ] shows In fact, we have seen in Example 64 that the matrix A = is not diagonalizable since is 0 an eigenvalue of multiplicity, while dime = Thus the second condition of Theorem 8 does not holds, and one can easily see that lim n An does [ ] n not exist since A n = 0 The following theorem summarizes what have been discussed so far about a stochastic matrix Theorem 8 Let A be a stochastic matrix with positive entries Then () The multiplicity of the eigenvalue λ = of A is () lim k Ak = L exists (that is, the diagonalizability of A is not necessary for this), and L is also a stochastic matrix (3) AL = LA = L (4) The columns of L are identical In fact, each column of L is equal to the unique eigenvector u of A belonging to λ = which is also a probability vector (5) For any probability vector x, lim k Ak x = u Proof: () By the proof of Theorem 8, each Jordan block of is, while dime = by Corollary 80: ie, has only one Jordan block

25 83 Application: Discrete dynamical systems II 357 () The first part follows directly from part (), Corollary 80, and Theorem 8 For the second part, since A k are stochastic matrices, each entry of A k is nonnegative for k =,, Thus [L] j i = lim k [Ak j ] i 0 for 0 i, j n Moreover, for j n, n [L]i j = i= n [ ] lim k [Ak j ] i i= = lim k [ n i= ] [A k j ] i = lim = k (3) Trivial, since A( lim k Ak ) = ( lim k Ak )A = lim k Ak+ = L (4) Since AL = L, each column of L is an eigenvector of A belonging to λ = But dime = means they columns of L are the same eigenvector belonging to λ =, and each of them is a probability vector by part () (5) The vector y = Lx is also a probability vector by () and Lemma 77, and Ay = ALx = Lx = y That is, y is also an eigenvector of A belonging to λ =, and so y = u Note that (5) in Theorem 8 means that the eventual distribution of the objects depends only on the sum of all the components of the probability vector, but not the initial distribution The vector u in Theorem 8 is called the stationary vector or the steady state of A Remark: Actually, in Theorem 8, the positiveness of the entries of a stochastic matrix may be weakened: That is, Theorem 8 still holds if some power of the stochastic matrix A has only positive entries For example, for the matrix A = , A has positive entries To prove this assertion, it is good enough to reprove Corollary 80 In fact, suppose that there is d > such that A d has only positive entries Then it is clear that the entries of A d+ = (A d )A are all positive since A is a stochastic matrix If λ is an eigenvalue of A such that λ =, then λ d and λ d+ are eigenvalues of A d and A d+, respectively, with absolute value Thus, by Corollary 80, λ d = λ d+ =, which means λ = Note that E (A) E (A d ), and dime (A d ) =, which means E (A) = E (A d ) and dime (A) =

26 358 Chapter 8 The Jordan canonical forms 834 Linear differential equations II Now, we go back to a system of linear differential equations y = Ay with an initial condition y(0) = y 0 Its solution is known as y(t) = e ta y 0 (see Theorem 76) In particular, if A is diagonalizable with n eigenvectors v,,v n belonging to the eigenvalues λ i s, this solution can be written as y(t) = e ta y 0 = c e λ t v + c e λ t v + + c n e λnt v n For an arbitrary square matrix A (not necessarily diagonalizable), find its Jordan canonical form Q AQ = J Then, the solution y(t) = e ta y 0 is e ta y 0 = Qe tj Q y 0 e tj 0 0 = [ u u u n] 0 e tj 0 c 0 e tjs e λt e tn 0 0 = [ u u u n] 0 e λt e tn 0 c c n 0 e λst e tns where Q y 0 = (c,, c n ) and the u i s are generalized eigenvectors of A In particular, if J is just a Jordan block with corresponding generalized eigenvectors u i of order k, then the solution becomes: e ta y 0 = e λt Qe tn Q y 0 t t n t! (n )! = e λt [ u u u n] 0 t t n c (n )! c c n t 0 (( n ) ( = e λt t k n ) c k+ u t k + c k+ )u + + c n u n k! k! k=0 k=0 c c c n,

27 83 Application: Linear differential equations II 359 Example 8 Solve the linear differential equation y = Ay with initial condition y(0) = y 0, where 4 3 A = 0, y 0 = 3 4 Solution: () The characteristic polynomial of A is det(λi A) = λ 3 7λ + 6λ = (λ 3)(λ ) Since x = (,, ) and x 3 = (,, ) are linearly independent eigenvectors belonging to λ = and λ = 3, respectively, one can compute the Jordan canonical form of A as follows: where J = J = Q AQ = [ [ J 0 = 0 J ], J = [3], and Q = ], 0 () Let y = Qx Then the given system changes to x = Jx with 5 x(0) = Q y(0) = 0 = 5, 0 4 and the solution of this new system is given by [ ] x(t) = e tj e tj 5 e 0 t te t 0 x(0) = 0 e tj 5 = 0 e t e 3t since (3) Thus, we get e tj = e t [ t 0 y(t) = Qx(t) = ] 0 = e t (5 + 5t) and e tj = e 3t + 5 e t te t 0 0 e t e 3t 0 + e 3t ,

28 360 Chapter 8 The Jordan canonical forms Example 83 Solve a system of linear differential equations y (t) = Ay(t), where 5 3 A = Solution: In Example 86 or 87, we have found a transition matrix Q = 0 [x x x 3 ] = so that [ ] 0 [ ] Q AQ = J = 0 0 = 0 J 0 = 0 J 0 0 [] Thus, the solution y(t) = e ta y 0 is [ ] e ta e J 0 y 0 = Q 0 e J Q y 0 = e t Q = e t te t 0 0 e t e t t Q y c c c 3 = c 4te t + e t 8te t 4te t + c 3te t e t 6te t 3te t + c 3 te t 4te t te t + e t Or, if we set Q y 0 = (d, d, d 3 ), then d y(t) = e t [x tx + x x 3 ] d d 3 = e t ((d + d t)x + d x + d 3 x 3 ) Problem 87 Solve the system of linear differential equations y = Ay with the initial condition y(0) = y 0, where A = 3, y 0 = 9 3 4

29 84 Cayley-Hamilton theorem Cayley-Hamilton theorem As we saw in earlier chapters, the association of the characteristic polynomial with each matrix is very useful in studying matrices In this section, using this association of the polynomials with matrices we prove one more useful theorem, called the Cayley-Hamilton theorem, which makes the calculation of matrix polynomials simple, and has many applications to real problems Let f(x) = a m x m + a m x m + + a x + a 0 be a polynomial, and let A be an n n square matrix The matrix defined by f(a) = a m A m + a m A m + + a A + a 0 I n is called a [ matrix ] polynomial of A For example, if f(x) = x x + and A =, then f(a) = A A + I [ ] [ 5 4 = 4 5 ] [ ] = [ Problem 88 Let λ be an eigenvalue of A and x an eigenvector belonging to λ If f(x) is any polynomial, then f(λ) is an eigenvalue of the matrix polynomial f(a) ] Theorem 83 (Cayley-Hamilton) For any n n matrix A, if f(λ) = det(λi A) is the characteristic polynomial of A, then f(a) = 0 Proof: By Theorem 8, any square matrix A is similar to the Jordan J 0 canonical form J = = Q AQ, or A = QJQ Then 0 J s clearly f(a) = Qf(J)Q, and J k 0 f(j ) 0 J k = and f(j) = 0 Js k 0 f(j s ) On the other hand, the characteristic polynomial f of A is factorized by the characteristic polynomials f j s of the Jordan block J j s: In fact, by Exercise 4, det(λi A) = det(λi J) = det(λi J ) det(λi J s ) or f(λ) = f (λ) f s (λ),

30 36 Chapter 8 The Jordan canonical forms where f j (λ) = det(λi J j ) is the characteristic polynomial of J j Thus, it is good enough to show that g(j) = 0 for the characteristic polynomial g(λ) of a single Jordan block J = ai + N with an eigenvalue a of multiplicity m Note that N m = 0 and g(λ) = det(λi J) = (λ a) m Hence g(j) = (J ai) m = (ai + N ai) m = N m = 0 Thus, f(j j ) = f (J j ) f j (J j ) f s (J j ) = 0 for each j =,,s Example 84 The characteristic polynomial of A = is f(λ) = det(λi A) = λ 3 + λ 6λ, and f(a) = A 3 + A 6A = = Remark: It is interesting that the last part of the proof of the Cayley- Hamilton Theorem 83 can be replaced by a direct computation of f(j) for a Jordan block J of order m as follows: Recall that m ( ) k J k = λ k j N j j j=0 ( ) ( ) k k = λ k I + λ k N + + λ k (m ) N m m λ k ( k ( ) λ k k ) ( λ k k ) m λ k m+ 0 λ k ( k ( ) λ k k ) m λ k m+ = λ k ( k ) λ k 0 0 λ k

31 84 Application: The inverse matrices 363 Therefore, for any polynomial p(λ) = b l λ l + b l λ l + + b λ + b 0, p(j) = b l J l + b l J l + + b J + b 0 p(λ) p p (λ) (m ) (λ) (m )! p = 0 p(λ) (m ) (λ) (m )!, 0 0 p(λ) where p (i) (λ) is the i-th derivative of p(λ) In particular, if p(λ) is the characteristic polynomial f(λ) of A, and a is an eigenvalue of multiplicity m, then f(λ) = (λ a) m and so f (i) (a) = 0 for i = 0,,,m Thus f(j) = 0 84 Application to inverse matrices The Cayley-Hamilton theorem can be used to find the inverse of a nonsingular matrix If f(λ) = λ n + a n λ n + + a λ + a 0 is the characteristic polynomial of a matrix A, then or 0 = f(a) = A n + a n A n + + a A + a 0 I, a 0 I = (A n + a n A n + + a I)A Since a 0 = f(0) = det(0i A) = det( A) = ( ) n deta, A is nonsingular if and only if a 0 = ( ) n deta 0 Therefore, if A is nonsingular, A = a 0 (A n + a n A n + + a I) Example 85 The characteristic polynomial of the matrix A = is f(λ) = det(λi 3 A) = λ 3 8λ + 7λ 0, and the Cayley-Hamilton theorem yields A 3 8A + 7A = 0I 3

32 364 Chapter 8 The Jordan canonical forms Hence A = 0 (A 8A + 7I 3 ) = = Problem 89 Let A and B be square matrices, not necessarily of the same size, and let f(λ) = det(λi A) be the characteristic polynomial of A Show that f(b) is invertible if and only if A has no eigenvalue in common with B 84 Application to matrix polynomials The Cayley-Hamilton theorem can also be used to simplify the calculation of matrix polynomials Let p(λ) be any polynomial and let f(λ) be the characteristic polynomial of an n n matrix A By the Euclidean division algorithm of polynomials, one can find two polynomials q(λ) and r(λ) such that p(λ) = q(λ)f(λ) + r(λ), where the degree of r(λ) less than that of f(λ) Then p(a) = q(a)f(a) + r(a) By the Cayley-Hamilton theorem, f(a) = 0 and p(a) = r(a) Thus the problem of evaluating a polynomial of an n n matrix A, in particular A k, can be reduced to the problem of evaluating a polynomial of degree less than n [ ] Example 86 The characteristic polynomial of the matrix A = is f(λ) = λ λ 3 Let p(λ) = λ 4 7λ 3 3λ + λ + 4 be a polynomial A straightforward calculation shows that p(λ) = (λ 5λ 0)f(λ) 34λ 6

33 84 Application: Computations of A k and e A II 365 Therefore p(a) = (A 5A + 0)f(A) 34A 6I = 34A 6I [ ] [ 0 = Problem 80 For the matrix A = A 5 + 3A 4 + A 3 A + 4A + 6I ] = [ ], evaluate the matrix polynomial 843 Computations of A k and e A II Once we have the Cayley-Hamilton Theorem, we can use this to compute A k or e A without finding the Jordan canonical form of A or a transition matrix Q Let f(λ) = λ n + a n λ n + + a λ + a 0 be the characteristic polynomial of A Then f(a) = 0 implies that A n = a n A n a A a 0 I Thus, for any k n, the power A k can be computed by a matrix polynomial of degree less than n This fact implies that the computation of e A = k=0 Ak k!, an infinite series of powers of A, can also be reduced to that of a matrix polynomial g(λ) of degree at most n However, the computation of the coefficients of g might not be easy This problem can be handled if we consider a matrix polynomial g(λ) of the smallest degree such that g(a) = 0 If we require the coefficient of the highest power of λ in g to be, such a polynomial exists uniquely, and is called the minimal polynomial of A By the Cayley-Hamilton Theorem, the degree of g is less than or equal to that of f Moreover, by the definition of the minimal polynomial g, it is easy to see that g(λ) is a factor of f(λ) The following facts are well known: () If the eigenvalues of A are all distinct, the minimal polynomial is the characteristic polynomial: g(λ) = f(λ) = (λ λ ) (λ λ n ), where λ i s are all distinct n eigenvalues of A

34 366 Chapter 8 The Jordan canonical forms () For each eigenvalue λ i with multiplicity m i >, g(λ) has a factor (λ λ i ) p i of highest power p i for λ i, where p i is the order of the largest Jordan block among the Jordan blocks with λ i on the diagonal Thus, if dime λi = m i : ie, λ i has m i linearly independent eigenvectors so that p i =, then (λ λ i ) is a factor of the highest power in λ i of g(λ) In particular, if A is diagonalizable, then the minimal polynomial is g(λ) = (λ λ ) (λ λ s ), where λ i s are distinct eigenvalues of A (3) In general, if A has s distinct eigenvalues λ,,λ s with multiplicities m,,m s, respectively, and p i m i, i =,,s, denote the orders of the largest Jordan blocks belonging to λ i, then the minimal polynomial is g(λ) = (λ λ ) p (λ λ s ) ps so that the degree of g(λ) is q = s i= p i n Lat g(λ) = λ q + a q λ q + a λ + a 0 be the minimal polynomial of A Then g(a) = 0 implies that, for any k q, we may set up A k = x 0 I + x A + + x q A q, or e A = x 0 I + x A + + x q A q, for some coefficients x i s We can now determine those coefficients x i s as follows: Let {v,,v p } be a maximal chain of generalized eigenvectors belonging to an eigenvalue λ Then Av p = λv p + v p,, Av = λv + v, Av = λv From the last equation, A k v = λ k v = (x 0 I + x A + + x q A q )v Thus, we get λ k = x 0 + λx + + λ q x q Similarly from the second to the last equation, A k v = λ k v +kλ k v On the other hand, by a direct computation we get: A k v = (x 0 I + x A + + x q A q )v = (x 0 + λx + + λ q x q )v +(x + λx + + (q )λ q x q )v Since v and v are linearly independent, we have λ k = x 0 + λx + λ x + + λ q x q kλ k = x + λx + + (q )λ q x q

35 84 Application: Computations of A k and e A II 367 Note that the second equation is just the derivative of the first one with respect to λ By the same computations with A k v 3,,A k v p, one gets the following additional p equations: ( k ) λ k =! x + 3 λx 3 + +! ( q ) λ q 3 x q ( ) k λ k (p ) = p (p )! (p )! x p + p! (p )! λx p + + ( ) q λ q p x q p Note also that these equations are the higher derivatives of the first equation Thus, for each eigenvalue λ of multiplicity m, we get a p equations, and so, all together, a system of q = s i= p i equations in q unknowns x j s It is easy to see that the coefficient matrix is invertible so that the unknowns x 0, x,,x q are uniquely determined The same computation can be applied to e A = x 0 I + x A + + x q A q = h(a) For a maximal chain of generalized eigenvectors {v,,v p } belonging to an eigenvalue λ, by computing both sides of e A v j = (x 0 I + x A + + x q A q )v j, for j =,,p, separately, we get the following p equations e λ = h(λ) = x 0 + λx + + λ q x q e λ = h (λ) = x + + (q )λ q x q (p )! eλ = h (p ) (λ) = ( ) p λx p + + p ( ) q λ q p x q p With the same argument as the case of A k, one can determine the coefficients x j s uniquely Example 87 Compute A, A k and e A for A =

36 368 Chapter 8 The Jordan canonical forms Solution: The characteristic polynomial of A is f(λ) = (λ ) 4 Since, by a direct computation, (A I) 3 = 0 and (A I) 0, the Jordan canonical form J of A must be of the form J = Thus the minimal polynomial of A is g(λ) = (λ ) 3 = λ 3 6λ + λ 8, and so 6 A = 8 (A 6A + I) = Let A k = x 0 I + x A + x A Then, for λ =, we get The solution is so that k = x 0 + x + x, k k = x + x, k(k ) k = x x 0 = k (k )(k ), x = k( k) k, x = k(k ) k 3, A k = k (k )(k )I + k( k) k A + k(k ) k 3 A ( k) k k k k 3 (3k + k ) k k = 0 k k k k 0 k k k k k(k ) k 3 (k + ) k Similarly, let e A = x 0 I + x A + x A Then, for λ =, e = x 0 + x + x e = x + x e = x

37 85 Exercises 369 The solution is x 0 = e, x = e, x = e, so that e A = e (I A + A ) = e Exercises 8 Show that if A nonsingular, then A has the same block structure in its Jordan canonical form as A does 8 Find the number of linearly independent eigenvectors for each of the following matrices: () , () , (3) Solve the system of linear equations { ( i)x + ( + i)y = i ( + i)x + ( + i)y = + 3i 84 Find the Jordan-canonical form for A = 85 Let A = [ 3 3 ] and y 0 = [ 0 () Solve y n = Ay n with y 0 () Solve y = Ay with y(0) = y Let A = 8 4 and y 0 = 6 () Solve y n = Ay n with y 0 () Solve y = Ay with y() = y 0 ] [ 0 87 Solve the initial value problem y = y +y 3, y (0) = y = y +y y 3, y (0) = 0 y 3 = y +3y 3, y 3 (0) = 0 ], and compute e A

38 370 Chapter 8 The Jordan canonical forms [ a b 88 Consider a matrix A = c d ] () Find a necessary and sufficient condition for A to be diagonalizable () The characteristic polynomial for A is f(t) = t (a + d)t + (ad bc) Show that f(a) = 0 89 For each of the following matrices, find a polynomial of which the matrix is a root [ ] [ ] (), (), (3) Verify that each of the matrices below satisfies its own characteristic polynomial and from these results compute A, if it exists [ ] [ ] 0 0 (), (), (3) For f(x) = 3x 3 + x x + 3, compute f(a) for 0 () A = 3 4 0, () A = Show that a Jordan block J is similar to its transpose, J T = P JP, by the permutation matrix P = [e n e ] Deduce that every matrix is similar to its transpose 83 For any square matrix A = [a i j ], define A = max{ a i j : i,j n} Prove the followings for any square matrices A, B and c C: () A 0 and equality holds if and only if A is the zero matrix () ca = c A (3) A + B A + B (4) AB n A B 84 Let A be a stochastic matrix, and Q AQ = J be the Jordan canonical form of A () Prove A k for any positive integer k () Deduce that { J k : k =,,} is bounded (3) Prove that each Jordan block belonging to the eigenvalue λ = of A is (4) Prove that lim k Ak exists if and only if any eigenvalue λ of A such that λ = is in fact λ = 85 Compute A, A k and e A for [ ] 0 i 0 (), () 0 i 0 0, (3)

39 85 Exercises Determine whether the following statements are true or false, in general, and justify your answers () Any square matrix is similar to a triangular matrix () If a matrix A has exactly k linearly independent eigenvectors, then the Jordan canonical form of A has k Jordan blocks (3) If a matrix A has k distinct eigenvalues, then the Jordan canonical form of A has k Jordan blocks (4) If a 4 4 matrix A has eigenvalues and, each of multiplicity, such that dim E = and dime =, then the Jordan canonical form of A has three Jordan blocks (5) If λ,,λ k are k distinct eigenvalues of A with multiplicities m i and dim E λi m i, then A is not diagonalizable (6) For any Jordan block J with eigenvalue λ, dete J = e λ (7) For any square matrix A, A and A T have the same Jordan canonical forms (8) The inverse of any invertible matrix A can be written as a polynomial in A

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