Jordan Canonical Form

Transcription

1 Jordan Canonical Form Massoud Malek Jordan normal form or Jordan canonical form (named in honor of Camille Jordan) shows that by changing the basis, a given square matrix M can be transformed into a certain normal form J... J 2... J = ,... J p where each block J i is a square matrix of the form λ i. J i = λ.. i λ i Note that λ i s are the repeated eigenvalues of M.. Given an eigenvalue λ i of an n n matrix M, its geometric multiplicity is the dimension of Ker(M λ i I n ), and it is the number of Jordan blocks corresponding to λ i. 2. The sum of the sizes of all Jordan blocks corresponding to an eigenvalue λ i is its algebraic multiplicity.. M is diagonalizable if and only if, for any eigenvalue λ of M, its geometric and algebraic multiplicities coincide. A non diagonalizable matrix is sometimes called a defective matrix. To obtain a Jordan normal form of an n n diagonalizable matrix M, we use n linearly independent eigenvectors to construct a matrix P, where the diagonal matrix will be the Jordan normal form. P MP = J(M) If λ is an eigenvalue of a defective n n matrix A with an algebraic multiplicity greater than one and a geometric multiplicity less than its algebraic multiplicity, then any nonzero vector v satisfying: is called a generalized eigenvector. [M λ I n ] k v = θ for k = 2,,... To construct the Jordan canonical form of M, we form a sequence of generalized eigenvectors that satisfy: [M λ I n ] v = u, [M λ I n ] v 2 = v, [M λ I n ] v = v 2, The eigenvectors and generalized eigenvectors of M form the columns of the invertible matrix P which gives us the Jordan normal form J(M).

2 Massoud Malek Jordan Canonical Form 2 Suppose the geometric multiplicity of λ is m and its algebraic multiplicity is larger than m. First, we need to find M u = λu, M u 2 = λu 2,, M u m = λu m ; then we obtain v jk s, the generalized eigenvectors produced by u j s. The matrix P will be formed as follows: P = [u v v 2... v r u 2 v 2 v v 2s u......]. If we are unable to produce generalized eigenvectors with u j s, then we need to find other linearly independent eigenvectors. Repeated Eigenvalues. Consider the following matrices: A = , B = 6 7 5, C = 6 9, and D = with the characteristic polynomials: K A (λ) = K B (λ) = K C (λ) = K D (λ) = λ 6λ + 96λ 2 256λ = (λ ). Thus λ = λ 2 = λ = λ =. Case. The geometric multiplicity of the matrix A is one, so there is only one Jordan block The rank of the matrix  = A I = is, so the vector u(a) = which 5 2 is the solution of the homogeneous system Âu = θ is the only linearly independent eigenvector of A. The vectors v(a) =, w(a) = are the generalized eigenvectors of A, obtain by solving 9, and z(a) = Âv = u(a), Âw = v(a), and Âz = w(a) respectively. By constructing the matrix P = [u(a) v(a) w(a) z(a)] = P = , 5 2 with

3 Massoud Malek Jordan Canonical Form we obtain the Jordan canonical form of A as follows: J(A) = P AP = Case 2a. The geometric multiplicity of the matrix B is two, so there are two blocks in the Jordan normal matrix. Since the rank of the matrix B = B I = is 2, the vectors u (B) = and u 2(B) = 2 which are the solution of the homogeneous system Bu = θ are linearly independent eigenvectors of B. To obtain the Jordan normal form of B, we need to obtain two generalized eigenvectors of B by solving the systems Bv = u (B) and Bv = u 2 (B). The generalized eigenvectors are v (B) = and v 2(B) = 5. The Jordan normal form of B is obtain from the matrix Q = [u (B) v (B) u 2 (B) v 2 (B)] = Q = with We have J(B) = Q BQ =. Case 2b. The geometric multiplicity of the matrix C is also two, so there are two blocks in the Jordan normal form. Since the rank of the matrix Ĉ =C I =

4 Massoud Malek Jordan Canonical Form is 2, the vectors u (C)= and u 2(C)= are linearly independent eigenvectors of C. By solving the system Ĉv = u (C), we obtain two linearly independent generalized eigenvectors v (C) = and v 2(B) =. 2 but the vectors u (C), u 2 (C), v (C) and v 2 (C) are linearly dependent; so we should use one of the two options:. Remove one of the generalized eigenvectors and solve the system Ĉv = u 2(C). 2. Solve either Ĉw = v (C) or Ĉw = v 2(C). The system Ĉv = u 2(C) is inconsistent, so we use the second option. The system Ĉw = v (C) is also inconsistent, but Ĉw = v 2(C) has the vector w 2 (C) = 8 as a solution. Now by choosing u (C), u 2 (C), v 2 (C) and w 2 (C), we may construct the matrix R = [ u (C) v 2 (C) w 2 (C) u 2 (C) ] = 8, with R = The Jordan canonical form of C is as follows: J(C) = R C R =. 6 9 Case. The rank of the matrix D = D I = is one, so there are three blocks 6 9 in the Jordan normal form. The three linearly independent eigenvectors of D are as follows: u (D) =, u 2(D) = 2, and u (D) = The generalized eigenvector v (D) = is a solution of the system Dv = u (D). Note that the eigenvectors 2,, and 2 are linearly independent but do not produce any

5 Massoud Malek Jordan Canonical Form 5 generalized eigenvector since all the rows of D are identical but components of these three vectors are not all equal. The matrix S = [u (D) v (D) u 2 (D) u (D)] = 2 with S = will produce the Jordan normal form as follows: J (B) = Q BQ =. Remark. The following vectors u (D) = 2(D) = 2 (D) = are also linearly independent eigenvectors of the matrix D; but none of them could produce a generalized eigenvector v which is needed to obtain J(D). Since D 2 is the zero matrix, it follows that any vector is a generalized eigenvector of D; for example the first three columns of the invertible matrix T = 2 are the eigenvectors of D, and the forth column is a generalized eigenvector in the generalized sense. Unfortunately 9 T D T = 9 is not the Jordan canonical form of D. In this case, we need to go back to our eigenvectors and choose the ones that will produce a Jordan canonical form. Example. Consider the matrix with the characteristic polynomial 7 7 M = K M (λ) = λ 9λ + 29λ 2 9λ + 8. The eigenvalues are λ =, λ 2 = 2,, and λ = λ =. with the corresponding eigenvectors: u =, u 2 =, and u = u =.

6 Massoud Malek Jordan Canonical Form 6 The equation (M I )v = u produces the generalized eigenvector v = P = 2 with P = Thus we have Hence the Jordan canonical form J(M) of M is obtained as follows: P M P = = 2 = J(M) System of Linear Differential Equations. Consider the system X (t) = A X(t) with P AP = J(A). Let X(t) = P Y (t), then X (t) = P Y (t), P X(t) = Y (t), and P X (t) = Y ( t). We have: Y (t) = P X (t) = P A X (t) = P A P Y (t) = J(A) Y (t). Thus by solving Y (t) = J(A) Y (t), we may obtain the solution of the original system as follows: X (t) = P Y (t) = P [J(A)Y (t)] = P [ P A P Y (t) ] = [ P P ] A P Y (t) = A [P Y (t)] = A X(t). Consider the following matrices A = and B =. 2 2 Step. The characteristic polynomial of A and B are as follows: 5 λ 9 K A (λ) = det ( A λi ) = λ = (2 λ) 5 λ 9 2 λ λ = (2 λ) [ (5 λ) ( λ) + 9 ] = (2 λ) and λ 8 K B (λ) = det(b λi ) = λ = (2 λ) λ 2 λ λ = (2 λ) [ λ( λ) + ] = (2 λ). By setting K A (λ) to zero, we obtain λ = λ 2 = λ = 2. Hence the multiplicity of 2 in the K A (λ) is. We obtain the same eigenvalues by setting K B (λ) to zero; the multiplicity of 2 in the K A (λ) is also. Step 2. Next, we are going to see if A and B are diagonalizable. We have: A 2 I = and B 2 I =.

7 Massoud Malek Jordan Canonical Form 7 The nullity of A 2 I is ; but the nullity of B 2 I is 2. Hence there is one eigenvector for A, corresponding to the eigenvalue λ = 2; but two linearly independent eigenvectors for B, corresponding to the eigenvalue λ = 2. Therefore neither A nor B are diagonalizable. Step. To find the eigenvectors of A and B, we need to solve the following homogeneous linear systems: [ A 2 I ] = and [ B 2 I ] =. By using some row operations, we obtain: [A 2 I ] = and [B 2 I ] =. By solving the above systems, we find u(a) =, u (B) = Notice that u (B) and u 2 (B) are linearly independent. 2, and u 2 (B) =. Step. Next, we need to find generalized eigenvectors for A and B, corresponding to the eigenvalue λ = 2. From 9 [A 2 I u(a)] =, we obtain the linearly independent generalized eigenvectors: v (A) =, and v 2 (A) = /. But the matrix P = [ u(a), v (A), v 2 (A) ] = / is singular, so we need to find a second generation generalized eigenvector. For this, we solve the following system, using v (A) (we could also use v 2 (A)): [ A 2 I v ( A) ] = 9 the vector w = is a solution and the matrix Q = [ u(a), v (A), w (A) ] =,

8 Massoud Malek Jordan Canonical Form 8 has an inverse Q =. The Jordan normal form of A is 2 J(A) = Q A Q = 2. 2 To find the Jordan normal form of B, we only need to find one generalized eigenvector; using either u (B) or u 2 (B); we just choose u (B). From [ B 2 I u(a)] = we obtain the linearly independent generalized eigenvector v (B) =,. Note that v 2 (B) = is another generalized eigenvector, but does not produce a non-singular matrix. b The matrix has an inverse R = R = [ u (B), v (B), u 2 (B) ] = 2 2. The Jordan normal form of A is 2 2 J(A) = R A R = 2. 2 Exercise. Consider, the following matrices: A =, B =, with Characteristic Polynomials: 2 2 C =, D = K A (λ) = K B (λ) = K C (λ) = K D (λ) = (λ 2). (i) Find the Jordan Canonical forms J(A), J(B), J(C), and J(D)and invertible matrices P A, P B, P C, and P D ;

9 Massoud Malek Jordan Canonical Form 9 such that (ii) Let P A A P A = J(A), P B B P B = J(B), P C C P C = J(C), and P D D P D = J(D). x (t) X(t) = x 2 (t) x (t). x (t) Solve the following systems of linear differential equations: X (t) = A X(t), X (t) = B X(t), X (t) = C X(t), and X (t) = D X(t). Consider the following matrices 5 A = 5 8 and B = 8. Then find the Jordan canonical form of A and B.