Computing Generalized Eigenvectors. Let λ be an eigenvalue of multiplicity m. Let n be the smallest integer such that rank (nullspace ((A − λI) n )) = m. Let N k denote the number of eigenvalues of rank k. These have the value: N k = rank  nullspace  (A − λI) k  − rank  nullspace  (A − λI) k−1  . One can compute the generalized eigenvectors of a matrix by looping through the following three steps until all the the N k are zero: 1. Select the largest k for which N k is positive. Find a generalized eigenvector x k of rank k which is linearly independent of all the generalized eigenvectors found thus far. 2. From x k generate the chain of eigenvectors {x 1 , x 2 , . . . , x k }. Add this chain to the known generalized eigenvec- tors. 3. Decrement each positive N k by one. Example 15.3.1 Consider the matrix A =   1 1 1 2 1 −1 −3 2 4   . The characteristic polynomial of the matrix is χ(λ) =       1 − λ 1 1 2 1 − λ −1 −3 2 4 − λ       = (1 −λ) 2 (4 − λ) + 3 + 4 + 3(1 −λ) − 2(4 − λ) + 2(1 −λ) = −(λ − 2) 3 . 854 Thus we see that λ = 2 is an eigenvalue of multiplicity 3. A − 2I is A − 2I =   −1 1 1 2 −1 −1 −3 2 2   The rank of the nullspace space of A − 2I is less than 3. (A − 2I) 2 =   0 0 0 −1 1 1 1 −1 −1   The rank of nullspace((A −2I) 2 ) is less than 3 as well, so we have to take one more step. (A − 2I) 3 =   0 0 0 0 0 0 0 0 0   The rank of nullspace((A − 2I) 3 ) is 3. Thus there are generalized eigenvectors of ranks 1, 2 and 3. The generalized eigenvector of rank 3 satisfies: (A − 2I) 3 x 3 = 0   0 0 0 0 0 0 0 0 0   x 3 = 0 We choose the solution x 3 =   1 0 0   . 855 Now to compute the chain generated by x 3 . x 2 = (A − 2I)x 3 =   −1 2 −3   x 1 = (A − 2I)x 2 =   0 −1 1   Thus a set of generalized eigenvectors corresponding to the eigenvalue λ = 2 are x 1 =   0 −1 1   , x 2 =   −1 2 −3   , x 3 =   1 0 0   . Jordan Block. A J ordan block is a square matrix which has the constant, λ, on the diagonal and ones on the first super-diagonal:           λ 1 0 ··· 0 0 0 λ 1 ··· 0 0 0 0 λ . . . 0 0 . . . . . . . . . . . . . . . . . . 0 0 0 . . . λ 1 0 0 0 ··· 0 λ           856 Jordan Canonical Form. A matrix J is in Jordan canonical form if all the elements are zero except for Jordan blocks J k along the diagonal. J =         J 1 0 ··· 0 0 0 J 2 . . . 0 0 . . . . . . . . . . . . . . . 0 0 . . . J n−1 0 0 0 ··· 0 J n         The Jordan canonical form of a matrix is obtained with the similarity transformation: J = S −1 AS, where S is the matrix of the generalized eigenvectors of A and the generalized eigenvectors are grouped in chains. Example 15.3.2 Again consider the matrix A =   1 1 1 2 1 −1 −3 2 4   . Since λ = 2 is an eigenvalue of multiplicity 3, the Jordan canonical form of the matrix is J =   2 1 0 0 2 1 0 0 2   . In Example 15.3.1 we found the generalized eigenvectors of A. We define the matrix with generalized eigenvectors as columns: S =   0 −1 1 −1 2 0 1 −3 0   . 857 We can verify that J = S −1 AS. J = S −1 AS =   0 −3 −2 0 −1 −1 1 −1 −1     1 1 1 2 1 −1 −3 2 4     0 −1 1 −1 2 0 1 −3 0   =   2 1 0 0 2 1 0 0 2   Functions of Matrices in Jordan Canonical Form. The function of an n × n Jordan block is the upper- triangular matrix: f(J k ) =            f(λ) f  (λ) 1! f  (λ) 2! ··· f (n−2) (λ) (n−2)! f (n−1) (λ) (n−1)! 0 f(λ) f  (λ) 1! ··· f (n−3) (λ) (n−3)! f (n−2) (λ) (n−2)! 0 0 f(λ) . . . f (n−4) (λ) (n−4)! f (n−3) (λ) (n−3)! . . . . . . . . . . . . . . . . . . 0 0 0 . . . f(λ) f  (λ) 1! 0 0 0 ··· 0 f(λ)            The function of a matrix in Jordan canonical form is f(J) =         f(J 1 ) 0 ··· 0 0 0 f(J 2 ) . . . 0 0 . . . . . . . . . . . . . . . 0 0 . . . f(J n−1 ) 0 0 0 ··· 0 f(J n )         The Jordan canonical form of a matrix satisfies: f(J) = S −1 f(A)S, 858 where S is the matrix of the generalized eigenvectors of A. This gives us a convenient method for computing functions of matrices. Example 15.3.3 Consider the matrix exponential function e A for our old friend: A =   1 1 1 2 1 −1 −3 2 4   . In Example 15.3.2 we showed that the Jordan canonical form of the matrix is J =   2 1 0 0 2 1 0 0 2   . Since all the derivatives of e λ are just e λ , it is especially easy to compute e J . e J =   e 2 e 2 e 2 /2 0 e 2 e 2 0 0 e 2   We find e A with a similarity transformation of e J . We use the matrix of generalized eigenvec tors found in Example 15.3.2. e A = S e J S −1 e A =   0 −1 1 −1 2 0 1 −3 0     e 2 e 2 e 2 /2 0 e 2 e 2 0 0 e 2     0 −3 −2 0 −1 −1 1 −1 −1   e A =   0 2 2 3 1 −1 −5 3 5   e 2 2 859 15.4 Using the Matrix Exponential The homogeneous differential equation x  (t) = Ax(t) has the solution x(t) = e At c where c is a vector of constants. The solution subject to the initial condition, x(t 0 ) = x 0 is x(t) = e A(t−t 0 ) x 0 . The homogeneous differential equation x  (t) = 1 t Ax(t) has the solution x(t) = t A c ≡ e A Log t c, where c is a vector of constants. The solution subject to the initial condition, x(t 0 ) = x 0 is x(t) =  t t 0  A x 0 ≡ e A Log(t/t 0 ) x 0 . The inhomogeneous problem x  (t) = Ax(t) + f(t), x(t 0 ) = x 0 has the solution x(t) = e A(t−t 0 ) x 0 + e At  t t 0 e −Aτ f(τ) dτ. Example 15.4.1 Consider the system dx dt =   1 1 1 2 1 −1 −3 2 4   x. 860 The general solution of the system of differential equations is x(t) = e At c. In Example 15.3.3 we found e A . At is just a constant times A. The eigenvalues of At are {λ k t} where {λ k } are the eigenvalues of A. The generalized eigenvectors of At are the same as those of A. Consider e Jt . The derivatives of f(λ) = e λt are f  (λ) = t e λt and f  (λ) = t 2 e λt . Thus we have e Jt =   e 2t t e 2t t 2 e 2t /2 0 e 2t t e 2t 0 0 e 2t   e Jt =   1 t t 2 /2 0 1 t 0 0 1   e 2t We find e At with a similarity transformation. e At = S e Jt S −1 e At =   0 −1 1 −1 2 0 1 −3 0     1 t t 2 /2 0 1 t 0 0 1   e 2t   0 −3 −2 0 −1 −1 1 −1 −1   e At =   1 − t t t 2t − t 2 /2 1 −t + t 2 /2 −t + t 2 /2 −3t + t 2 /2 2t −t 2 /2 1 + 2t − t 2 /2   e 2t The solution of the system of differential equations is x(t) =   c 1   1 − t 2t − t 2 /2 −3t + t 2 /2   + c 2   t 1 − t + t 2 /2 2t − t 2 /2   + c 3   t −t + t 2 /2 1 + 2t −t 2 /2     e 2t 861 Example 15.4.2 Consider the Euler equation system dx dt = 1 t Ax ≡ 1 t  1 0 1 1  x. The solution is x(t) = t A c. Note that A is almost in Jordan canonical form. It has a one on the sub-diagonal instead of the super-diagonal. It is clear that a function of A is defined f(A) =  f(1) 0 f  (1) f(1)  . The function f(λ) = t λ has the derivative f  (λ) = t λ log t. Thus the solution of the system is x(t) =  t 0 t log t t  c 1 c 2  = c 1  t t log t  + c 2  0 t  Example 15.4.3 Consider an inhomogeneous system of differential e quations. dx dt = Ax + f(t) ≡  4 −2 8 −4  x +  t −3 −t −2  , t > 0. The general solution is x(t) = e At c + e At  e −At f(t) dt. First we find homogeneous solutions. The characteristic equation for the matrix is χ(λ) =     4 − λ −2 8 −4 − λ     = λ 2 = 0 λ = 0 is an eigenvalue of multiplicity 2. Thus the Jordan canonical form of the matrix is J =  0 1 0 0  . 862 [...]... = 5−λ 3 −2 8 −5 − λ −4 = − 3 + 3 2 − 3 + 1 = −(λ − 1 )3 −4 3 3−λ λ = 1 is an eigenvalue of multiplicity 3 The rank of the null space of A − I is 2 (The second and third rows are multiples of the first.)   4 3 −2 A − I =  8 −6 −4 −4 3 2 887 Thus there are two eigenvectors   4 3 −2 ξ1  8 −6 −4 ξ2  = 0 3 −4 3 2     1 0 (1) 0 , xi(2) =  2  xi = 2 3  Two linearly independent solutions... is x= 3 + 4t −3t e 2 + 4t Both coordinates tend to zero as t → ∞ Method 2 Use the Exponential Matrix The Jordan canonical form of the matrix is J= 3 1 0 3 882 The solution of the initial value problem is x = eAt x0 x = eAt x0 = S eJt S−1 x0 = e−3t t e−3t e−3t 0 1 1/4 1 0 x= 0 1 4 −4 3 2 3 + 4t −3t e 2 + 4t Solution 15.9 We consider an initial value problem   −1 0 0 x = Ax ≡ −4 1 0 x, 3 6 2... for the solution whose second component vanishes to 886 simplify the algebra (A − 2I)ζ = η     1 −1 1 1 ζ1  2 −1 −1  0  = 0 1 3 2 2 3  −ζ1 + 3 = 1, 2ζ1 − 3 = 0,   1 0 ζ= 2 3 1 + 2 3 = 1 A third linearly independent solution is      1 1 0  1  (t2 /2) e2t + 0 t e2t + 0 e2t = 1 2 −1  x (3) 2 (a) We compute the eigenvalues of the matrix χ(λ) = 5−λ 3 −2 8 −5 − λ −4 = − 3. .. x(0) = x0 ≡ 1 3 The Jordan canonical form of the matrix is J= −1 0 0 3 The solution of the initial value problem is x = eAt x0 x = eAt x0 = S eJt S−1 x0 = e−t 0 0 e3t 1 1 1 5 = 1 2 1 4 5 −1 −1 1 1 3 e−t + e3t e−t +5 e3t x= 1 2 1 −t 1 e + 1 2 1 3t e 5 Solution 15.5 We consider an initial value problem   1 1 2 x = Ax ≡  0 2 2 x, −1 1 3   2 x(0) = x0 ≡ 0 1 The Jordan canonical form of the matrix... obtained in part (iii) Solve the third equation, and write down a third independent solution of the system (15.2) 2 Consider the system   5 3 −2 x = Ax =  8 −5 −4 x −4 3 3 (15 .3) (a) Show that λ = 1 is an eigenvalue of multiplicity 3 of the coefficient matrix A, and that there are only two linearly independent eigenvectors, which we may take as xi(1)   1 = 0 , 2  xi(2)  0 = 2  3 Find two... sin( 2t) e−t +c3  2 cos( 2t) + sin( 2t)  e−t √ 1 3 cos( 2t) 3 sin( 2t)  8 73 We apply the initial condition to determine the constants √      2 2 √2 c1 1 −2 −1 2 c2  = 0 c3 0 1 3 0 1 1 5 c 1 = , c 2 = − , c3 = √ 3 9 9 2 The solution subject to the initial condition is √ √ √     2 cos( √ − 4 2 sin( 2t) 2t) √ 2 √ 1 1 x = −2 e−2t +  4 cos( √2t) + √sin( √  e−t 2 2t) 3 6 1 −2 cos(... η is only determined up to a linear combination of xi1 and xi2 Thus we can take the first two components of η to be zero        4 3 −2 0 1 0  8 −6 −4  0  = c1 0 + c2  2  3 2 3 −4 3 2 −2 3 = c1 , −4 3 = 2c2 , c1 = c2 , 888 2 3 = 2c1 − 3c2 c1 3 = − 2 ... e−2t +c2 −1 + ı 2 e(−1−ı 2)t +c3 −1 − ı 2 e(−1+ı 2)t 1 3 3  We can take the real and imaginary parts of the second or third solution to obtain two real-valued solutions √ √ √  √ √  √   √ 2 cos( √2t) + √2 sin( √2t) 2 cos( √ − 2 sin( 2t) 2t) 2+ı √ 2 √ √ √ −1 + ı 2 e(−1−ı 2)t = − cos( 2t) + 2 sin( 2t) e−t +ı  2 cos( 2t) + sin( 2t)  e−t √ √ 3 3 cos( 2t) 3 sin( 2t) √ √ √  √ √  √  ... system (15 .3) Hint, Solution 867 Exercise 15.11 (mathematica/ode/systems/systems.nb) Consider the system of ODE’s dx = Ax, x(0) = x0 dt where A is the constant 3 × 3 matrix   1 1 1 1 −1 A= 2 −8 −5 3 1 Find the eigenvalues and associated eigenvectors of A [HINT: notice that λ = −1 is a root of the characteristic polynomial of A.] 2 Use the results from part (a) to construct eAt and therefore the... constant times xi Thus we look for the solution whose second component vanishes to simplify the algebra (A − 2I)η = xi     −1 1 1 η1 0  2 −1 −1  0  =  1  3 2 2 3 −1  −η1 + 3 = 0, 2η1 − 3 = 1,   1 0 η= 1 3 1 + 2 3 = −1 A second linearly independent solution is  x(2)    0 1 =  1  t e2t + 0 e2t −1 1 (d) To find a third solution we substutite the form x = xi(t2 /2) e2t +ηt e2t . 2I) 3 =   0 0 0 0 0 0 0 0 0   The rank of nullspace((A − 2I) 3 ) is 3. Thus there are generalized eigenvectors of ranks 1, 2 and 3. The generalized eigenvector of rank 3 satisfies: (A − 2I) 3 x 3 =. satisfies: (A − 2I) 3 x 3 = 0   0 0 0 0 0 0 0 0 0   x 3 = 0 We choose the solution x 3 =   1 0 0   . 855 Now to compute the chain generated by x 3 . x 2 = (A − 2I)x 3 =   −1 2 3   x 1 = (A. convenient method for computing functions of matrices. Example 15 .3. 3 Consider the matrix exponential function e A for our old friend: A =   1 1 1 2 1 −1 3 2 4   . In Example 15 .3. 2 we showed