Generalized eigenvector - Wikipedia, the free encyclopedia

1 of 30 18/03/2013 20:00 Generalized eigenvector From Wikipedia, the free encyclopedia In linear algebra, for a matrix A, there may not always exist a full set of linearly independent eigenvectors that form a complete basis a matrix may not be diagonalizable. This happens when the algebraic multiplicity of at least one eigenvalue λ is greater than its geometric multiplicity (the nullity of the matrix, or the dimension of its nullspace). In such cases, a generalized eigenvector of A is a nonzero vector v, which is associated with λ having algebraic multiplicity k 1, satisfying The set of all generalized eigenvectors for a given λ, together with the zero vector, form the generalized eigenspace for λ. Ordinary eigenvectors and eigenspaces are obtained for k=1. Contents 1 For defective matrices 2 Examples 2.1 Example 1 2.2 Example 2 3 Other meanings of the term 4 The Nullity of (A λ I) k 4.1 Introduction 4.2 Existence of Eigenvalues 4.3 Constructive proof of Schur's triangular form 4.4 Nullity Theorem's Proof 5 Motivation of the Procedure 5.1 Introduction 5.2 Notation 5.3 Preliminary Observations 5.4 Recursive Procedure 5.5 Generalized Eigenspace Decomposition 5.6 Powers of a Matrix 5.6.1 using generalized eigenvectors 5.6.2 the minimal polynomial of a matrix 5.6.3 using confluent Vandermonde matrices 5.6.4 using difference equations 5.7 Chains of generalized eigenvectors 5.7.1 Differential equations y = Ay 5.7.2 Revisiting the powers of a matrix 5.8 Ordinary linear difference equations 6 References For defective matrices Generalized eigenvectors are needed to form a complete basis of a defective matrix, which is a matrix in which there are fewer linearly independent eigenvectors than eigenvalues (counting multiplicity). Over an

2 of 30 18/03/2013 20:00 algebraically closed field, the generalized eigenvectors do allow choosing a complete basis, as follows from the Jordan form of a matrix. In particular, suppose that an eigenvalue λ of a matrix A has an algebraic multiplicity m but fewer corresponding eigenvectors. We form a sequence of m eigenvectors and generalized eigenvectors that are linearly independent and satisfy for some coefficients, for. It follows that The vectors can always be chosen, but are not uniquely determined by the above relations. If the geometric multiplicity (dimension of the eigenspace) of λ is p, one can choose the first p vectors to be eigenvectors, but the remaining m p vectors are only generalized eigenvectors. Examples Example 1 Suppose Then there is one eigenvalue λ=1 with an algebraic multiplicity m of 2. There are several ways to see that there will be one generalized eigenvector necessary. Easiest is to notice that this matrix is in Jordan normal form, but is not diagonal, meaning that this is not a diagonalizable matrix. Since there is 1 superdiagonal entry, there will be one generalized eigenvector (or you could note that the vector space is of dimension 2, so there can be only one generalized eigenvector). Alternatively, you could compute the dimension of the nullspace of to be p=1, and thus there are m-p=1 generalized eigenvectors. Computing the ordinary eigenvector is left to the reader (see the eigenvector page for examples). Using this eigenvector, we compute the generalized eigenvector by solving Writing out the values: This simplifies to This simplifies to

3 of 30 18/03/2013 20:00 And has no restrictions and thus can be any scalar. So the generalized eigenvector is, where the * indicates that any value is fine. Usually picking 0 is easiest. Example 2 The matrix has eigenvalues of and with algebraic multiplicities of and, but geometric multiplicities of and. The generalized eigenspaces of are calculated below.

4 of 30 18/03/2013 20:00 This results in a basis for each of the generalized eigenspaces of. Together they span the space of all 5 dimensional column vectors. The Jordan Canonical Form is obtained. where Other meanings of the term The usage of generalized eigenfunction differs from this; it is part of the theory of rigged Hilbert spaces, so that for a linear operator on a function space this may be something different. One can also use the term generalized eigenvector for an eigenvector of the generalized eigenvalue problem The Nullity of (A λ I) k Introduction In this section it is shown, when is an eigenvalue of a matrix with algebraic multiplicity, then the null space of has dimension. Existence of Eigenvalues Consider a nxn matrix A. The determinant of A has the fundamental properties of being n linear and alternating. Additionally det(i) = 1, for I the nxn identity matrix. From the determinant's definition it can be seen that for a triangular matrix T = (t ij ) that

5 of 30 18/03/2013 20:00 det(t) = (t ii ). There are three elementary row operations, scalar multiplication, interchange of two rows, and the addition of a scalar multiple of one row to another. Multiplication of a row of A by α results in a new matrix whose determinant is α det(a). Interchange of two rows changes the sign of the determinant, and the addition of a scalar multiple of one row to another does not affect the determinant. The following simple theorem holds, but requires a little proof. Theorem: The equation A x = 0 has a solution x 0, if and only if det(a) = 0. proof: Given the equation A x = 0 attempt to solve it using the elementary row operations of addition of a scalar multiple of one row to another and row interchanges only, until an equivalent equation U x = 0 has been reached, with U an upper triangular matrix. Since det(u) = ±det(a) and det(u) = (u ii ) we have that det(a) = 0 if and only if at least one u ii = 0. The back substitution procedure as performed after Gaussian Elimination will allow placing at least one non zero element in x when there is a u ii = 0. When all u ii 0 back substitution will require x = 0. Theorem: The equation A x = λ x has a solution x 0, if and only if det( λ I A) = 0. proof: The equation A x = λ x is equivalent to ( λ I A) x = 0. Constructive proof of Schur's triangular form The proof of the main result of this section will rely on the similarity transformation as stated and proven next. Theorem: Schur Transformation to Triangular Form Theorem For any n n matrix A, there exists a triangular matrix T and a unitary matrix Q, such that A Q = Q T. (The transformations are not unique, but are related.) Proof: Let λ 1, be an eigenvalue of the n n matrix A and x be an associated eigenvector, so that A x = λ 1 x. Normalize the length of x so that x = 1. Q should have x as its first column and have its columns an orthonormal basis for C n. Now, A Q = Q U 1, with U 1 of the form:

6 of 30 18/03/2013 20:00 Let the induction hypothesis be that the theorem holds for all (n-1) (n-1) matrices. From the construction, so far, it holds for n = 2. Choose a unitary Q 0, so that U 0 Q 0 = Q 0 U 2, with U 2 of the upper triangular form: Define Q 1 by: Now:

7 of 30 18/03/2013 20:00 Summarizing, U1 Q1 = Q1 U3 with: Now, A Q = Q U 1 and U 1 Q 1 = Q 1 U 3, where Q and Q 1 are unitary and U 3 is upper triangular. Thus A Q Q 1 = Q Q 1 U 3. Since the product of two unitary matrices is unitary, the proof is done. Nullity Theorem's Proof Since from A Q = Q U, one gets A = Q U Q T. It is easy to see (x I A) = Q (x I U) Q T and hence det(x I A) = det(x I U). So the characteristic polynomial of A is the same as that for U and is given by p(x) = (x λ 1 )(x λ 2 )... (x λ n ). (Q unitary) Observe, the construction used in the proof above, allows choosing any order for the eigenvalues of A that will end up as the diagonal elements of the upper triangular matrix U obtained. The algebraic mutiplicity of an eigenvalue is the count of the number of times it occurs on the diagonal.

8 of 30 18/03/2013 20:00 Now. it can be supposed for a given eigenvalue λ, of algebraic multiplicity k, that U has been contrived so that λ occurs as the first k diagonal elements. Place (U λ I) in block form as below. The lower left block has only elements of zero. The β i = λ i λ 0 for i = k+1,..., n. It is easy to verify the following.

9 of 30 18/03/2013 20:00 Where B is the kxk sub triangular matrix, with all elements on or below the diagonal equal to 0, and T is the (n-k)x(n-k) upper triangular matrix, taken from the blocks of (U λ I), as shown below. Now, almost trivially! That is B k has only elements of 0 and T k is triangular with all non zero diagonal elements. Just observe that if a column vector v = <v 1, v 2,..., v k > T, is mutiplied by B, then after the first multiplication the last, k'th, component is zero. After the second multiplication the second to last, k-1'th component is zero, also, and so on. The conclusion that (U λ I) k has rank n-k and nullity k follows. It is only left to observe, since (A λ I) k = Q (U λ I) k Q T, that (A λ I) k has rank n-k and nullity k, also. A unitary, or any other similarity transformation by a non-singular matrix preserves rank. The main result is now proven. Theorem: If λ is an eigenvalue of a matrix A with algebraic multiplicity k, then the null space of (A λ I) k has dimension k. An important observation is that raising the power of (A λ I) above k will not affect the rank and nullity any further.

10 of 30 18/03/2013 20:00 Motivation of the Procedure Introduction In the section Existence of Eigenvalues it was shown that when a nxn matrix A, has an eigenvalue λ, of algebraic multiplicity k, then the null space of (A λ I) k, has dimension k. The Generalized Eigenspace of A, λ will be defined to be the null space of (A λ I) k. Many authors prefer to call this the kernel of (A λ I) k. Notice that if a nxn matrix has eigenvalues λ 1, λ 2,..., λ r with algebraic multiplicities k 1, k 2,..., k r, then k 1 + k 2 +... + k r = n. It will turn out that any two generalized eigenspaces of A, associated with different eigenvalues, will have a trivial intersection of {0}. From this it follows that the generalized eigenspaces of A combined span C n, the set of all n dimensional column vectors of complex numbers. The motivation for using a recursive procedure starting with the eigenvectors of A and solving for a basis of the generalized eigenspace of A, λ using the matix (A λ I), will be expounded on. Notation Some notation is introduced to help abbreviate statements. C n is the vector space of all n dimensional column vectors of complex numbers. The Null Space of A, N(A) = {x: A x = 0}. V W will mean V is a subset of W. V W will mean V is a proper subset of W. A(V) = {y: y = A x, for some x V}. W \ V will mean { x: x W and x is not in V}. The Range of A is A(C n ) and will be denoted by R(A). dim(v) will stand for the dimension of V. {0} will stand for the trivial subspace of C n. Preliminary Observations Throughout this discussion it is assumed that A is a nxn matrix of complex numbers. Since A m x = A (A m-1 x), the inclusions N(A) N(A 2 )... N(A m-1 ) N(A m ), are obvious. Since A m x = A m-1 (A x), the inclusions R(A) R(A 2 )... R(A m-1 ) R(A m ), are clear too.

11 of 30 18/03/2013 20:00 Theorem: When the more trivial case N(A 2 ) = N(A), does not hold, there exists k 2, such that the inclusions, N(A) N(A 2 )... N(A k-1 ) N(A k ) = N(A k+1 ) =..., and R(A) R(A 2 )... R(A k-1 ) R(A k ) = R(A k+1 ) =..., are proper. proof: 0 dim(r(a m+1 )) dim(r(a m )) so eventually dim(r(a m+1 )) = dim(r(a m )), for some m. From the inclusion R(A m+1 ) R(A m ) it is seen that a basis for R(A m+1 ) is a basis for R(A m ) too. That is R(A m+1 ) = R(A m ). Since R(A m+1 ) = A(R(A m )), when R(A m+1 ) = R(A m ), it will be R(A m+2 ) = A(R(A m+1 )) = A(R(A m )) = R(A m+1 ). By the rank nullity theorem, it will also be the case that dim(n(a m+2 )) = dim(n(a m+1 )) = dim(n(a m )), for the same m. From the inclusions N(A m+2 ) N(A m+1 ) N(A m ), it is clear that a basis for N(A m+2 ) is also a basis for N(A m+1 ) and N(A m ). So N(A m+2 ) = N(A m+1 ) = N(A m ). Now, k is the first m for which this happens. Since certain expressions will occur many times in the following, some more notation will be introduced. A λ, k = (A λ I) k N λ, k = N((A λ I) k ) = N(A λ, k ) R λ, k = R((A λ I) k ) = R(A λ, k ) From the inclusions N λ, 1 N λ, 2... N λ, k-1 N λ, k = N λ, k+1 =..., N λ, k \ {0} = (N λ, m \ N λ, m-1 ), for m = 1,..., k and N λ, 0 = {0}, follows. When λ is an eigenvalue of A, in the statement above, k will not exceed the algebraic multiplicity of λ, and can be less. In fact when k would only be 1 is when there is a full set of linearly independent eigenvectors. Let's consider when k 2. Now, x N λ, m \ N λ, m-1, if and only if A λ, m x = 0, and A λ, m-1 x 0. Make the observation that A λ, m x = 0, and A λ, m-1 x 0, if and only if A λ, m-1 A λ, 1 x = 0, and A λ, m-2 A λ, 1 x 0. So, x N λ, m \ N λ, m-1, if and only if A λ, 1 x N λ, m-1 \ N λ, m-2. Recursive Procedure Consider a matrix A, with an eigenvalue λ of algebraic multiplicity k 2, such that there are not k linearly independent eigenvectors associated with λ. It is desired to extend the eigenvectors to a basis for N λ, k. That is a basis for the generalized eigenvectors associated with λ.

12 of 30 18/03/2013 20:00 There exists some 2 r k, such that N λ, 1 N λ, 2... N λ, r-1 N λ, r = N λ, r+1 =..., N λ, r \ {0} = (N λ, m \ N λ, m-1 ), for m = 1,..., r and N λ, 0 = {0},. The eigenvectors are N λ, 1 \ {0}, so let x 1,..., x r1 be a basis for N λ, 1 \ {0}. Note that each N λ, m is a subspace and so a basis for N λ, m-1 can be extended to a basis for N λ, m. Because of this we can expect to find some r 2 = dim(n λ, 2 ) dim(n λ, 1 ) linearly independent vectors x r1 + 1,..., x r1 + r 2 such that x 1,..., x r1, x r1 + 1,..., x r1 + r 2 is a basis for N λ, 2 Now, x N λ, 2 \ N λ, 1, if and only if A λ, 1 x N λ, 1 \ {0}. Thus we can expect that for each x {x r1 + 1,..., x r1 + r 2 }, A λ, 1 x = α 1 x 1 +... + α r1 x r1, for some α 1,..., α r1, depending on x. Suppose we have reached the stage in the construction so that m-1 sets, {x 1,..., x r1 }, {x r1 + 1,..., x r1 + r 2 },..., {x r1 +... + r m-2 + 1,..., x r1 +... + r m-1 } such that x 1,..., x r1, x r1 + 1,..., x r1 + r 2,..., x r1 +... + r m-2 + 1,..., x r1 +... + r m-1 is a basis for N λ, m-1, have been found. We can expect to find some r m = dim(n λ, m ) dim(n λ, m-1 ) linearly independent vectors x r1 +... + r m-1 + 1,..., x r1 +... + r m such that x 1,..., x r1, x r1 + 1,..., x r1 + r 2,..., x r1 +... + r m-1 + 1,..., x r1 +... + r m is a basis for N λ, m Again, x N λ, m \ N λ, m-1, if and only if A λ, 1 x N λ, m-1 \ N λ, m-2. Thus we can expect that for each x {x r1 +... + r m-1 + 1,..., x r1 +... + r m }, A λ, 1 x = α 1 x 1 +... + α r1 +... + r m-1 x r1 +... + r m-1, for some α 1,..., α r1 +... + r m-1, depending on x. Some of the {α r1 +... + r m-2 + 1,..., α r1 +... + r m-1 }, will be non zero, since A λ, 1 x must lie in N λ, m-1 \ N λ, m-2. The procedure is continued until m = r.

13 of 30 18/03/2013 20:00 The α i are not truly arbitrary and must be chosen, accordingly, so that sums α 1 x 1 + α 2 x 2 +... are in the range of A λ, 1. Generalized Eigenspace Decomposition As was stated in the Introduction, if a nxn matrix has eigenvalues λ 1, λ 2,..., λ r with algebraic multiplicities k 1, k 2,..., k r, then k 1 + k 2 +... + k r = n. When V 1 and V 2 are two subspaces, satisfying V 1 V 2 = {0}, their direct sum,, is defined and notated by V 1 V 2 = {v 1 + v 2 : v 1 V 1 and v 2 V 2 }. V 1 V 2 is also a subspace and dim(v 1 V 2 ) = dim(v 1 ) + dim(v 2 ). Since dim(n λi, k i ) = k i, for i = 1, 2,..., r, after it is shown that N λi, k i N λj, k j = {0}, for i j, we have the main result. Theorem: Generalized Eigenspace Decomposition Theorem C n = N λ1, k 1 N λ2, k 2... N λr, k r. This follows easily after we prove the theorem below. Theorem: Let λ be an eigenvalue of A and β λ. Then A β, r (N λ, m \ N λ, m-1 ) = N λ, m \ N λ, m-1, for any positive integers m and r. proof: If x N λ, 1 \ {0}, A λ, 1 x = (A λ I)x = 0, then A x = λ x and A β, 1 x = (A β I)x = (λ β)x. So A β, 1 x N λ, 1 \ {0} and A β, 1 (λ β) 1 x = x. It holds A β, 1 (N λ, 1 \ {0}) = N λ, 1 \ {0}. Now, x N λ, m \ N λ, m-1, if and only if A λ, m x = (A λ I)A λ, m-1 x = 0, and A λ, m-1 x 0. In the case, x N λ, m \ N λ, m-1, A λ, m-1 x N λ, 1 \ 0, and A β, 1 A λ, m-1 x = (λ β) A λ, m-1 x 0. The operators A β, 1 and A λ, m-1 commute. Thus A λ, m (A β, 1 x) = 0 and A λ, m-1 (A β, 1 x) 0, which means A β, 1 x N λ, m \ N λ, m-1. Now, let our induction hypothesis be, A β, 1 (N λ, m \ N λ, m-1 ) = N λ, m \ N λ, m-1,. The relation A β, 1 x = (λ β) x + A λ, 1 x holds.

14 of 30 18/03/2013 20:00 For y N λ, m+1 \ N λ, m, let x = (λ β) -1 y + z. Then A β, 1 x = y + (λ β) -1 A λ, 1 y + (λ β) z + A λ, 1 z = y + (λ β) -1 A λ, 1 y + A β, 1 z. Now, A λ, 1 y N λ, m \ N λ, m-1 and, by the induction hypothesis, there exists z N λ, m \ N λ, m-1 that solves A β, 1 z = (λ β) -1 A λ, 1 y. It follows x N λ, m+1 \ N λ, m and solves A β, 1 x = y. So A β, 1 (N λ, m+1 \ N λ, m ) = N λ, m+1 \ N λ, m,. Repeatedly applying A β, r = A β, 1 A β, r-1 finishes the proof. In fact, from the theorem just proved, for i j, A λi, k i (N λj, k j ) = N λj, k j,. Now, suppose that N λi, k i N λj, k j {0}, for some i j. Choose x N λi, k i N λj, k j 0. Since x N λi, k i, it follows A λi, k i x = 0. Since x N λj, k j, it follows A λi, k i x 0, because A λi, k i preserves dimension on N λj, k j. So it must be N λi, k i N λj, k j = {0}, for i j. This concludes the proof of the Generalized Eigenspace Decomposition Theorem. Powers of a Matrix using generalized eigenvectors Assume A is a nxn matrix with eigenvalues λ 1, λ 2,..., λ r of algebraic multiplicities k 1, k 2,..., k r. For notational convenience A λ, 0 = I. Note that A β, 1 = (λ β)i + A λ, 1. and apply the binomial theorem. A β, s = ((λ β)i + A λ, 1 ) s = s m = 0 ( s m ) (λ β) s m A λ, m When λ is an eigenvalue of algebraic multiplicity k, and x N λ, k, then A λ, m x = 0, for m k, so in this case:

15 of 30 18/03/2013 20:00 A β, s x = min(s, k 1) m = 0 ( s m ) (λ β) s m A λ, m x Since C n = N λ1, k 1 N λ2, k 2... N λr, k r, any x in C n can be expressed as x = x 1 + x 2 +... + x r, with each x i N λi, k i. Hence: A β, s x = r min(s, k i 1) ( s m ) (λ i β) s m A λi, m x i. i = 1 m = 0 The columns of A β, s are obtained by letting x vary across the standard basis vectors. The case A 0, s is the power A s of A. the minimal polynomial of a matrix Assume A is a nxn matrix with eigenvalues λ 1, λ 2,..., λ r of algebraic multiplicities k 1, k 2,..., k r. For each i define α(λ i ), the null index of λ i, to be the smallest positive integer α such that N λi, α = N λi, k i. It is often the case that α(λ i ) < k i. Then p(x) = (x λ i ) α(λ i) is the minimal polynomial for A. To see this note p(a) = A λi,α(λ i ) and the factors can be commuted in any order. So p(a) (N λj, k j ) = {0}, because A λj,α(λ j ) (N λj, k j ) = {0}. Being that C n = N λ1, k 1 N λ2, k 2... N λr, k r, it is clear p(a) = 0. Now p(x) can not be of less degree because A β, 1 (N λj, k j ) = N λj, k j, when β λ j, and so A λj,α(λ j ) must be a factor of p(a), for each j. using confluent Vandermonde matrices An alternative strategy is to use the characteristic polynomial of matrix A. Let p(x) = a 0 + a 1 x + a 2 x 2 +... + a n-1 x n-1 + x n be the characteristic polynomial of A. The minimal polynomial of A can be substituted for p(x) in this discussion, if it is known, and different, to reduce the degree n and the multiplicities of the eigenvalues.

16 of 30 18/03/2013 20:00 Then p(a) = 0 and A n = (a 0 I + a 1 A + a 2 A 2 +... + a n-1 A n-1 ). So A n+m = b m, 0 I + b m, 1 A + b m, 2 A 2 +... + b m, n-1 A n-1, where the b m, 0, b m, 1, b m, 2,..., b m, n-1, satisfy the recurrence relation b m, 0 = a 0 b m-1, n-1, b m, 1 = b m-1, 0 a 1 b m-1, n-1, b m, 2 = b m-1, 1 a 2 b m-1, n-1,..., b m, n-1 = b m-1, n-2 a n-1 b m-1, n-1 with b 0, 0 = b 0, 1 = b 0, 2 =... = b 0, n-2 = 0, and b 0, n-1 = 1. This alone will reduce the number of multiplications needed to calculate a higher power of A by a factor of n 2, as compared to simply multiplying A n+m by A. In fact the b m, 0, b m, 1, b m, 2,..., b m, n-1, can be calculated by a formula. Consider first when A has distinct eigenvalues λ 1, λ 2,..., λ n. Since p(λ i ) = 0, for each i, the λ i satisfy the recurrence relation also. So: The matrix V in the equation is the well studied Vandermonde's, for which formulas for it's determinant and inverse are known. In the case that λ 2 = λ 1, consider instead when λ 1 is near λ 2, and subtract row 1 from row 2, which does not affect the determinant.

17 of 30 18/03/2013 20:00 After dividing the second row by (λ 2 λ 1 ) the determinant will be affected by the removal of this factor and still be non-zero. Taking the limit as λ 1 λ 2, the new system has the second row differentiated. The new system has determinant: In the case that λ 3 = λ 2, also, consider like before when λ 2 is near λ 3, and subtract row 1 from row 3, which does not affect the determinant. Next divide row three by (λ 3 λ 2 ) and then subtract row 2 from the new row 3 and follow by dividing the resulting row 3 by (λ 3 λ 2 ) again. This will affect the determinant by removing a factor of (λ 3 λ 2 ) 2. Each element of row 3 is now of the form

18 of 30 18/03/2013 20:00 The effect is to differentiate twice and multiply by one half. The new system has determinant: If it were that the multiplicity of the eigenvalue was even higher, then the next row would be differentiated three times and mutiplied by 1/3!. The progression is 1/s! f (s), with the constant coming from the coefficients of the derivatives in the Taylor expansion. This being done for each eigenvalue of algebraic multiplicity greater than 1. example The matrix has characteristic polynomial p(x) = (x 1) 2 (x 2) 3. The b m, 0, b m, 1, b m, 2, b m, 3, b m, 4, for which A 5+m = b m, 0 I + b m, 1 A + b m, 2 A 2 + b m, 3 A 3 + b m, 4 A 4,

19 of 30 18/03/2013 20:00 satisfy the confluent Vandermonde system next. using difference equations Returning to the recurrence relation for b m, 0, b m, 1, b m, 2,..., b m, n-1, b m, 0 = a 0 b m-1, n-1, b m, 1 = b m-1, 0 a 1 b m-1, n-1, b m, 2 = b m-1, 1 a 2 b m-1, n-1,..., b m, n-1 = b m-1, n-2 a n-1 b m-1, n-1 with b 0, 0 = b 0, 1 = b 0, 2 =... = b 0, n-2 = 0, and b 0, n-1 = 1. Upon substituting the first relation into the second, b m, 1 = a 0 b m-2, n-1 a 1 b m-1, n-1, and now this one into the next b m, 2 = b m-1, 1 a 2 b m-1, n-1, b m, 2 = a 0 b m-3, n-1 a 1 b m-2, n-1 a 2 b m-1, n-1,..., and so on, the following difference equation is found. b m, n-1 = a 0 b m-n, n-1 a 1 b m-n+1, n-1 a 2 b m-n+2, n-1... a n-2 b m-2, n-1 a n-1 b m-1, n-1 with b 0, n-1 = b 1, n-1 = b 2, n-1 =... = b n-2, n-1 = 0, and b n-1, n-1 = 1. See the subsection on linear difference equations for more explanation. Chains of generalized eigenvectors Some notation and results from previous sections are restated. A is a nxn matrix of complex numbers. A λ, k = (A λ I) k N λ, k = N((A λ I) k ) = N(A λ, k ) For V 1 V 2 = {0}, V 1 V 2 = {v 1 + v 2 : v 1 V 1 and v 2 V 2 }.

20 of 30 18/03/2013 20:00 Assume A has eigenvalues λ 1, λ 2,..., λ r of algebraic multiplicities k 1, k 2,..., k r. For each i define α(λ i ), the null index of λ i, to be the smallest positive integer α such that N λi, α = N λi, k i. It is always the case that α(λ i ) k i. When α(λ) 2, N λ, 1 N λ, 2... N λ, α-1 N λ, α = N λ, α+1 =..., N λ, α \ {0} = (N λ, m \ N λ, m-1 ), for m = 1,..., α and N λ, 0 = {0}. x N λ, m \ N λ, m-1, if and only if A λ, 1 x N λ, m-1 \ N λ, m-2 Define a chain of generalized eigenvectors to be a set { x 1, x 2,..., x m } such that x 1 N λ, m \ N λ, m-1, and x i+1 = A λ, 1 x i. Then x m 0 and A λ, 1 x m = 0. When x 1 N λ, 1 \ {0}, {x 1 } can be, for the sake of not requiring extra terminology, considered trivially a chain. When a disjoint collection of chains combined form a basis set for N λ, α(λ), they are often referred to as Jordan chains and are the vectors used for the columns of a transformation matrix in the Jordan canonical form. When a disjoint collection of chains that combined form a basis set, is needed that satisfy β i+1 x i+1 = A λ, 1 x i, for some scalars β i, chains as already defined can be scaled for this purpose. What will be proven here is that such a disjoint collection of chains can always be constructed. Before the proof is started, recall a few facts about direct sums. When the notation V 1 V 2 is used, it is assumed V 1 V 2 = {0}. For x = v 1 + v 2 with v 1 V 1 and v 2 V 2, then x = 0, if and only if v 1 = v 2 = 0. In the discussion below δ i = dim(n λ, i ) dim(n λ, i 1 ), with δ 1 = dim(n λ, 1 ). First consider when N λ, 2 \ N λ, 1 {0}, Then a basis for N λ, 1 can be extended to a basis for N λ, 2. If δ 2 = 1, then there exists x 1 N λ, 2 \ N λ, 1, such that N λ, 2 = N λ, 1 span{x 1 }. Let x 2 = A λ, 1 x 1. Then x 2 N λ, 1 \ {0}, with x 1 and x 2 linearly independent. If dim(n λ, 2 ) = 2, since {x 1, x 2 } is a chain we are through. Otherwise x 1, x 2 can be extended to a basis x 1, x 2,..., x δ1 for N λ, 2. The sets {x 1, x 2 }, {x 3 },..., {x δ1 } form a disjoint collection of chains. In the case that δ 2 > 1, then there exist

21 of 30 18/03/2013 20:00 linearly independent x 1, x 2,..., x δ2 N λ, 2 \ N λ, 1, such that N λ, 2 = N λ, 1 span{x 1, x 2,..., x δ2 }. Let y i = A λ, 1 x i. Then y i N λ, 1 \ {0}, for i = 1, 2,..., δ 2. To see the y 1, y 2,..., y δ2 are linearly independent, assume that for some β 1, β 2,..., β δ2, that β 1 y 1 + β 2 y 2 +... + β δ2 y δ2 = 0, Then for x = β 1 x 1 + β 2 x 2 +... + β δ2 x δ2, x N λ, 1, and x span{x 1, x 2,..., x δ2 }, which implies that x = 0, and β 1 = β 2 =... = β δ2 = 0. Since span{y 1, y 2,..., y δ2 } N λ, 1, the vectors x 1, x 2,..., x δ2, y 1, y 2,..., y δ2 are a linearly independent set. If δ 2 = δ 1, then the sets {x 1, y 1 }, {x 2, y 2 },..., {x δ2, y δ2 } form a disjoint collection of chains that when combined are a basis set for N λ, 2. If δ 1 > δ 2, then x 1, x 2,..., x δ2, y 1, y 2,..., y δ2 can be extended to a basis for N λ, 2 by some vectors x δ2 +1,..., x δ1 in N λ, 1, so that {x 1, y 1 }, {x 2, y 2 },..., {x δ2, y δ2 }, {x δ2 +1},..., {x δ1 } forms a disjoint collection of chains. To reduce redundancy, in the next paragraph, when δ = 1 the notation x 1, x 2,..., x δ will be understood simply to mean just x 1 and when δ = 2 to mean x 1, x 2. So far it has been shown that, if linearly independent x 1, x 2,..., x δ2 N λ, 2 \ N λ, 1, are chosen, such that N λ, 2 = N λ, 1 span{x 1, x 2,..., x δ2 }, then there exists a disjoint collection of chains with each of the x 1, x 2,..., x δ2 being the first member or top of one of the chains. Furthermore, this collection of vectors, when combined, forms a basis for N λ, 2. Now, let the induction hypothesis be that, if linearly independent x 1, x 2,..., x δm N λ, m \ N λ, m 1, are chosen, such that N λ, m = N λ, m 1 span{x 1, x 2,..., x δm }, then there exists a disjoint collection of chains with each of the x 1, x 2,..., x δm being the first member or top of one of the chains. Furthermore, this collection of vectors, when combined, forms a basis for N λ, m. Consider m < α(λ). A basis for N λ, m can always be extended to a basis for N λ, m+1. So linearly independent x 1, x 2,..., x δm+1 N λ, m+1 \ N λ, m, such that N λ, m+1 = N λ, m span{x 1, x 2,..., x δm+1 }, can be chosen. Let y i = A λ, 1 x i. Then y i N λ, m \ N λ, m 1, for i = 1, 2,..., δ m+1. To see the y 1, y 2,..., y δm+1 are linearly independent, assume that for some β 1, β 2,..., β δm+1,

22 of 30 18/03/2013 20:00 that β 1 y 1 + β 2 y 2 +... + β δm+1 y δm+1 = 0, Then for x = β 1 x 1 + β 2 x 2 +... + β δm+1 x δm+1, x N λ, 1, and x span{x 1, x 2,..., x δm+1 }, which implies that x = 0, and β 1 = β 2 =... = β δm+1 = 0. In addition, span{y 1, y 2,..., y δm+1 } N λ, m 1 = {0}. To see this assume that for some β 1, β 2,..., β δm+1, that β 1 y 1 + β 2 y 2 +... + β δm+1 y δm+1 N λ, m 1 Then for x = β 1 x 1 + β 2 x 2 +... + β δm+1 x δm+1, x N λ, m, and x span{x 1, x 2,..., x δm+1 }, which implies that x = 0, and β 1 = β 2 =... = β δm+1 = 0. The proof is nearly done. At this point suppose that b 1, b 2,..., b dm 1 is any basis for N λ, m 1. Then B = span{b 1, b 2,..., b dm 1 } span{y 1, y 2,..., y δm+1 } is a subspace of N λ, m. If B N λ, m, then b 1, b 2,..., b dm 1, y 1, y 2,..., y δm+1 can be extended to a basis for N λ, m, by some set of vectors z 1, z 2,..., z (δm δ m+1 ), in which case N λ, m = N λ, m 1 span{y 1, y 2,..., y δm+1 } span{z 1, z 2,..., z (δm δ m+1 )}. If δ m = δ m+1, then N λ, m = N λ, m 1 span{y 1, y 2,..., y δm+1 } or if δ m > δ m+1, then N λ, m = N λ, m 1 span{z 1, z 2,..., z (δm δ m+1 ), y 1, y 2,..., y δm+1 } In either case apply the induction hypothesis to get that there exists a disjoint collection of chains with each of the y 1, y 2,..., y δm+1 being the first member or top of one of the chains. Furthermore, this collection of vectors, when combined, forms a basis for N λ, m. Now, y i = A λ, 1 x i, for i = 1, 2,..., δ m+1, so each of the chains beginning with y i can be extended upwards into N λ, m+1 \ N λ, m to a chain beginning with x i. Since N λ, m+1 = N λ, m span{x 1, x 2,..., x δm+1 }, the combined vectors of the new chains form a basis for N λ, m+1. Differential equations y = Ay Let A be a n n matrix of complex numbers and λ an eigenvalue of A, with associated eigenvector x. Suppose y(t) is a n dimensional vector valued function, sufficiently smooth, so that y (t) is continuous. The restriction that y(t) be smooth can be relaxed somewhat, but is not the main focus of this discussion. The solutions to the equation y (t) = Ay(t) are sought. The first observation is that y(t) = e λt x will be a solution. When A does not have n linearly independent

23 of 30 18/03/2013 20:00 eigenvectors, solutions of this kind will not provide the total of n needed for a fundamental basis set. In view of the existence of chains of generalized eigenvectors seek a solution of the form 'y(t) = e λt x 1 + t'e λt x 2, then y (t) =' λ e λt x 1 + e λt x 2 + λ t e λt x 2 =' e λt (λ x 1 + x 2 )'''' + t e λt (λx 2 ) and Ay(t) = e λt A x 1 + t e λt A x 2. In view of this, y(t) will be a solution to y (t) = Ay(t), when 'A x 1 = λ x 1 + x 2 ' and A x 2 = λ x 2. That is when (A λ I)x 1 = x 2 and (A λ I)x 2 = 0. Equivalently, when {x 1, x 2 } is a chain of generalized eigenvectors. Continuing with this reasoning seek a solution of the form y(t) = e λt x 1 + t e λt x 2 + t 2 e λt x 3, then y (t) = λ e λt x 1 + e λt x 2 + λ t e λt x 2 + 2 t e λt x 3 + λ t 2 e λt x 3 = e λt (λ x 1 + x 2 ) + t e λt (λ x 2 + 2 x 3 ) + t 2 e λt (λ x 3 )' and Ay(t) = e λt A x 1 + t e λt A x 2 + t 2 e λt A x 3. Like before, y(t) will be a solution to y (t) = Ay(t), when 'A x 1 = λ x 1 + x 2 ', 'A x 2 = λ x 2 + 2 x 3 ', and A x 3 = λ x 3. That is when (A λ I)x 1 = x 2, (A λ I)x 2 = 2 x 3, and (A λ I)x 3 = 0. Since it will hold (A λ I)(2 x 3 ) = 0, also, equivalently, when {x 1, x 2, 2 x 3 } is a chain of generalized eigenvectors. More generally, to find the progression, seek a solution of the form y(t) = e λt x 1 + t e λt x 2 + t 2 e λt x 3 + t 3 e λt x 4 +... + t m 2 e λt x m 1 + t m 1 e λt x m, then y (t) = λ e λt x 1 + e λt x 2 + λ t e λt x 2 ' + 2 t e λt x 3 ' + λ t 2 e λt x 3 ' + 3 t 2 e λt x 4 ' + λ t 3 e λt x 4 ' +...' + (m 2)t m 3 e λt x m 1 ' + λ t m 2 e λt x m 1 ' + (m 1)t m 2 e λt x m ' + λ t m 1 e λt x m =' e λt (λ x 1 + x 2 ) + t e λt (λ x 2 + 2 x 3 ) + t 2 e λt (λ x 3 + 3 x 4 ) + t 3 e λt (λ x 4 + 4 x 5 ) +... + t m 3 e λt (λ x m 2 + (m 2) x m 1 ) + t m 2 e λt (λ x m 1 + (m 1) x m ) + t m 1 e λt (λ x m )' and Ay(t) = e λt A x 1 + t e λt A x 2 + t 2 e λt A x 3 + t 3 e λt A x 4 +... + t m 2 e λt A x m 1 + t m 1 e λt A x m. Again, y(t) will be a solution to y (t) = Ay(t), when 'A x 1 = λ x 1 + x 2 ', A x 2 = λ x 2 + 2 x 3, A x 3 = λ x 3 + 3 x 4, A x 4 = λ x 4 + 4 x 5, A x m 2 = λ x m 2 + (m 2) x m 1, A x m 1 = λ x m 1 + (m 1) x m,

24 of 30 18/03/2013 20:00 and A x m = λ x m. That is when (A λ I)x 1 = x 2, (A λ I)x 2 = 2 x 3, (A λ I)x 3 = 3 x 4, (A λ I)x 4 = 4 x 5,..., (A λ I)x m 2 = (m 2) x m 1, (A λ I)x m 1 = (m 1) x m, and (A λ I)x m = 0. Since it will hold (A λ I)((m 1)! x 3 ) = 0, also, equivalently, when {x 1, 1! x 2, 2! x 3, 3! x 4,..., (m 2)! x m 1, (m 1)! x m } is a chain of generalized eigenvectors. Now, the basis set for all solutions will be found through a disjoint collection of chains of generalized eigenvectors of the matrix A. Assume A has eigenvalues λ 1, λ 2,..., λ r of algebraic multiplicities k 1, k 2,..., k r. For a given eigenvalue λ i there is a collection of s, with s depending on i, disjoint chains of generalized eigenvectors C i,1 = { 1 z 1, 1 z 2,..., 1 z j1 }, C i,2 = { 2 z 1, 2 z 2,..., 2 z j2 },..., C i,js(i) = { s z 1, s z 2,..., s z js }, that when combined form a basis set for N λi, k i. The total number of vectors in this set will be j1 + j2 +... + js = k i. Sets in this collection may have only one or two members so in this discussion understand the notation { β z 1, β z 2,..., β z jβ } will mean { β z 1 } when jβ = 1, and { β z 1, β z 2 } when jβ = 2, and so forth. Being that this notation is cumbersome with many indices, in the next paragraphs any particular C i,β, when more explanation is not needed, may just be notated as C = {z 1, z 2,..., z j }. For each such of these chain sets, C = {z 1, z 2,..., z j } the sets {x j }, {x j 1, x j }, {x j 2, x j 1, x j },..., {z 2, z 3,..., z j }, {z 1, z 2,..., z j } are also chains. This notation being understood to mean when C = {z 1 } just {z 1 }, when C = {z 1, z 2 } just {z 2 }, {z 1, z 2 } and when C = {z 1, z 2, z 2 } just {z 3 }, {z 2, z 3 }, {z 1, z 2, z 3 }, and so on. The conclusion of the top of the discussion was that y(t) = e λt x 1, is a solution when {x 1 } is a chain. y(t) = e λt x 1 + t e λt x 2, is a solution when {x 1, 1! x 2 } is a chain. y(t) = e λt x 1 + t e λt x 2 + t 2 e λt x 3, is a solution when {x 1, 1! x 2,, 2! x 3 } is a chain. The progression continues to y(t) = e λt x 1 + t e λt x 2 + t 2 e λt x 3 + t 3 e λt x 4 +... + t m 2 e λt x m 1 + t m 1 e λt x m,

25 of 30 18/03/2013 20:00 is a solution when {x 1, 1! x 2, 2! x 3, 3! x 4,..., (m 2)! x m 1, (m 1)! x m }, is a chain of generalized eigenvectors. In light of the preceding calculations, all that must be done is to provide the proper scaling for each of the chains arising from the set C = {z 1, z 2,..., z j }. The progression for the solutions is given by y(t) = e λt z j, for chain {z j } y(t) = e λt z j 1 + (1 1!) t e λt z j, for chain {z j 1, 1!(1 1!) z j } y(t) = e λt z j 2 + (1 1!) t e λt z j 1 + (1 2!) t 2 e λt z j, for chain {z j 2, 1!(1 1!) z j 1, 2!(1 2!) z j } y(t) = e λt z j 3 + (1 1!) t e λt z j 2 + (1 2!) t 2 e λt z j 1 + (1 3!) t 3 e λt z j, for chain {z j 3, 1!(1 1!) z j 2, 2!(1 2!) z j 1, 3!(1 3!) z j }, and so on until, y(t) = e λt z 1 + (1 1!) t e λt z 2 + (1 2!) t 2 e λt z 3 +... + (1 (j 1)!) t j 1 e λt z j, for the chain of generalized eigenvectors, {z 1, 1!(1 1!) z 2, 2!(1 2!) z 3,..., (j 2)!(1 (j 2)!) x j 1, (j 1)!(1 (j 1)!) z j }. What is left to show is that when all the solutions constructed from the chain sets, as described, are considered, they form a fundamental set of solutions. To do this it has to be shown that there are n of them and that they are linearly independent. Reiterating, for a given eigenvalue λ i there is a collection of s, with s depending on i, disjoint chains of generalized eigenvectors C i,1 = { 1 z 1, 1 z 2,..., 1 z j1(i) }, C i,2 = { 2 z 1, 2 z 2,..., 2 z j2(i) },..., C i,js(i) = { s(i) z 1, s(i) z 2,..., s(i) z js(i) }, that when combined form a basis set for N λi, k i. The total number of vectors in this set will be j1(i) + j2(i) +... + js(i) = k i. Thus the total number of all such basis vectors and so solutions is k 1 + k 2 +... + k r = n. Each solution is one of the forms y(t) = e λt x 1, y(t) = e λt x 1 + t e λt x 2, y(t) = e λt x 1 + t e λt x 2 + t 2 e λt x 3, y(t) = e λt x 1 + t e λt x 2 + t 2 e λt x 3 +... Now each basis vector v j, for j = 1, 2,..., n; of the combined set of generalized eigenvectors, occurs as x 1 in one of the expressions immediately above precisely once. That is, for each j, there is one y j (t) = e λt v j +... Since y j (0) = e λ0 v j = v j, the set of solutions are linearly independent at t = 0. Revisiting the powers of a matrix

As a notational convenience A λ, 0 = I. Note that A = λ I + A λ, 1. and apply the binomial theorem. A s = (λ I + A λ, 1 ) s = s r = 0 ( s r ) λ s r A λ, r Assume λ is an eigenvalue of A, and let { x 1, x 2,..., x m } be a chain of generalized eigenvectors such that x 1 N λ, m \ N λ, m-1, x i+1 = A λ, 1 x i,. x m 0, and A λ, 1 x m = 0. Then x r+1 = A λ, r x 1, for r = 0, 1,..., m-1. A s x 1 = s r = 0 ( s r ) λ s r A λ, r x 1 = s r = 0 ( s r ) λ s r x r+1 So for s m 1 A s x 1 = s r = 0 ( s r ) λ s r x r+1 and for s m 1, since A λ, m x 1 = 0, A s x 1 = m 1 r = 0 ( s r ) λ s r x r+1. Ordinary linear difference equations Ordinary linear difference equations are equations of the sort: y n = a y n 1 + b y n = a y n 1 + b y n 2 + c or more generally, y n = a m y n 1 + a m 1 y n 2 +... + a 2 y n m + 1 + a 1 y n m + a 0 with initial conditions y 0, y 1, y 2,..., y m 2, y m 1. A case with a 1 = 0 can be excluded, since it represents an equation of less degree. They have a characteristic polynomial p(x) = x m a m x m 1 a m 1 x m 2... a 2 x a 1. 26 of 30 18/03/2013 20:00

27 of 30 18/03/2013 20:00 To solve a difference equation it is first observed, if y n and z n are both solutions, then (y n z n ) is a solution of the homogeneous equation: y n = a m y n 1 + a m 1 y n 2 +... + a 2 y n m + 1 + a 1 y n m. So a particular solution to the difference equation must be found together with all solutions of the homogeneous equation to get the general solution for the difference equation. Another observation to make is that, if y n is a solution to the inhomogeneous equation, then z n = y n+1 y n is also a solution to the homogeneous equation. So all solutions of the homogeneous equation will be found first. When β is a root of p(x) = 0, then it is easily seen y n = β n is a solution to the homogeneous equation since y n a m y n 1 a m 1 y n 2... a 2 y n m + 1 a 1 y n m, becomes upon the substitution y n = β n, β n a m β n 1 a m 1 β n 2... a 2 β n m + 1 a 1 β n m = β n m (β m a m β m 1 a m 1 β m 2... a 2 β a 1 ) = β n m p(β) = 0. When β is a repeated root of p(x) = 0, then y n = nβ n 1 is a solution to the homogeneous equation since nβ n 1 a m (n 1)β n 2 a m 1 (n 2)β n 3... a 2 (n m + 1)β n m a 1 (n m)β n m 1 = (n m)β n m 1 (β m a m β m 1 a m 1 β m 2... a 2 β a 1 ) + β n m 1 (mβ m 1 (m 1)a m β m 2 (m 2)a m 1 β m 3... 2a 3 β a 2 ) == (n m)β n m 1 p(β) + β n m 1 p (β) == 0. After reaching this point in the calculation the mystery is solved. Just notice when β is a root of p(x) = 0 with mutiplicity k, then for s = 1, 2,..., k 1 d s (β n m p(β))/dβ s = 0. Referring this back to the original equation β n a m β n 1 a m 1 β n 2... a 2 β n m + 1 a 1 β n m it is seen that y n = d s (β n )/dβ s are solutions to the homogeneous equation. For example, if β is a root of multiplicity 3, then y n = n(n 1)β n 2 is a solution. In any case this gives m linearly independent solutions to the homogeneous equation. To look for a particular solution first consider the simpliest equation.

28 of 30 18/03/2013 20:00 y n = a y n 1 + b. It has a particular solution y p,n given by y p,0 = 0, y p,1 = b, y p,2 = (1 + a)b,..., y p,n = (1 + a + a 2 +... + a n 1 )b,...,. It's homogeneous equation y n = a y n 1 has solutions y n = a n y 0. So z n = y n+1 y n = a n b can be telescoped to get y n = (y n y n 1 ) + (y n 1 y n 2 ) +... + (y 2 y 1 ) + (y 1 y 0 ) + y 0 = z n 1 + z n 2 +... + z 1 + z 0 + y 0 = (1 + a + a 2 +... + a n 1 )b, the particular solution with y 0 = 0. Now, returning to the general problem, the equation y n = a m y n 1 + a m 1 y n 2 +... + a 2 y n m + 1 + a 1 y n m + a 0. When y p,n is a particular solution with y p,0 = 0, then z n = y p,n+1 y p,n is a solution to the homogeneous equation with z 0 = y p,1. So z n = y p,n+1 y p,n can be telescoped to get y p,n = (y p,n y p,n 1 ) + (y p,n 1 y p,n 2 ) +... + (y p,2 y p,1 ) + (y p,1 y p,0 ) + y p,0 = z n 1 + z n 2 +... + z 1 + z 0 Considering y p,m = a m y p,m 1 + a m 1 y p,m 2 +... + a 2 y p,1 + a 1 y p,0 + a 0. and rewriting the equation in the z i z m 1 + z m 2 +... + z 1 + z 0 = (a m ) ( z m 2 + z m 3 +... + z 1 + z 0 ) + (a m 1 ) ( z m 3 + z m 4 +... + z 1 + z 0 ) + (a m 2 ) ( z m 4 + z m 5 +... + z 1 + z 0 ) + + (a 3 ) ( z 1 + z 0 ) + (a 2 ) ( z 0 ) + (a 0 ) and z m 1 = (a m 1) z m 2 + (a m + a m 1 1) z m 3 + (a m + a m 1 + a m 2 1) z m 4 + + (a m + a m 1 +... + a 4 + a 3 1) z 1 + (a m + a m 1 +... + a 3 + a 2 1) z 0 + (a 0 ). Since a solution of the homogeneous equation can be found for any initial conditions z 0, z 1, z 2,..., z m 2, z m 1. reasoning conversely find such z i satisfying the equation, just before and define y p,n by the relation y p,0 = 0, y p,n = z n 1 + z n 2 +... + z 1 + z 0

One choice is, for example, z m 1 = a 0, z 0 = z 1 = z 2 =... = z m 2 = 0. This solution solves the problem for all initial values equal to zero. The general solution to the inhomogeneous equation is given by y n = y p,n + γ 1 w(1) n + γ 2 w(2) n +... + γ m 1 w(m 1) n + γ m w(m) n where w(1) n, w(2) n,..., w(m 1) n, w(m) n are a basis for the homogeneous equation, and γ 1, γ 2,..., γ m 1, γ m are scalars. example y n = 8 y n 1 25 y n 2 + 38 y n 3 28 y n 4 + 8 y n 5 + 1 with initial conditions y 0 = 0, y 1 = 0, y 2 = 0, y 3 = 0, and y 4 = 0. The characteristic polynomial for the equation is p(x) = x 5 8x 4 + 25x 3 38x 2 + 28x 8 = (x 1) 2 (x 2) 3. The homogeneous equation has independent solutions w1 n = 1 n = 1, w2 n = n 1 n 1 = n, and w3 n = 2 n, w4 n = n 2 n 1, w5 n = n(n 1) 2 n 2. The solution to the homogeneous equation z n = 3 w1 n w2 n + 3 w3 n 2 w4 n + ½ w5 n satisfies the initial conditions z 4 = 1, z 0 = z 1 = z 2 = z 3 = 0. A particular solution can be found by y p,0 = 0, y p,n = z n 1 + z n 2 +... + z 1 + z 0. Calculating sums: w1 = w1 n 1 + w1 n 2 +... + w1 1 + w1 0 = n. w2 = w2 n 1 + w2 n 2 +... + w2 1 + w2 0 = (n 1)n / 2. w3 = w3 n 1 + w3 n 2 +... + w3 1 + w3 0 = 2 n 1. Sums of these kinds are found by differentiating (x n 1) / (x 1). w4 = w4 n 1 + w4 n 2 +... + w4 1 + w4 0 = (n 2)2 n 1 + 1. w5 = w5 n 1 + w5 n 2 +... + w5 1 + w5 0 = (n 2 5n + 8)2 n 2 2. Now, y p,n = 3 w1 n w2 n + 3 w3 n 2 w4 n + ½ w5 n solves the initial value problem of this example. 29 of 30 18/03/2013 20:00

30 of 30 18/03/2013 20:00 At this point it is worthwhile to notice that all the terms that are combinations of scalar multiples of basis elements can be removed. These are any multiples of 1, n, 2 n, n 2 n 1, and n 2 2 n 2. So instead the particular solution next, may be preferred. y p,n = ½ n 2. This solution has non zero initial values, which must be taken into account. y 0 = 0, y 1 = 1 2, y 2 = 2, y 3 = 9 2, and y 4 = 8. References Axler, Sheldon (1997). Linear Algebra Done Right (2nd ed.). Springer. ISBN 978-0-387-98258-8. Retrieved from "http://en.wikipedia.org/w/index.php?title=generalized_eigenvector&oldid=544346744" Categories: Linear algebra This page was last modified on 15 March 2013 at 11:25. Text is available under the Creative Commons Attribution-ShareAlike License; additional terms may apply. See Terms of Use for details. Wikipedia is a registered trademark of the Wikimedia Foundation, Inc., a non-profit organization.