205 Math Supplement
c 2006, Steven H. Weintraub
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Mathematics 205 Supplement
A. Jordan Canonical Form In order to understand and be able to use Jordan Canonical Form, we must introduce a new concept, that of a generalized eigenvector. Definition A.1. If v = 0 is a vector such that, for some , (A - I)k (v) = 0 for some positive integer k, then v is a generalized eigenvector of A associated to the eigenvalue . The smallest k with (A - I)k (v) = 0 is the index of the generalized eigenvector v. Let us note that if v is a generalized eigenvector of index 1, then (A - I)(v) = 0 (A)v = (I)v Av = v and so v is an (ordinary) eigenvector. Recall that, for an eigenvalue of A, E is the eigenspace of , E = {v | Av = v} = {v | (A - I)v = 0}. ~ We let E denote the generalized eigenspace of , ~ E = {v | (A - I)k (v) = 0 for some k}. ~ It is easy to check that E is a subspace. Since every eigenvector is a generalized eigenvector, we see that ~ E E . The following result (which we shall not prove) is an important fact about generalized eigenspaces. ~ Proposition A.2. Let be an eigenvalue of A of multiplicity m. Then E is a subspace of dimension m.
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Example A.3. Let A be the matrix A =
0 1 1 . Then, as you can check, if u= , -4 4 2 then (A - 2I)u = 0, so u is an eigenvector of A with associated eigenvalue 2 (and hence a generalized eigenvector of index 1 of A with associated eigenvalue 2). On the other hand, 1 if v = , then (A - 2I)2 v = 0 but (A - 2I)v = 0, so v is a generalized eigenvector of 0 index 2 of A with associated eigenvalue 2. In this case, as you can check, the vector u is a basis for the eigenspace E2 , so E2 = { cu | c R} is 1-dimensional. On the other hand, u and v are both generalized eigenvectors associated to the eigenvalue 2, and are linearly independent (the equation c1 u+c2 v = 0 only has the solution c1 = c2 = 0, ~ ~ as you can readily check), so E2 has dimension at least 2. Since E2 is a subspace of R2 , it ~ must have dimension exactly 2, and E2 = R2 (and { u, v } is indeed a basis for R2 ).
Let us next consider a generalized eigenvector vk of index k associated to an eigenvalue , and set vk-1 = (A - I)vk . We claim that vk-1 is a generalized eigenvector of index k - 1 associated to the eigenvalue . To see this, note that (A - I)k-1 vk-1 = (A - I)k-1 (A - I)vk = (A - I)k vk = 0 but (A - I)k-2 vk-1 = (A - I)k-2 (A - I)vk = (A - I)k-1 vk = 0. Proceeding in this way, we may set vk-2 = (A - I)vk-1 = (A - I)2 vk vk-3 = (A - I)vk-2 = (A - I)2 vk-1 = (A - I)3 vk . . . v1 = (A - I)v2 = = (A - I)k-1 vk and note that each vi is a generalized eigenvector of index i associated to the eigenvalue . A collection of generalized eigenvectors obtained in this way gets a special name.
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Definition A.4. If { v1 , . . . , vk } is a set of generalized eigenvectors associated to the eigenvalue of A, such that vk is a generalized eigenvector of index k and also vk-1 =(A - I)vk , , vk-2 = (A - I)vk-1 , vk-3 = (A - I)vk-2 ,
v2 = (A - I)v3 ,
v1 = (A - I)v2 ,
then { v1 , . . . , vk } is called a chain of generalized eigenvectors of length k. The vector vk is called the top of the chain and the vector v1 (which is an ordinary eigenvector) is called the bottom of the chain. Example A.5. Let us return to example 3. We saw there that v= eigenvector of index 2 of A= 1 is a generalized 0
0 1 associated to the eigenvalue 2. Let us set -4 4 v2 = v = 1 . 0
Then v1 = (A - 2I)v2 = -2 -4
is a generalized eigenvector of index 1 (i.e., an ordinary eigenvector), and { v1 , v2 } is a chain of length 2. Remark A.6. It is important to note that a chain of generalized eigenvectors { v1 , . . . , vk } is entirely determined by the vector vk at the top of the chain. For once we have chosen vk , there are no other choices to be made: the vector vk-1 is determined by the equation vk-1 = (A-I)vk ; then the vector vk-2 is determined by the equation vk-2 = (A-I)vk-1 ; etc. With this concept in hand, let us return to Jordan Canonical Form. As we have seen, a matrix J in Jordan Canonical Form has a number of blocks B1 , B2 , . . . , Bm , called Jordan blocks, along the diagonal. Let us begin our analysis with the case when J consists of a single Jordan block. So suppose J is a k-by-k matrix
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1 1 0 1 . J = .. .. . . 0 1 Then 0 1 0 1 0 1 . J - I = .. .. . . 0 1 0 1 0 0 0 0 1 0 0 0, e2 = 0, e3 = 1, . . . , ek = 0. Let e1 = . . . . . . . . . . . . 0 0 0 1 Then direct calculation shows: (J - I)ek = ek-1 (J - I)ek-1 = ek-2 . . . (J - I)e2 = e1 (J - I)e1 = 0 and so we see that { e1 , . . . , ek } is a chain of generalized eigenvectors. We also note that { e1 , . . . , ek } is a basis for Rk , and so ~ E = Rk .
We first see that the situation is very analogous when we consider any k-by-k matrix with a single chain of generalized eigenvectors of length k. Proposition A.7. Let { v1 , . . . , vk } be a chain of generalized eigenvectors of length k associated to the eigenvalue of a matrix A. Then { v1 , . . . , vk } is linearly independent.
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Proof. Suppose we have a linear combination c1 v1 + c2 v2 + + ck-1 vk-1 + ck vk = 0. We must show each ci = 0. By the definition of a chain, vk-i = (A - I)i vk for each i, so we may write this equation as c1 (A - I)k-1 vk + c2 (A - I)k-2 vk + + ck-1 (A - I)vk + ck vk = 0. Now let us multiply this equation on the left by (A - I)k-1 . Then we obtain the equation c1 (A - I)2k-2 vk + c2 (A - I)2k-3 vk + + ck-1 (A - I)k vk + ck (A - I)k-1 vk = 0. Now (A - I)k-1 vk = v1 = 0. However, (A - I)k vk = 0, and then also (A - I)k+1 vk = (A - I)(A - I)k vk = (A - I)(0) = 0, and then similarly (A - I)k+2 vk = 0, . . . , (A - I)2k-2 vk = 0, so every term except the last one is zero and this equation becomes ck v1 = 0. Since v1 = 0, this shows ck = 0, so our linear combination is c1 v1 + c2 v2 + + ck-1 vk-1 = 0. Repeat the same argument, this time multiplying by (A - I)k-2 instead of (A - I)k-1 . Then we obtain the equation ck-1 v1 = 0. and, since v1 = 0, this shows that ck-1 = 0 as well. Keep going to get c1 = c2 = = ck-1 = ck = 0, so { v1 , . . . , vk } is linearly independent.
Theorem A.8. Let A be a k-by-k matrix and suppose that Rk has a basis { v1 , . . . , vk } consisting of a single chain of generalized eigenvectors of length k associated to an eigenvalue a. Then A = P JP -1
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where
a 1 a 1 a 1 J = .. .. . . a 1 a is a matrix consisting of a single Jordan block and P = v1 v2 . . . vk is a matrix whose columns are generalized eigenvectors forming a chain. Proof. Let P be the given matrix. We will first show by direct computation that AP = P J. It will be convenient to write J = j1 j2 . . . jk and we see that ji , the i - th column of J, is the vector 0 . . . 1 ji = a 0 . . . with 1 in the (i - 1) - st position, a in the i - th position, and 0 elsewhere. We show that AP = P J by showing that their corresponding columns are equal. Now AP = A v1 v2 . . . vk so the i - th column of AP is Avi . But Avi = (A - aI + aI)vi = (A - aI)vi + aIvi = vi-1 + avi . On the other hand, PJ = v1 v2 . . . vk J
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and the i - th column of P J is P ji , P ji = v1 v2 . . . vk ji .
Remembering what the vector ji is, and multiplying, we see that P ji = vi-1 + avi as well. Thus every column of AP is equal to the corresponding column of P J, so AP = P J. But Proposition 7 shows that the columns of P are linearly independent, so P is invertible. Multiplying on the right by P -1 , we see that A = P JP -1 .
Example A.9. Applying Theorem 8 to the matrix A= see that 0 1 -2 1 = -4 4 -4 0 2 1 0 2
0 1 of examples 3 and 5, we -4 4
-1
-2 1 -4 0
.
Now in general the Jordan Canonical Form of a matrix A will not consist of a single block, but will have a number of blocks, of varying sizes and associated to varying eigenvalues. But in this situation we merely have to "assemble" the various blocks (to get the matrix) and the various chains of generalized eigenvectors (to get a basis). Actually, the word "merely" is a bit misleading, as the proof that we can to do is in fact a subtle one. Thus we shall not give the proof here. Instead, we shall merely illustrate the situation. In fact, in order to avoid complicated notation we shall merely illustrate the situation for 2-by-2 and 3-by-3 matrices. Theorem A.10. Let A be a 2-by-2 matrix. Then one of the following situations applies: (i) A has distinct eigenvalues a and b. Let u be an eigenvector associated to the eigenvalue a and let v be an eigenvector associated to the eigenvalue b. Then A = P JP -1
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with J= a 0 0 b and P = u v .
(Note in this case A is diagonalizable) (ii) A has a single eigenvalue a of multiplicity 2. (a) A has two linearly independent eigenvectors u and v. Then A = P JP -1 with J= a 0 0 a and P = u v .
(Note in this case A is diagonalizable. In fact, in this case Ea = R2 and A itself is a 0 the matrix .) 0 a (b) A has a single chain { v1 , v2 } of generalized eigenvectors. Then A = P JP -1 with J= a 1 0 a and P = v1 v2 .
Theorem A.11. Let A be a 3-by-3 matrix. Then one of the following situations applies: (i) A has distinct eigenvalues a, b, and c. Let u be an eigenvector associated to the eigenvalue a, let v be an eigenvector associated to the eigenvalue b, and let w be an eigenvector associated to the eigenvalue c. Then A = P JP -1 with a 0 0 J = 0 b 0 and P = u v w . 0 0 c (Note in this case A is diagonalizable.) (ii) A has an eigenvalue a of multiplicity two and an eigenvalue b of multiplicity one. (a) A has two linearly independent eigenvectors u and v associated to the eigenvalue a. Let w be an eigenvector associated to the eigenvalue b. Then A = P JP -1 with a 0 0 J = 0 a 0 and P = u v w . 0 0 b (Note in this case A is diagonalizable.) (b) A has a single chain { u1 , u2 } of generalized eigenvectors associated to the eigenvalue a. Let v be an eigenvector associated to the eigenvalue b. Then A = P JP -1
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with
a 1 0 J = 0 a 0 0 0 b
and
P = u1 u2 v .
(iii) A has a single eigenvalue a of multiplicity three. (a) A has three linearly independent eigenvectors u, v, and w. Then A = P JP -1 with a 0 0 J = 0 a 0 and P = u v w . 0 0 a (Note in this case a 0 0 a the matrix 0 0 A diagonalizable. In fact, in this case Ea = R3 and A itself is is 0 0.) a
(b) A has a chain { u1 , u2 } of generalized eigenvectors and an eigenvector v with { u1 , u2 , v } linearly independent. Then A = P JP -1 with a 1 0 J = 0 a 0 and P = u1 u2 v . 0 0 a (c) A has a single chain { u1 , u2 , u3 } of generalized eigenvectors. Then A = P JP -1 with a 1 0 J = 0 a 1 and P = u1 u2 u3 . 0 0 a
Now we would like to apply Theorems 10 and 11. In order to do so, we need to have an effective method to determine which of the cases we are in, and we give that here (without proof). Definition A.12. Let be an eigenvalue of A. Then for any positive integer i,
i E = { v | (A - I)i (v) = 0 }
= ker((A - I)i ).
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i Note that E consists of generalized eigenvectors of index at most i (and the 0 vector),
and is a subspace. Note also that
1 2 ~ E = E E . . . E .
In general, the Jordan canonical form of A is determined by the dimensions of all the i spaces E , but this determination can be a bit complicated. For eigenvalues of multiplicity at most 3, however, the situation is simpler we need only consider the eigenspaces E . This is a consequence of the following general result: Proposition A.13. Let be an eigenvalue of A. Then the number of blocks in the Jordan Canonical Form of A corresponding to is equal to dim E . Idea of Proof. Suppose there are k such blocks. Since each block corresponds to a chain of generalized eigenvectors, there are k such chains. Now the bottom of the chain is an (ordinary) eigenvector, so we get k eigenvectors in this way. It can be shown that these k eigenvectors are always linearly independent and that they always span E , i.e., that they are a basis of E . Thus E has a basis consisting of k vectors, so dim E = k.
We can now determine the Jordan Canonical Forms of 1-by-1, 2-by-2, and 3-by-3 matrices, using the following consequences of this proposition.
1 Corollary A.14. Let be an eigenvalue of A of multiplicity 1. Then dim E = 1 and the
submatrix of the Jordan Canonical Form of A corresponding to the eigenvalue is a single 1-by-1 block. Corollary A.15. Let be an eigenvalue of A of multiplicity two. Then there are the following possibilities:
1 (a) dim E = 2. In this case, the submatrix of the Jordan Canonical Form of A corre-
sponding to the eigenvalue consists of two 1-by-1 blocks.
1 2 (b) dim E = 1, dim E = 2. In this case, the submatrix of the Jordan Canonical Form of A corresponding to the eigenvalue consists of a single 2-by-2 block.
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Corollary A.16. Let be an eigenvalue of A of multiplicity three. Then there are the following possibilities:
1 (a) dim E = 3. In this case, the submatrix of the Jordan Canonical Form of A corre-
sponding to the eigenvalue consists of three 1-by-1 blocks.
2 1 (b) dim E = 2, dim E = 3. In this case, the submatrix of the Jordan Canonical Form
of A corresponding to the eigenvalue consists of a 2-by-2 block and a 1-by-1 block. 3 2 1 (c) dim E = 1, dim E = 1, dim E = 3. In this case, the submatrix of the Jordan Canonical Form of A corresponding to the eigenvalue consists of a single 3-by-3 block. Now we shall do several examples. We will only do examples where A is not diagonalizable as examples where A is diagonalizable were done earlier in the text. 2 1 1 Example A.17. A= 2 1 -2. -1 0 -2 A has characteristic polynomial det (I - A) = (+1)2 (-3). Thus A has an eigenvalue 1 of multiplicity two and an eigenvalue 3 of multiplicity one. Computation shows that the -1 eigenspace E-1 = ker(A - (-I)) has basis 2 , so dim E-1 = 1 and we are in 1 2 Corollary case (b). Then we further compute that E-1 = ker((A - (-I))2 ) has basis 15 0 -1 2 , 0 , so is 2-dimensional, as we expect. More to the point, we may choose any 0 1 2 1 generalized eigenvector of index i.e., any vector in E-1 that is not in E-1 ,as the top 2, 0 1 of a chain. We choose u2 = 0, and then we have u1 = (A - (-I))u2 = -2, and 1 -1 { u1 , u2 } form a chain. -5 -6 . We also compute that, for the eigenvalue 3, the eigenspace E3 has basis v = 1 Hence we see that -1 2 1 1 1 0 -5 -1 1 0 1 0 -5 2 1 -2 = -2 0 -6 0 -1 0 -2 0 -6 . -1 0 2 -1 1 1 0 0 3 -1 1 1
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2 1 1 Example A.18. A= -2 -1 2. 1 1 2 A has characteristic polynomial det (I - A) = ( - 1)3 , so A has one eigenvalue 1 of -1 -1 0 , 1 , multiplicity three. Computation shows that E1 = ker (A - I)) has basis 1 0 2 so dim E1 = 2 and we are in Corollary 16 case (b). Computation then that dim E1 = shows 0 0 1 2 3 (i.e., (A - I)2 = 0 and E1 is all of R3 ) with basis 0 , 1 , 0 . We may choose 0 0 1 1 2 1 u2 to be any vector in E1 that is not in E1 , and we shall u2 choose = 0. Then u1 = 0 1 (A - I)u2 = -2, and { u1 , u2 } form a chain. For the third vector v we may choose any 1 -1 vector in E1 such that { u1 , v } is linearly independent. We choose v = 0 . Hence we 1 see that -1 2 1 1 1 1 -1 1 1 0 1 1 -1 -2 -1 2 = -2 0 0 0 1 0 -2 0 0 . 1 1 2 1 0 1 0 0 1 1 0 1
5 0 1 Example A.19. A= 1 1 0. -7 1 0 A has characteristic polynomial det (I - A) = ( - 2)3 , so A has one eigenvalue 2 -1 of multiplicity three. Computation shows that E1 = ker (A - 2I)) has basis -1 , 3 1 so dim E2 = 1 and we are in Corollary case (c). Then 16 computation shows that E2 = -1 -1 -1 -1 -1 ker (A - 2I)2 has basis 0 , 2 . (Note that -1 = 3/2 0 + 1/2 2 .) 2 0 3 2 0 3 3 3 3 Computation then shows that dim E1 = 3 (i.e., (A - 2I) = 0 and E1 is all of R ) with
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0 0 1 2 0 , 1 , 0 . We may choose u3 to be any vector in R3 that is not in E1 , basis 0 0 1 1 3 0. Then u2 = (A - 2I)u3 = 1 and u1 = (A - 2I)u2 = and we shall choose u3 = 0 -7 2 2 , and then { u1 , u2 , u3 } form a chain. Hence we see that -6 -1 2 1 0 2 3 1 5 0 1 2 3 1 1 1 0 = 2 1 0 . 1 0 0 2 1 2 -6 -7 0 0 0 2 -6 -7 0 -7 1 0
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A. Exercises. For each matrix A, write A = P JP -1 with P an invertible matrix and J a matrix in Jordan canonical form. 1. A = 75 56 , -90 -67 -50 99 , -20 39 -18 9 , -49 24 1 1 , -16 9 2 1 , -25 12 -15 9 , -25 15 det(I - A) = ( - 3)( - 5).
2. A =
det(I - A) = ( + 6)( + 5).
3. A =
det(I - A) = ( - 3)2 .
4. A =
det(I - A) = ( - 5)2 .
5. A =
det(I - A) = ( - 7)2 .
6. A =
det(I - A) = 2 .
1 0 0 7. A = 1 2 -3, det(I - A) = ( + 1)( - 1)( - 3). 1 -1 0 3 0 2 8. A = 1 3 1, det(I - A) = ( - 1)( - 2)( - 4). 0 1 1 5 8 16 1 8 , det(I - A) = ( + 3)2 ( - 1). 9. A = 4 -4 -4 -11 4 2 3 10. A = -1 1 -3, det(I - A) = ( - 3)2 ( - 8). 2 4 9 5 2 1 11. A = -1 2 -1, det(I - A) = ( - 4)2 ( - 2). -1 -2 3
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12. A =
13. A =
14. A =
15. A =
16. A =
17. A =
18. A =
19. A =
20. A =
8 -3 -3 4 0 -2, det(I - A) = ( - 2)2 ( - 7). -2 1 3 -3 1 -1 -7 5 -1, det(I - A) = ( + 2)2 ( - 4). -6 6 -2 3 0 0 9 -5 -18, det(I - A) = ( - 3)2 ( - 4). -4 4 12 -6 9 0 -6 6 -2, det(I - A) = 2 ( - 3). 9 -9 3 5 6 2 0 -1 -8, det(I - A) = ( - 3)2 ( + 4). 1 0 -2 -1 1 0 -4 3 0, det(I - A) = ( - 1)3 . -6 3 1 0 -4 1 2 -6 1, det(I - A) = ( + 2)3 . 4 -8 0 -4 1 2 -5 1 3, det(I - A) = 3 . -7 2 3 -4 -2 5 -1 -1 1, det(I - A) = ( + 1)3 . -2 -1 2
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A. Answers to odd-numbered exercises -7 4 1. A = 9 -5 -21 1 3. A = -49 0 -5 1 5. A = -25 0 3 0 0 5 3 1 0 3 7 1 0 7 -7 4 9 -5 -21 1 -49 0 -5 1 -25 0
-1
.
-1
.
-1
.
-1 0 2 0 -1 0 0 0 2 0 7. A = 1 1 -3 0 1 0 1 1 -3 . 1 1 1 0 0 3 1 1 1 -1 -1 -2 -2 -3 0 0 -1 -2 -2 0 -1 0 -3 0 1 0 -1 . 9. A = 1 0 1 1 0 0 1 0 1 1 -1 -1 -2 -1 4 0 0 -1 -2 -1 1 1 0 4 0 0 1 1 . 11. A = 0 1 0 1 0 0 2 1 0 1 -1 -1 0 0 -2 1 0 -1 0 0 13. A = -1 0 1 0 -2 0 -1 0 1 . 0 1 1 0 0 4 0 1 1 -1 -3 -1 -2 0 1 0 -3 -1 -2 15. A = -2 -1 -2 0 0 0 -2 -1 -2 . 3 1 3 0 0 3 3 1 3 -1 1 0 0 1 1 0 1 0 0 17. A = 2 1 0 0 1 0 2 1 0 . 3 0 1 0 0 1 3 0 1 -1 1 2 0 0 1 0 1 2 0 19. A = 2 3 0 0 0 1 2 3 0 . 1 3 1 0 0 0 1 3 1
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B. Homogeneous Systems with Constant Coefficients: The Nondiagonalizable Case We will now see how to use Jordan Canonical Form to solve systems X = AX where the coefficient matrix A is not diagonalizable. Our approach here is both simpler and more illuminating than the approach in the book. The key to understanding systems is to investigate a system Y = JY where J is a matrix consisting of a single Jordan block. Theorem B.1. Let J be a k-by-k Jordan block with eigenvalue a, a 1 a 1 0 a 1 J = ... ... . 0 a 1 a Then the system Y = JY has the solution 1 t t2 /2! t3 /3! 1 t t2 /2! 1 t Y = eat ... tk-1 /(k - 1)! tk-2 /(k - 2)! tk-3 /(k - 3)! C . . . t
1 c1 c2 where C = . is a vector of arbitrary constants c1 , c2 , . . . , ck . . . ck Proof. We will prove this in the cases k = 1, 2, and 3, which illustrate the pattern. As you will see, the proof is a simple application of the technique for solving first-order linear differential equations developed in the text. The case k = 1: Here we are considering the system [y ] = [a][y1 ] which is nothing other than the differential equation y1 = ay1 .
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The solution to this differential equation is very familiar. It is y1 = c1 eat which we can certainly write as [y1 ] = eat [1][c1 ]. The case k = 2: Here we are considering the system y1 a 1 = y2 0 a y1 y2
which is nothing other than the pair of differential equations y1 = ay1 + y2 y2 = ay2 .
We recognize the second equation as having the solution y2 = c2 eat and we substitute this into the first equation to get y1 = ay1 + c2 eat . To solve this, we rewrite this as y1 - ay1 = c2 eat and recognize that this differential equation has integrating factor e-at . Multiplying by this factor, we find e-at (y1 - ay1 ) = c2 (e-at y1 ) = c2 e-at y1 = so y1 = eat (c1 + c2 t). Thus our solution is y1 = eat (c1 + c2 t) y2 = eat (c2 t) c2 dt = c1 + c2 t
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which we see we can rewrite as y1 1 t = eat y2 0 1 The case k = 3: Here we are considering y1 a y 2 = 0 0 y3 c1 . c2
the system y1 1 0 a 1 y2 y3 0 a
which is nothing other than the triple of differential equations y1 = ay1 + y2 y2 = y3 = ay2 + y3 ay3 .
If we just concentrate on the last two equations, we see we are in the k = 2 case. Referring to that case, we see that our solution is y2 = eat (c2 + c3 t) y3 = eat (c3 t).
Substituting the value of y2 into the equation for y1 , we obtain y1 = ay1 + eat (c2 + c3 t). To solve this, we rewrite this as y1 - ay1 = eat (c2 + c3 t) and recognize that this differential equation has integrating factor e-at . Multiplying by this factor, we find e-at (y1 - ay1 ) = c2 + c3 t (e-at y1 ) = c2 + c3 t e-at y1 = so y1 = eat (c1 + c2 t + c3 (t2 /2)). (c2 + c3 t) dt = c1 + c2 t + c3 (t2 /2)
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Thus our solution is y1 = eat (c1 + c2 t + c3 (t2 /2)) y2 = y3 = eat (c2 + c3 t) eat (c3 t)
which we see we can rewrite as c1 1 t t2 /2 y1 y2 = eat 0 1 t c2 . 0 0 1 c3 y3
To apply Theorem 1 in general we make the following observation. Remark B.2. Suppose that Y = JY where J is a matrix in Jordan canonical form, but one consisting of several blocks, not just one block. We can see that this systems decomposes into several systems, one corresponding to each block, and that these systems are "uncoupled" (i.e., have nothing to do with each other), so we may solve them each separately, using Theorem 1, and then simply assemble these individual solutions together to obtain a solution of the general system.
Now consider a matrix system X = AX. Our method of solution is entirely analogous to that in the diagonalizable case (see the text). Step 1. Write A = P JP -1 with J in Jordan Canonical form, so the system becomes X = (P JP -1 )X X = P J(P -1 X) P -1 X = J(P -1 X) (P -1 X) = J(P -1 X) Step 2. Set Y = P -1 X, so this system becomes Y = JY
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and use Theorem 1 and Remark 2 to solve this system for Y . Step 3. Since Y = P -1 X, we have that X = PY is the solution to our original system.
We now illustrate this (confining our illustrations to the case that A is not diagonalizable, as the case where A is diagonalizable was handled in the text). Example B.3. Consider the system X = AX where A= 0 1 . -4 4
We saw in section A example 9 that A = P JP -1 with P = Then Y = JY has solution Y = e2t and so X = P Y , i.e., X= = = = -2 1 2t 1 t e -4 0 0 1 -2 1 -4 0 e2t te2t 0 e2t c1 c2 c1 c2 c1 c2 1 t 0 1 c1 e2t te2t = c2 0 e2t c1 c e2t + c2 te2t = 1 c2 c2 e2t -2 1 -4 0 and J= 2 1 . 0 2
-2e2t -2te2t + e2t -4e2t -4te2t
(-2c1 + c2 )e2t - 2c2 te2t . -4c1 e2t - 4c2 te2t 2 1 1 A = 2 1 -2 . -1 0 -2
Example B.4. Consider the system X = AX where
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We saw in section A example 17 that A = P JP -1 with 1 0 -5 -1 1 0 P = -2 0 -6 and J = 0 -1 0 . 1 1 1 0 0 3 Then Y = JY has solution -t c1 e te-t 0 0 e-t 0 c2 Y = 0 0 e3t c3 and so X = P Y , i.e., -t 1 0 -5 e te-t 0 c1 -2 0 -6 0 e-t 0 c2 X= 1 1 1 0 0 e3t c3 -t e te-t -5e3t c1 -t -t 3t -2e -2te -6e c2 = e-t te-t + e-t e3t c3 -t c1 e + c2 te-t - 5c3 e3t -t = -2c1 e - 2c2 te-t - 6c3 e3t . (c1 + c2 )e-t + c2 te-t + c3 e3t Example B.5. Consider the system X = AX where 2 1 1 A = -2 -1 2 . 1 1 2
We saw in section A example 18 that 1 1 -2 0 P = 1 0 Then Y = JY has solution
A = P JP -1 with 1 1 1 0 0 and J = 0 1 0 . 1 0 0 1
et tet 0 c1 0 et 0 c2 Y = c3 0 0 et and so X = P Y , i.e.,
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t 1 1 1 e tet 0 c1 -2 0 0 0 et 0 c2 X= 1 0 1 0 0 et c3 t e tet + et et c1 t -2e -2tet 0 c2 = et tet et c3 (c1 + c2 + c3 )et + c2 tet - 2c2 tet . = -2c1 et (c1 + c3 )et + c2 tet Example B.6. Consider the system X = AX where 5 0 1 A = 1 1 0 . -7 1 0
We saw in section A example 19 that A = P JP -1 with 2 3 1 1 0 P = 2 -6 -7 0 Then Y = JY has solution 2t e te2t (t2 /2)e2t c1 2t 0 e2t c2 te Y = 0 0 e2t c3 and so X = P Y , i.e., 2t 2 3 1 e te2t (t2 /2)e2t c1 1 0 0 e2t te2t c2 X= 2 -6 -7 0 0 0 e2t c3 2t 2t 2t 2 2t 2t c1 2e 2te + 3e t e + 3te + e2t 2te2t + e2t t2 e2t + te2t c2 = 2e2t c3 -6e2t -6te2t - 7e2t -3t2 e2t - 7te2t (2c1 + 3c2 + c3 )e2t + (2c2 + 3c3 )te2t + c3 t2 e2t + (2c2 + c3 )te2t + c3 t2 e2t . = (2c1 + c2 )e2t 2t (-6c1 - 7c2 )e + (-6c2 - 7c3 )te2t - 3c3 t2 e2t 2 1 0 and J = 0 2 1 . 0 0 2
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We conclude this section by showing how to solve initial value problems. This is just one more step, given what we have already done. Example B.7. Consider the initial value problem X = AX where A= 0 1 , -4 4 and X(0) = 3 . -8
In example 3 we saw that this system has the general solution X= (-2c1 + c2 )e2t - 2c2 te2t . -4c1 e2t - 4c2 te2t
Applying the initial condition (i.e., substituting t = 0 in this matrix), gives 3 -2c1 + c2 = X(0) = -8 -4c1 c1 2 = . c2 7 Substituting these values in the above matrix gives X= 3e2t - 14te2t . -8e2t - 28te2t with solution
Example B.8. Consider the initial value 2 X = AX where A = 2 -1
problem 1 1 1 -2 , 0 2
2 and X(0) = -20 . 10
In example 4 we saw that this system has the general solution -t c1 e + c2 te-t - 5c3 te3t X = -2c1 e-t - 2c2 te-t - 6c3 e3t . (c1 + c2 )e-t + c2 te-t + c3 e3t Applying the initial condition (i.e., substituting t = 0 in this matrix), gives 2 c1 - 5c3 -20 = X(0) = -2c1 - 6c3 c1 + c2 + c3 10 with solution 7 c1 c2 = 2 . c3 1
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Substituting these values in the above matrix gives 7e-t + 2te-t - 5e3t X = -14e-t - 4te-t - 6e3t . 9e-t + 2te-t + e3t
Remark B.9. There is an alternate way of solving initial value problems. It is actually less effective than the method we have given for solving a single initial value problem, but has the advantage of expressing the solution directly in terms of the initial conditions. This makes it more effective if the same system X = AX is to be solved for a variety of initial conditions. Let us write our solution of Y = JY as Y (t) = M (t)C where C is a vector of constants. The key observation is that M (0) = I, the identity matrix. Thus, if we wish to solve the initial value problem Y = JY, we find that, in general, Y (t) = M (t)C and in particular Y0 = Y (0) = M (0)C = IC = C so the solution to this initial value problem is Y (t) = M (t)Y0 . Now suppose we wish to solve the initial value problem X = AX, X(0) = X0 . Y (0) = Y0
Then, if A = P JP -1 , we set Y = P -1 X, just as before, to obtain the equation Y = JY , with solution Y (t) = M (t)Y0 . But Y = P -1 X, i.e., Y (t) = P -1 X(t), and in particular Y0 = Y (0) = P -1 X(0) = P -1 X0 . Substituting, we find P -1 X(t) = M (t)(P -1 X0 )
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or X(t) = (P M (t)P -1 )X0 . . Let us now use this technique. Example B.10. Consider the initial value problem X = AX where A= 0 1 , -4 4 and X(0) = a1 a2
As we have seen in example 3, A = P JP -1 with P = M (t) = e2t te2t and 0 e2t P M (t)P -1 = = so X(t) = = In particular, if X(0) = e2t - 2te2t te2t -4te2t e2t + 2te2t -2 1 -4 0 e2t te2t 0 e2t
-2 1 2 1 and J = . Then -4 0 0 2
-2 1 -4 0
-1
e2t - 2te2t te2t 2t 2t -4te e + 2te2t a1 a2
a1 e2t + (-2a1 + a2 )te2t . a2 e2t + (-4a1 + 2a2 )te2t
3 3e2t - 14te2t , then X(t) = , recovering the result of -8 -8e2t - 28te2t 2 2e2t + te2t -4 , then X(t) = , 2t 2t , and if X(0) = 5 5e + 2te 15
example 8. But also, if X(0) = then X(t) = -4e2t + 23te2t , etc. 15e2t + 46te2t
Remark B.11. Mathematicians usually set M (t) = eJt and P M (t)P -1 = eAt but we shall not explain this (or use this notation) here.
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B. Exercises. For each exercise, see the corresponding exercise in the section on Jordan Canonical form. In each problem: a) Solve the system X = AX. b) Solve the initial value problem X = AX, X(0) = X0 . 75 56 1 1. A = and X0 = . -90 -67 -1
2. A =
-50 99 7 and X0 = . -20 39 3
3. A =
-18 9 41 and X0 = . -49 24 98
4. A =
1 1 7 and X0 = . -16 9 16
5. A =
2 1 -10 and X0 = . -25 12 -75
6. A =
-15 9 50 and X0 = . -25 15 100
1 0 0 6 7. A = 1 2 -3 and X0 = -10. 1 -1 0 10 3 0 2 0 1 3 1 and X0 = 3. 8. A = 0 1 1 3
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5 8 16 0 1 8 and X0 = 2 . 9. A = 4 -4 -4 -11 -1 3 4 2 3 -1 1 -3 and X0 = 2. 10. A = 1 2 4 9 5 2 1 -3 -1 2 -1 and X0 = 2 . 11. A = -1 -2 3 9 8 -3 -3 5 0 -2 and X0 = 8. 12. A = 4 -2 1 3 7 -3 1 -1 -1 13. A = -7 5 -1 and X0 = 3 . -6 6 -2 6 3 0 0 2 9 -5 -18 and X0 = -1. 14. A = -4 4 12 1 -6 9 0 1 -6 6 -2 and X0 = 3 . 15. A = 9 -9 3 -6 5 6 2 1 16. A = 0 -1 -8 and X0 = 7. 1 0 -2 4
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-1 1 0 3 17. A = -4 3 0 and X0 = 11. -6 3 1 16 2 0 -4 1 2 -6 1 and X0 = 5. 18. A = 8 4 -8 0 -4 1 2 6 -5 1 3 and X0 = 11. 19. A = -7 2 3 9 -4 -2 5 9 20. A = -1 -1 1 and X0 = 5. -2 -1 2 8
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B. Answers to odd-numbered exercises. 1a. X = -7c1 e3t + 4c2 e5t 9c1 e3t - 5c2 e5t b. X = -7e3t + 8e5t 9e3t - 10e5t b. X = 41e3t + 21te3t 98e3t + 49te3t -10e7t - 25te7t -75e7t - 125te7t
3a. X =
(-21c1 + c2 )e3t - 21c2 te3t -49c1 e3t - 49c2 te3t (-5c1 + c2 )e7t - 5c2 te7t -25c1 e7t - 25c2 te7t
5a. X =
b. X =
6et 2c2 et 7a. X = c1 e-t + c2 et - 3c3 e3t b. X = 2e-t + 3et - 15e3t 2e-t + 3et + 5e3t c1 e-t + c2 et + c3 e3t (-c1 - 2c2 )e-3t - 2c3 et 0 c1 e-3t - c3 et b. X = 2e-3t 9a. X = c2 e-3t + c3 et -e-3t (-c1 - 2c2 )e4t - c3 e2t 2e4t - 5e2t c2 e4t + c3 e2t b. X = -3e43t + 5e2t 11a. X = c1 e4t + c3 e2t 4e4t + 5e2t -2t -c1 e-2t - c2 te-2t -e - 2te-2t 13a. X = -c1 e-2t - c2 te-2t + c3 e4t b. X = -e-2t - 2te-2t + 4e4t c2 e-2t + c3 e4t 2e-2t + 4e4t (-3c1 - c2 ) - 3c2 t - 2c3 e3t -9 - 9t + 10e3t 15a. X = (-2c1 - c2 ) - 2c2 t - 2c3 e3t b. X = -7 - 6t + 10e3t (3c1 + c2 ) + 3c2 t + 3c3 e3t 9 + 9t - 15e3t t c1 et + c2 tet 3e + 5tet 17a. X = (2c1 + c2 )et + 2c2 tet b. X = 11et + 10tet (3c1 + c3 )et + 3c2 tet 16et + 15tet 6 + 5t + t2 (c1 + 2c2 ) + (c2 + 2c3 )t + (c3 /2)t2 + (2c2 + 3c3 )t + c3 t2 b. X = 11 + 8t + 2t2 19a. X = (2c1 + 3c2 ) 9 + 7t + t2 (c1 + 3c2 + c3 ) + (c2 + 3c3 )t + (c3 /2)t2