Generalized Eigenvectors and Jordan Form

Home , Generalized eigenvector, Jordan matrix

We have seen that an n × n matrix A is diagonalizable precisely when the dimensions of its eigenspaces sum to n. So if A is not diagonalizable, there is at least one eigenvalue with a geometric multiplicity (dimension of its eigenspace) which is strictly less than its algebraic multiplicity. In this handout, we will discuss how one can make up for this deficiency of eigenvectors by finding what are called generalized eigenvectors, which can in turn be used to find the Jordan form of the matrix A. First consider the following non-diagonalizable system.

Example 1. 3 The matrix · ¸ 3 1 A = 0 3 2 has characteristic polynomial (λ − 3)µ·, so¸¶ it has only one eigenvalue λ = 3, and the cor- 1 responding eigenspace is E = span . Since dim(E ) = 1 < 3, the matrix A is not 3 0 3 diagonalizable. Nevertheless, it is still possible to solve the system dy = Ay dt without much difficulty. Writing out the two equations dy 1 = 3y + y dt 1 2 dy 2 = 3y dt 2 3t we see that the second equation has general solution y2 = c1e . Plugging this into the first equation gives dy 1 = 3y + c e3t dt 1 1 This is a first order linear equation that can be solved by the method of integrating factors. 3t 3t Its solution is y1 = c1te + c2e . Thus the general solution of the system is · ¸ · ¸ µ · ¸ · ¸¶ · ¸ 3t 3t y1 c1te + c2e 3t 1 0 3t 1 y = = 3t = c1e t + + c2e y2 c2e 0 1 0 3 The matrix A in the previous example is said to be in Jordan form.

Deﬁnition 1. A 2 × 2 matrix of the form · ¸ λ 1 J = 0 λ

for some number λ is called a Jordan form matrix.

1 The following theorem generalizes the calculations of the previous example.

Theorem 1. The general solution of the system · ¸ dy λ 1 = Jy,J = dt 0 λ

is µ · ¸ · ¸¶ · ¸ 1 0 1 y = c eλt t + + c eλt 1 0 1 2 0

In general, suppose A is a 2×2 matrix with a single repeated eigenvalue λ with dim(Eλ) = 1. Then, as we have seen before, making the change of variables y = Cx transforms the system dy = Ay dt into the system dx = C−1ACx. dt We now want to ﬁnd a matrix C such that the coeﬃcient matrix C−1AC for the new system is a Jordan form matrix J. That is, we want AC = CJ. Writing   | | · ¸ λ 1 C = v v  ,J = , 1 2 0 λ | | we have     | | | | AC = Av1 Av2 ,CJ = λv1 λv2 + v1 | | | | Therefore the columns of C must satisfy

Av1 = λv1

Av2 = λv2 + v1

Thus the vector v1 is an eigenvector with eigenvalue λ. Rewriting these equations

(A − λI)v1 = 0

(A − λI)v2 = v1

2 2 it follows that (A − λI) v2 = (A − λI)v1 = 0. Thus the vector v2 must be in ker(A − λI) .

2 Deﬁnition 2. A nonzero vector v which satisﬁes (A−λI)pv = 0 for some positive integer p is called a generalized eigenvector of A with eigenvalue λ.

The vectors v1 and v2 form a generalized eigenvector chain, as the following diagram illustrates: v2 −→ v1 −→ 0 A−λI A−λI Therefore, to ﬁnd the columns of the matrix C that puts A in Jordan form, we must ﬁnd a chain of generalized eigenvectors, as follows: 2 • Find a nonzero vector v2 in ker(A − λI) that is not in ker(A − λI).

• Set v1 = (A − λI)v2. dy Having found such vectors, we can then write down the solution of the system dt = Ay. dx From above, we know that the solution of the system dt = Jx is µ · ¸ · ¸¶ · ¸ 1 0 1 x = c eλt t + + c eλt 1 0 1 2 0 so multiplying this by C we get λt λt y = Cx = c1e (tv1 + v2) + c2e v1. In summary, we have the following result.

Theorem 2. Let A be a 2 × 2 matrix and suppose v2 → v1 → 0 is a chain of generalized dy eigenvectors of A with eigenvalue λ. Then the general solution of the system dt = Ay is

λt λt y(t) = c1e (tv1 + v2) + c2e v1.

Example 2. 3 Let · ¸ 2 1 A = −1 4 The characteristic polynomial of A is λ2 − 6λ + 9 = (λ − 3)2, so λ = 3 is the only eigenvalue of A. Next, we compute · ¸ · ¸ −1 1 0 0 A − 3I = , (A − 3I)2 = −1 1 0 0

2 Now we choose· v¸2 to be any vector in ker(A − 3I) that is not in ker(·A¸− 3I). One such 0 1 vector is v = . With this choice, we than have v = (A − 3I)v = . The solution of 2 1 1 2 1 dy the system dt = Ay is therefore µ · ¸ · ¸¶ · ¸ 1 0 1 y = c e3t t + + c e3t . 1 1 1 2 1 3

3 For larger systems it turns out that any chain of generalized eigenvectors of the coeﬃcient matrix can be used to write down a linearly independent set of solutions.

Theorem 3. Suppose {v1, v2,..., vk} is a chain of generalized eigenvectors of A with eigenvalue λ. That is,

vk −→ vk−1 −→ vk−2 −→ · · · −→ v2 −→ v1 −→ 0. A−λI A−λI A−λI A−λI A−λI A−λI

Then

λt y1(t) = e v1 λt y2(t) = e (v2 + tv1) · ¸ t2 y (t) = eλt v + tv + v 3 3 2 2! 1 . . · ¸ t2 tk−2 tk−1 y (t) = eλt v + tv + v + ··· + v + v k k k−1 2! k−2 (k − 2)! 2 (k − 1)! 1

dy is a linearly independent set of solutions of dt = Ay.

Proof. We show that yk is a solution. The proof that the others are solutions is similar. Diﬀerentiating gives · dy t2 k = eλt (λv + v ) + t(λv + v ) + (λv + v ) + ··· dt k k−1 k−1 k−2 2! k−2 k−3 ¸ tk−2 tk−1 + (λv + v ) + (λv ) . (k − 2)! 2 1 (k − 1)! 1 Multiplying by A gives · ¸ t2 tk−2 tk−1 Ay = eλt (Av ) + t(Av ) + (Av ) + ··· + (Av ) + (Av ) . k k k−1 2! k−2 (k − 2)! 2 (k − 1)! 1

The two expressions above are equal, since Av1 = λv1, Av2 = λv2 + v1, and so on. Next we show that the chain {v1, v2,..., vk} is necessarily linearly independent. Suppose

c1v1 + c2v2 + ··· + ckvk = 0. (1) k−1 The multiplying by (A − λI) sends v1 through vk to zero, and vk to v1, so we are left with ckv1 = 0. Since v1 is nonzero, this implies ck = 0. Similar reasoning shows that the remaining coeﬃcients must also be zero. Thus the chain of generalized eigenvectors is linearly independent. Now suppose

c1y1(t) + c2y2(t) + ··· + cnyn(t) = 0

for all t. Then since yj(0) = vj, at t = 0 this equation is precisely equation (1) above, which we just showed implies c1 = c2 = ··· = ck = 0.

4 The next theorem guarantees that we can always ﬁnd enough solutions of this form to generate a fundamental set of solutions.

Theorem 4. Let A be an n×n matrix, and suppose λ is an eigenvalue of A with algebraic multiplicity m. Then there is some integer p ≤ m such that dim(ker(A − λI)p) = m.

Example 3. 3 Let   5 1 −4 A = 4 3 −5 3 1 −2

3 The characteristic polynomial of A is pA(λ) = (λ − 2) . Since     3 1 −4 1 0 −1 A − 2I = 4 1 −5 −→ 0 1 −1 , 3 1 −4 rref 0 0 0

the eigenspace E2 = ker(A − 2I) has dimension 1. Next, since     1 0 −1 1 0 −1 (A − 2I)2 = 1 0 −1 −→ 0 0 0  , 1 0 −1 rref 0 0 0

the space ker(A − 2I)2 has dimension 2. Finally, since   0 0 0 (A − 2I)3 = 0 0 0 , 0 0 0

the space ker(A − 2I)3 has dimension 3, which matches the algebraic multiplicity of the eigenvalue λ = 2. We can now form a chain of 3 generalized eigenvectors by choosing a vector v3 in ker(A− 3 2 2I) and deﬁning v2 = (A − 2I)v3 and v1 = (A − 2I)v2 = (A − 2I) v3. To ensure that v2 2 and v1 are both non-zero, we need v3 to not be in ker(A − 2I) (which in turn implies that 3 3 v3 is not in ker(A − 2I)). Since ker(A − 2I) = R , we can choose v3 to be any vector not in ker(A − 2I)2. Since the ﬁrst column of (A − 2I)2 is non-zero, we may choose       1 3 1 v3 = 0 =⇒ v2 = (A − 2I)v3 = 4 =⇒ v1 = (A − 2I)v2 = 1 . 0 3 1

5 Then   1 2t y1(t) = e 1 1      1 3 2t y2(t) = e t 1 + 4 1 3        1 3 1 t2 y (t) = e2t  1 + t 4 + 0 3 2! 1 3 0

dy is a fundamental set of solutions of dt = Ay. 3 There may in general be more than one chain of generalized eigenvectors corresponding to a given eigenvalue. Since the last vector in each chain is an eigenvector, the number of chains corresponding to an eigenvalue λ is equal to the dimension of the eigenspace Eλ. Example 4. 3 Let   4 1 1 A = −2 1 −2 . 1 1 4 3 The characteristic polynomial of A is pA(λ) = (λ − 3) . Since     1 1 1 1 1 1 A − 3I = −2 −2 −2 −→ 0 0 0 , 1 1 1 rref 0 0 0

the eigenspace E3 = ker(A − 3I) has dimension 2, so there will be two chains. Next, since   0 0 0 (A − 3I)2 = 0 0 0 , 0 0 0 the space ker(A − 3I)2 has dimension 3, which matches the algebraic multiplicity of λ = 3. Thus one of the chains will have length 2, so the other must have length 1. 2 We now form a chain of 2 generalized eigenvectors by choosing v2 in ker(A − 3I) such 2 that v2 is not in ker(A − 3I). Since every vector is in ker(A − 3I) and the ﬁrst column of A − 3I is non-zero,we may again choose     1 1 v2 = 0 =⇒ v1 = (A − 3I)v2 = −2 0 1 To form a basis for R3, we now need one additional chain of one generalized eigenvector. This vector must be an eigenvector independent from v1. Since     −1 −1 E3 = span  1  ,  0  , 0 1

6 and neither of these spanning vectors is itself a scalar multiple of v1, we may choose either one of them. So let   −1 w1 =  1  0 Then we have the two chains

v2 −→ v1 −→ 0

w1 −→ 0.

The ﬁrst chain generates the two solutions   1 3t y1(t) = e −2 1      1 1 3t y2(t) = e t −2 + 0 1 0 and the second chain generates a third solution   −1 3t y3(t) = e  1  0

dy and together y1, y2 and y3 form a fundamental set of solutions of dt = Ay. 3

Example 5. 3 Let   5 1 3 2 0 5 0 −3 A =   0 0 5 1  0 0 0 5 The characteristic polynomial of A is (λ − 5)4. Since     0 1 3 2 0 1 3 0 0 0 0 −3 0 0 0 1 A − 5I =   −→   , 0 0 0 1  rref 0 0 0 0 0 0 0 0 0 0 0 0 the eigenspace E5 = ker(A − 5I) has dimension 2, and there are two chains. Since   0 0 0 0 0 0 0 0 (A − 5I)2 =   0 0 0 0 0 0 0 0 the dimension of the space ker(A − 5I)2 is 4, matching the algebraic multiplicity of λ = 5.

7 2 To form a chain of length 2, we ﬁrst choose a vector v2 in ker(A − 5I) which is not in ker(A − 5I). Let     0 1 1 0 v =   =⇒ v = (A − 5I)v =   . 2 0 1 2 0 0 0 We now need one more chain of length 2. To ﬁnd a second chain, we need to choose another 2 vector w2 in ker(A − 5I) which is not in ker(A − 5I), in such a way that w2 is independent of v2 and w1 = (A − 5I)w2 is independent of v1. One such choice is     0 2 0 −3 w =   =⇒ w = (A − 5I)w =   . 2 0 1 2  1  1 0 Thus        1 1 0 0  0 1 y (t) = e5t   y (t) = e5t t   +   1 0 2  0 0 0 0 0        2 2 0 −3  −3 0 y (t) = e5t   y (t) = e5t t   +   3  1  4   1  0 0 0 1

dy is a fundamental set of solutions of dt = Ay. 3

Jordan Form for n × n Matrices

Deﬁnition 3. A matrix of the form   λ 1    λ 1   . .   .. ..   .   .. 1 λ consisting of λ in each entry along the main diagonal, 1 in each entry directly above the main diagonal, and 0 elsewhere, is called an elementary Jordan block. A matrix of the form   J1    J2   .   .. 

8 where each Ji is an elementary Jordan block (with possibly diﬀerent values of λ) is called a Jordan form matrix. Example 6. 3 The following are examples of Jordan form matrices:         2 1 0 2 1 0 −3 1 0 2 0 0 0 2 0 , 0 2 1 ,  0 −3 0  , 0 1 0 0 0 3 0 0 2 0 0 −3 0 0 3 3 Theorem 5. For any n × n matrix A, there is a matrix C and a Jordan form matrix J such that C−1AC = J. The Jordan matrix J is determined by the number and length of the generalized eigenvector chains for A. Example 7. 3 Let A be the matrix in Example 3. Since

Av1 = 2v1

Av2 = v1 + 2v2

Av2 = v2 + 2v3,

letting   | | | C = v1 v2 v3 | | | implies       | | | | | | 2 1 0 AC = 2v1 v1 + 2v2 v2 + 2v3 = v1 v2 v3 0 2 1 = CJ, | | | | | | 0 0 2 so C−1AC = J consists of a single elementary Jordan block. 3 Example 8. 3 Let A be the matrix in Example 4. Since

Av1 = 3v1

Av2 = v1 + 3v2

Aw1 = 3w1,