ECEN 605 LINEAR SYSTEMS. Lecture 7 Solution of State Equations 1/77

1/77 ECEN 605 LINEAR SYSTEMS Lecture 7 Solution of State Equations

Solution of State Space Equations Recall from the previous Lecture note, for a system: ẋ(t) = A x(t) + B u(t) y(t) = C x(t) + D u(t), (1a) (1b) the solutions, x(t) and y(t), of the state equation are and t x(t) = e At x(0) + e A(t τ) Bu(τ)dτ, (2) 0 y(t) = C e At x(0) + C t 0 e A(t τ) Bu(τ)dτ + D u(t). 2/77

State Transition Matrix In the expression of the solution x(t) of the state equation, the term e At is called state transition matrix and is commonly notated as Φ(t) := e At. (3) Let us examine some properties of the state transition matrix that will be used in later chapters. Assume zero input, u(t) = 0. Then the solution of the system becomes ẋ(t) = Ax(t) (4) x(t) = e At x(0) = Φ(t)x(0). (5) At t = 0, we know Φ(0) = I. Differentiating eq. (5), we have ẋ(t) = Φ(t)x(0) = Ax(t). (6) 3/77

State Transition Matrix (cont.) At t = 0, ẋ(0) = Φ(0)x(0) = Ax(0) (7) leads us to have Φ(0) = A. (8) Therefore, the state transition matrix Φ(t) has the following properties: Φ(0) = I and Φ(0) = A. (9) 4/77

State Transition Matrix (cont.) In the following, we summarize the property of the state transition matrix. (a) Φ(0) = e A0 = I (b) Φ(t) = e At = ( e At) 1 = Φ 1 ( t) (c) Φ 1 (t) = Φ( t) (d) Φ(a + b) = e A(a+b) = e Aa e Ab = Φ(a)Φ(b) (e) Φ n (t) = ( e At) n = e Ant = Φ(nt) (f) Φ(a b)φ(b c) = Φ(a)Φ( b)φ(b)φ( c) = Φ(a)Φ( c) = Φ(a c) 5/77

State Transition Matrix (cont.) If A = λ 1 λ 2... λn, e At = e λ1t e λ2t.... e λnt 6/77

State Transition Matrix (cont.) If e At = A = λ 1 0 0 0 λ 1 0 0 0 λ 1 0 0 0 λ µ 1 0 µ, e λt te λt 1 2 t2 e λt 1 6 t3 e λt 0 e λt te λt 1 2 t2 e λt 0 0 e λt te λt 0 0 0 e λt e µt te µ 0 e µt The state transition matrix contains all the information about the free motions of the system in eq. (4). However, computing the state transition matrix is in general not easy. To develop some approaches to computation, we first study functions of a square matrix., 7/77

Functions of a Square Matrix Consider a state space model: ẋ(t) = Ax(t) + Bu(t) y(t) = Cx(t) + Du(t). Taking the Laplace transform, we have sx (s) x(0) = AX (s) + BU(s) Y (s) = CX (s) + DU(s) and Y (s) = = C [ (si A) 1 BU(s) + (si A) 1 x(0) ] + DU(s) C(sI A) 1 B + D ] U(s) + C(sI A) 1 x(0). 8/77

Functions of a Square Matrix (cont.) The transfer function G(s) is obtained by letting the initial condition x(0) = 0. Y u (s) = [ C(sI A) 1 B + D ] U(s). (10) }{{} G(s) This shows the relationship between a state space representation of the system and its corresponding transfer function. The following example illustrates how to compute the transfer function from the given state space representation of the system. 9/77

Functions of a Square Matrix (cont.) Example Consider a 3rd order system with 2 inputs and outputs. 1 2 0 1 0 ẋ(t) = 1 4 1 x(t) + 0 1 0 0 1 1 0 y(t) = [ 0 1 0 1 0 1 ] x(t) + [ 1 0 0 0 u(t) ] u(t) 10/77

Functions of a Square Matrix (cont.) Example (cont.) To compute the transfer function, (si A) 1 = = s + 4 2 s 2 + 5s + 6 s 2 + 5s + 6 1 s + 1 s 2 + 5s + 6 s 2 + 5s + 6 0 0 2 s + 2 1 s + 3 1 s + 2 + 1 s + 3 2 s + 2 2 s + 3 2 s 2 + 5s + 6 1 s 2 + 5s + 6 1 s + 1 1 s + 2 + 2 s + 3 0 0 1 s + 1 2 s + 2 + 1 s + 3 1 s + 2 + 1 s + 3 1 s + 1. 11/77

Functions of a Square Matrix (cont.) Example (cont.) Therefore, the state transition matrix is 2e 2t e 3t 2e 2t 2e 3t e t 2e 2t + e 3t e At = e 2t + e 3t e 2t + 2e 3t e 2t + e 3t 0 0 e t. The transfer function is G(s) = C(sI A) 1 B+D = 2 s + 2 + 2 s + 3 + 1 1 s + 2 + 2 s + 3 2 s + 1 2 s + 2 2 s + 3 12/77

Functions of a Square Matrix (cont.) Example (cont.) To solve for x(t) with x(0) = 1 0 1 and u(t) = [ U(t) 0 ], we have X (s) = (si A) 1 x(0) + (si A) 1 BU(s) 1 s + 1 = 2 s + 2 + 2 s + 3 1 s + 1 + 1 s + 1 1 s 1 s + 1 2 s + 2 + 3 1 s 1 s + 1 2 s 3 s + 3 13/77

Functions of a Square Matrix (cont.) Example (cont.) so that x(t) = = e t + 1 e t 2e 2t + 2e 3t 1 3 + e 2t 2 3 e 3t e t + 1 e t 1 1 3 e 2t + 4 3 e 3t 1. (11) 14/77

Leverrier s Algorithm This algorithm determines (si A) 1 without symbolic calculation. First write (si A) 1 = Adj[sI A] det[si A] = T (s) a(s) (12) where a(s) = det[si A] = s n + a n 1 s n 1 + + a 1 s + a 0 T (s) = Adj[sI A] = T n 1 s n 1 + T n 2 s n 2 + + T 1 s + T 0, T i R n n. Thus, (si A) 1 is determined if the matrices T i and the coefficients a j are found. Leverrier s algorithm does that as follows. Let a(s) = (a λ 1 )(s λ 2 ) (s λ n ) (13) 15/77

Leverrier s Algorithm (cont.) where the λ i are complex numbers and possibly repeated. Then it is easily seen that a (s) := da(s) ds = (s λ 2 ) (s λ n ) + (s λ 1 )(s λ 3 ) (s λ n ) + This leads to the following result. Lemma (Leverrier I) To proceed, we note that +(s λ 1 ) (s λ n 1 ). (14) a (s) a(s) = 1 + 1 + + 1. (15) s λ 1 s λ 2 s λ n (si A)(sI A) 1 = I (16) 16/77

Leverrier s Algorithm (cont.) or si (si A) 1 A(sI A) 1 = I (17) and taking the trace of both sides, we have strace [ (si A) 1] ( ) AT (s) Trace = n (18) a(s) or strace [ (si A) 1] ( AT0 + AT 1 s + + AT n 1 s n 1 ) Trace = n. a(s) (19) Lemma (Leverrier II) Trace(sI A) 1 = a (s) a(s) (20) 17/77

Leverrier s Algorithm (cont.) Proof If J denotes the Jordan form of A λ 1 1... 1 0 λ 1 J = λ 2 1... 1 0 λ 2... (21) and for some T. A = T 1 JT (22) 18/77

Leverrier s Algorithm (cont.) Proof (cont.) Thus, and (si A) 1 = T (si J) 1 T 1 (23) Trace(sI A) 1 = Trace ( T (si J) 1 T 1) = Trace ( TT 1 (si J) 1) = Trace(sI J) 1 1 = + 1 + + 1 s λ 1 s λ 2 s λ n = a (s) (by Lemma (LeverrierI)). (24) a(s) 19/77

Leverrier s Algorithm (cont.) Proof (cont.) Then (19) becomes ( a ) ( (s) AT0 + AT 1 s + + AT n 1 s n 1 ) s Trace = n (25) a(s) a(s) or sa (s) Trace(AT 0 ) strace(at 1 ) s n 1 Trace(AT n 1 ) = na(s). (26) 20/77

Leverrier s Algorithm (cont.) Proof (cont.) Now note that sa (s) = na(s) a n 1 s n 1 2a n 2 s n 2 (n 2)a 2 s 2 (n 1)a 1 s na 0 (27) so that (26) reduces to Trace(AT 0 ) strace(at 1 ) s 2 Trace(AT 2 ) s n 1 Trace(AT n 1 ) = a 0 n + sa 1 (n 1) + s 2 a 2 (n 2) + + s n 1 a n 1. (28) 21/77

Leverrier s Algorithm (cont.) Proof (cont.) Equating coefficients in (28) we get a 0 = 1 n Trace(AT 0) a 1 = 1 n 1 Trace(AT 1). a k = 1 n k Trace(AT k) (29). a n 1 = Trace(AT n 1 ). 22/77

Leverrier s Algorithm (cont.) Proof (cont.) Now from (si A)(sI A) 1 = I we obtain a(s)i = (si A) ( T 0 + T 1 s + + T n 1 s n 1) = a 0 I + a 1 Is + + a n 1 Is + Is n or AT 0 + (T 0 AT 1 )s + (T 1 AT 2 )s 2 + + (T n 2 AT n 1 )s n 1 + T n 1 s n = a 0 I + a 1 Is + a 2 Is 2 + + a n 1 Is n 1 + Is n. 23/77

Leverrier s Algorithm (cont.) Proof (cont.) Equating the matrix coefficients, we get T n 1 = I T n 2 = AT n 1 + a n 1 I T n 3 = AT n 2 + a n 2 I. (30) T 0 = AT 1 + a 1 I 0 = AT 0 + a 0 I. 24/77

Leverrier s Algorithm (cont.) Proof (cont.) The relations (29) and (30) suggest the following sequence of calculations. T n 1 = I, a n 1 = Trace(AT n 1 ) T n 2 = AT n 1 + a n 1 I, T n 3 = AT n 2 + a n 2 I, a n 2 = 1 2 Trace(AT n 2) a n 3 = 1 3 Trace(AT n 3).. (31) T 1 = AT 2 + a 2 I, a 1 = 1 n 1 Trace(AT 1) T 0 = AT 1 + a 1 I, a 0 = 1 n Trace(AT 0) The computation sequence above constitutes Leverrier s algorithm. 25/77

Leverrier s Algorithm (cont.) Example Let A = 2 1 1 2 0 1 1 0 1 1 1 1 1 1 1 0 (si A) 1 = T 3s 3 + T 2 s 2 + T 1 s + T 0 s 4 + a 3 s 3 + a 2 s 2 + a 1 s + a 0 26/77

Leverrier s Algorithm (cont.) Example T 2 = AT 3 + a 3 I = A 4I = T 1 = AT 2 + a 2 I = T 0 = AT 1 + a 1 = 2 1 1 2 0 3 1 0 1 1 3 1 1 1 1 4 1 4 0 3 1 0 2 1 2 0 0 5 3 3 1 5 0 2 0 2 1 5 2 4 1 7 2 4 0 4 2 2 T 3 = I a 3 = Trace[A] = 4 a 2 = 1 2 Trace [AT 2] = 2 a 1 = 1 3 Trace [AT 1] = 5 a 0 = 1 4 Trace [AT 0] = 2 27/77

Leverrier s Algorithm (cont.) Example Therefore, det[si A] = s 4 4s 3 + 2s 2 + 5s + 2 2 1 1 2 Adj[sI A] = Is 3 + 0 3 1 0 1 1 3 1 s2 1 1 1 4 + 1 4 0 3 1 0 2 1 2 0 0 5 3 3 1 5 s + 0 2 0 2 1 5 2 4 1 7 2 4 0 4 2 2 28/77

Leverrier s Algorithm (cont.) Example and (si A) 1 = 1 s 4 4s 3 + 2s 2 + 5s + 2 s 3 2s 2 s s 2 + 4s + 2 s 2 2s 2 3s 2 s + 1 s 3 3s 2 + 5 s 2 2s 2 s 4 s 2 + 2s 1 s 2 7 s 3 3s 2 + 2 s 2 5s + 4 s 2 3s 2 s 2 3s + 4 s 2 s 2 s 3 4s 2 + 5s 2 29/77

Cayley-Hamilton Theorem An eigenvalue of the n n real matrix is a real or complex number λ such that Ax = λx with a nonzero vector x. Any nonzero vector x satisfying Ax = λx is called an eigenvector of A associated with eigenvalue λ. To find the eigenvalues of A, we solve (A λi )x = 0 (32) In order for eq. (32) to have a nonzero solution x, the matrix (A si ) must be singular. Thus, the eigenvalues of A are just the solutions of the following n th order polynomial equation: (s) = det(si A) = 0. (33) (s) is called the characteristic polynomial of A. 30/77

Cayley-Hamilton Theorem (cont.) Theorem (Cayley-Hamilton Theorem) Let A be a n n matrix and let (s) = det(si A) = s n + a n 1 s n 1 + + a 1 s + a 0 be the characteristic polynomial of A. Then (A) = A n + a n 1 A n 1 + + a 1 A + a 0 I = 0. (34) 31/77

Cayley-Hamilton Theorem (cont.) Proof (si A) 1 can always be written as (si A) 1 = 1 ( Rn 1 s n 1 + R n 2 s n 2 ) + + R 1 s + R 0 (s) where (s) = det(si A) = s n + a n 1 s n 1 + + a 1 s + a 0 and R i, i = 0, 1,, n 1 are constant matrices formed from the adjoint of (si A). 32/77

Cayley-Hamilton Theorem (cont.) Proof (cont.) Write (s)i = (si A) ( R n 1 s n 1 + R n 2 s n 2 + + R 1 s + R 0 ) = (si A)R n 1 s n 1 + (si A)R n 2 s n 2 + + (si A)R 1 s + (si A)R 0 = R n 1 s n + (R n 2 AR n 1 ) s n 1 + (R n 3 AR n 2 ) s n 2 + + (R 0 AR 1 ) s AR 0. Since (s)i = Is n + a n 1 Is n 1 + + a 1 Is + a 0 I, 33/77

Cayley-Hamilton Theorem (cont.) Proof (cont.) we have by matching the coefficients, R n 1 = I R n 2 = AR n 1 + a n 1 I R n 3 = AR n 2 + a n 2 I. R 0 = AR 1 + a 1 I 0 = AR 0 + a 0 I. 34/77

Cayley-Hamilton Theorem (cont.) Proof (cont.) Substituting R i s successively from the bottom, we have 0 = AR 0 + a 0 I = A 2 R 1 + a 1 A + a 0 I = A 3 R 2 + a 2 A 2 + a 1 A + a 0 I. = A n + a n 1 A n 1 + a n 2 A n 2 + + a 1 A + a 0 I = (A). Therefore, (A) = 0. 35/77

Cayley-Hamilton Theorem The Cayley-Hamilton theorem implies that a square matrix A n can be expressed as a linear combination of I, A, A 2,, A n 1. Furthermore, multiplying A to (34), we have A n+1 + a n 1 A n + a n 2 A n 1 + + a 1 A 2 + a 0 A = 0 which implies that A n+1 can also be expressed as a linear combination of I, A, A 2,, A n 1. In other words, any matrix polynomial f (A) of arbitrary degree can always be expressed as a linear combination of I, A, A 2,, A n 1, i.e., f (A) = β n 1 A n 1 + β n 2 A n 2 + + β 1 A + β 0 I with appropriate values of β i. 36/77

Application of C-H Theorem Computing A 1 Let (s) be the characteristic polynomial of a matrix A R n n. (s) = s n + a n 1 s n 1 + a n 2 s n 2 + + a 1 s + a 0 From the Cayley-Hamilton theorem, we have (A) = A n + a n 1 A n 1 + a n 2 A n 2 + + a 1 A + a 0 I = 0. (35) Multiplying A 1 both sides, we have after rearrangement, A 1 = 1 ( A n 1 + a n 1 A n 2 + + a 2 A + a 1 I ). (36) a 0 37/77

Application of C-H Theorem (cont.) Computing A 1 Example A = [ 3 1 1 2 Then the characteristic polynomial is (s) = (s 3)(s 2) 1 = s 2 5s + 5. ] Therefore, A 1 = 1 5 (A 5I ) = 1 5 ([ 3 1 1 2 ] [ 5 0 0 5 ]) [ 2 = 5 1 5 1 5 3 5 ]. 38/77

Application of C-H Theorem (cont.) Computing A 1 Proceeding forward, we consider any polynomial f (s), then we can write as f (s) = q(s) (s) + h(s). It implies that f (λ i ) = h(λ i ), for all eigenvalues of A. Applying Cayley-Hamilton theorem, (A) = 0, we have f (A) = q(a) (A) + h(a) = h(a). 39/77

Application of C-H Theorem (cont.) Computing A 1 Example Using A given in the previous example, compute f (A) = A 4 + 3A 3 + 2A 2 + A + I. Calculation can be simplified by following. Consider f (s) = q(s) (s) + h(s) = ( s 2 + 8s + 37 ) (s) + (146s 184). }{{} h(s) Thus, f (A) = h(a) = 146A 184. 40/77

Application of C-H Theorem (cont.) Computing A 1 Remark An obvious way of computing h(s) is to carry out long division. However, when long division requires lengthy computation, h(s) can be directly calculated by letting h(s) = β n 1 s n 1 + β n 2 s n 2 + + β 1 s + β 0. The unknowns β i can be easily calculated by using the relationship f (s) s=λ1 = h(s) s=λi, for i = 1, 2,, n where λ i are eigenvalues of A. This idea is extended to any analytic function f (s) and A with eigenvalues with m multiplicities. 41/77

Application of C-H Theorem (cont.) Computing A 1 Theorem Let A be a n n square matrix with characteristic polynomial (λ) = Π m i=1 (s λ i ) n i where n = Π m i=1 n i. Let f (s) be any function and h(s) be a polynomial of degree n 1 then h(s) = β n 1 s n 1 + β n 2 s n 2 + + β 1 s + β 0. f (A) = h(a) if the coefficients, β i s, of h(s) are chosen such that d l f (s) ds l = d l h(s) ds l, for l = 0, 1,, n i 1; i = 1, 2,, m. s=λi s=λi 42/77

Application of C-H Theorem (cont.) Computing A 1 Example Compute A 100 with [ 0 1 A = 1 2 ]. Define f (s) = s 100 h(s) = β 1 s + β 0. Note that (s) = s(s + 2) + 1 = s 2 + 2s + 1 = (s + 1) 2 = 0 and the eigenvalues of A are { 1, 1}. 43/77

Application of C-H Theorem (cont.) Computing A 1 Example (cont.) To determine β i, f ( 1) = h( 1) ( 1) 100 = β 1 + β 0 f ( 1) = h ( 1) 100( 1) 99 = β 1. Thus, we have β 1 = 100 and β 0 = 1 100 = 99, and h(s) = 100s 99. Therefore, from Cayley-Hamilton theorem [ A 100 = h(a) = 100A 99I = 0 100 100 200 ] [ 99 0 0 99 ] [ 99 100 = 100 101 ]. 44/77

Application of C-H Theorem (cont.) Computing A 1 Example A = 0 2 2 0 1 0 1 1 3 To compute e At, we let f (s) = e st and h(s) = β 2 s 2 + β 1 s + β 0. Since the eigenvalues of A are {1, 1, 2}, we evaluate the following. f (s) s=1 = h(s) s=1 e t = β 2 + β 1 + β 0 f (s) s=1 = h (s) s=1 te t = 2β 2 + β 1 f (s) s=2 = h(s) s=2 e 2t = 4β 2 + 2β 1 + β 0. 45/77

Application of C-H Theorem (cont.) Computing A 1 Example (cont.) From these, we have β 0 = 2te t + e 2t β 1 = 3te t + 2e t 2e 2t β 2 = e 2t e t te t and h(s) = ( e 2t e t te t) s 2 + ( 3te t + 2e t 2e 2t) s 2te t + e 2t. 46/77

Application of C-H Theorem (cont.) Computing A 1 Example (cont.) Therefore, e At = f (A) = h(a) = ( e 2t e t te t) A 2 + ( 3te t + 2e t 2e 2t) A + ( 2te t + e 2t) I = e2t + 2e t 2te t 2e 2t + 2e t 0 e t 0. e 2t e t te t 2e 2t e t Remark Consider two square matrices A 1 and A 2 that are similar. Then f (A 1 ) = f (A 2 ). 47/77

Application of C-H Theorem (cont.) Computing A 1 Example Compute e At with A = λ 1 1 0 0 λ 1 0 0 0 λ 2. The characteristic polynomial is (s) = (s λ 1 ) 2 (s λ 2 ) and the eigenvalues of A are {λ 1, λ 1, λ 2 }. Define f (s) = e st and h(s) = β 2 s 2 + β 1 s + β 0. 48/77

Application of C-H Theorem (cont.) Computing A 1 Example (cont.) To determine β i, f (s) s=λ1 = h(s) s=λ1 e λ 1t = β 2 λ 2 1 + β 1 λ 1 + β 0 f (s) s=λ1 = h (s) s=λ1 te λ1t = 2β 2 λ 1 + β 1 f (s) s=λ2 = h(s) s=λ2 e λ2t = β 2 λ 2 2 + β 1 λ 2 + β 0. 49/77

Application of C-H Theorem (cont.) Computing A 1 Example (cont.) β 2 β 1 β 0 = = = λ2 1 λ 1 1 2λ 1 1 0 λ 2 2 λ 2 1 1 1 e λ 1 t λ 1 e λ 1 t e λ 2 t (λ 1 λ 2 ) 2 λ 1 λ 2 (λ 1 λ 2 ) 2 1 1 2λ 1 (λ 1 λ 2 ) 2 λ 1 + λ 2 2λ 1 λ 1 λ 2 (λ 1 λ 2 ) 2 2λ 1λ 2 λ 2 2 2 λ 1 λ 2 λ1 (λ 1 λ 2 ) 2 λ 1 λ 2 (λ 1 λ 2 ) 2 e λ 1 t + (λ 1 λ 2 )te λ 1 t + e λ 2 t (λ 1 λ 2 ) 2 2λ 1 e λ 1 t (λ 2 1 λ2 2 )teλ 1 t 2λ 1 e λ 2 t (λ 1 λ 2 ) 2 (2λ 1 λ 2 λ 2 2 )eλ 1 t + (λ 1 λ 2 )λ 1 λ 2 te λ 1 t + λ 2 1 eλ 2 t (λ 1 λ 2 ) 2 e λ 1 t te λ 1 t e λ 2 t 50/77

Application of C-H Theorem (cont.) Computing A 1 Example (cont.) h(s) = e λ 1 t + (λ 1 λ 2 )te λ 1 t + e λ 2 t s 2 + 2λ 1e λ 1 t (λ 2 1 λ 2 2 )teλ 1 t 2λ 1 e λ 2 t s (λ 1 λ 2 ) 2 (λ 1 λ 2 ) 2 + (2λ 1λ 2 λ 2 2 )eλ 1 t + (λ 1 λ 2 )λ 1 λ 2 te λ 1 t + λ 2 1 eλ 2 t (λ 1 λ 2 ) 2 51/77

Application of C-H Theorem (cont.) Computing A 1 Example (cont.) f (A) = e At = h(a) e λ 1 t + (λ 1 λ 2 )te λ 1 t + e λ 2 t = (λ 1 λ 2 ) 2 = + 2λ 1e λ 1 t (λ 2 1 λ2 2 )teλ 1 t 2λ 1 e λ 2 t (λ 1 λ 2 ) 2 λ 2 1 2λ 1 0 0 λ 2 1 0 0 0 λ 2 2 λ 1 1 0 0 λ 1 0 0 0 λ 2 + (2λ 1λ 2 λ 2 2 )eλ 1 t + (λ 1 λ 2 )λ 1 λ 2 te λ 1 t + λ 2 1 eλ 2 t (λ 1 λ 2 ) 2 e λ 1 t te λ 1 t 0 0 e λ 1 t 0 0 0 e λ 2 t. 0 1 0 1 0 0 0 0 1 52/77

Similarity Transformation The state variables describing a system are not unique and the state space representation of a given system depends on the choice of state variables. By choosing different bases for the underlying n ddimensional state space we obtain different nth order representations of the same system. Such systems are called similar. Similar systems have the same transfer function although their state space representations are different. This fact can be exploited to develop equivalent state space representations consists of matrices that display various structural properties or simplify computation. Now consider a state space representation ẋ(t) = Ax(t) + Nu(t) y(t) = Cx(t) + Du(t). 53/77

Similarity Transformation (cont.) Define x(t) = Tz(t) where T is a square invertible matrix. Then T ż(t) = ATz(t) + Bu(t) y(t) = CTz(t) + Du(t). or { ż(t) = T 1 ATz(t) + T 1 Bu(t) y(t) = CTz(t) + Du(t) { ż = An z(t) + B n u(t) y(t) = C n z(t) + Du(t). The transfer function of the system may be computed CT ( si T 1 AT ) T 1 B + D = CT [ T 1 (si A)T ] 1 T 1 B + D = CT [ T 1 (si A) 1 T ] T 1 B + D = CTT 1 (si A) 1 TT 1 B + D = C(sI A) 1 B + D which shows that the representations in eqs. (37) and (37) have the same transfer function. 54/77

Diagonalization Theorem Consider an n n matrix A with distinct eigenvalues λ 1, λ 2,, λ n. Let x i, i = 1, 2,, n be eigenvectors associated with λ i. Define a n n matrix T = [x 1 x 2 x 3 x n ]. Then T 1 AT = λ 1 λ 2.... λn 55/77

56/77 Diagonalization (cont.) Proof Since x i is an eigenvector of A associated with λ i, Ax i = λ i x i. So we write Ax i = λ i x i = [x 1 x 2 x i x n ] 0. 0 λ i 0. 0.

Diagonalization (cont.) Proof cont. Consequently, A [x 1 x 2 x i x n ] = [x 1 x 2 x i x n ] }{{}}{{} T T λ 1 λ 2... λn. and λ 1 T 1 λ 2 AT =.... λn 57/77

Jordan Form We have seen in the previous section that an n n matrix A can be converted through a similarity transformation into a diagonal matrix when the eigenvalues of A are distinct. When the eigenvalues of A are not distinct, that is, some are repeated, it is not always possible to diagonalize A, as before. In this subsection, we deal with this case, and develop the so-called Jordan form of A which is the maximally diagonal form that can be obtained. The characteristic polynomial of the n n real or complex matrix A, denoted Π(s), is: Π(s) = det(si A) Write (37) in the factored form = s n + a n 1 s n 1 + + a 1 s + a 0. (37) Π(s) = (s λ 1 ) n 1 (s λ 2 ) n 2 + + (s λ p ) np (38) 58/77

Jordan Form (cont.) where the λ i are distinct, that is, λ i λ j, for i j and n 1 + n 2 + + n p = n. It is easy to establish the following. 59/77

Jordan Form (cont.) Theorem There exists an n n nonsingular matrix T such that A 1 T 1 A 2 AT =... Ap where A i is n i n i and det (si A i ) = (s λ i ) n i, for i = 1, 2,, p. 60/77

Jordan Form (cont.) In the remaining part of the section, we develop the Jordan decomposition of the matrices A i. Therefore, we now consider, without loss of generality, an n n matrix A such that Π(s) = det(si A) = (s λ) n. (39) The Jordan structure of such an A matrix may be found by forming the matrices (A λi ) k, for k = 0, 1, 2,, n. (40) Let N k denote the null space of (A λi ) k : { } N k = x (A λi ) k x = 0 (41) and denote the dimension of N k by ν k : dimension N k = ν k, for k = 0, 1, 2,, n + 1. (42) 61/77

Jordan Form (cont.) We remark that 0 = ν 0 ν 1 ν n 1 ν n = n = ν n+1. (43) The following theorem gives the Jordan structure of A. 62/77

Jordan Form (cont.) Theorem Given the n n matrix A with det(si A) = (s λ) n, (44) there exists an n n matrix T with det(t ) 0, such that T 1 AT = A 1 A 2... A r where A j = λ 1 0 0 0 λ 1 0.... 0 λ 1 0 0 0 λ is called a Jordan block n j n j., for j = 1, 2,, r 63/77

Jordan Form (cont.) The number r and sizes n j of the Jordan blocks A i can be found from the formula given in the following Lemma. Lemma Under the assumption of the previous theorem, the number of Jordan blocks of size k k is given by 2ν k ν k 1 ν k+1, for k = 1, 2,, n. 64/77

Jordan Form (cont.) Example Let A be an 11 11 matrix with Suppose that det(si A) = (s λ) 11. ν 1 = 6, ν 2 = 9, ν 3 = 10, ν 4 = 11 = ν 5 = ν 6 = = v 11 = ν 12. 65/77

Jordan Form (cont.) Example (cont.) Then the number of Jordan blocks: a) of size 1 1 = 2ν 1 ν 0 ν 2 = 3 b) of size 2 2 = 2ν 2 ν 1 ν 3 = 2 c) of size 3 3 = 2ν 3 ν 2 ν 4 = 0 d) of size 4 4 = 2ν 4 ν 3 ν 5 = 1 e) of size 5 5 = 2ν 5 ν 4 ν 6 = 0. k) of size 11 11 = 2ν 11 ν 12 ν 10 = 0. 66/77

Jordan Form (cont.) Example (cont.) Therefore, the Jordan form of A is as shown below: λ 1 0 0 0 λ 1 0 0 0 λ 1 0 0 0 λ A = λ 1 0 λ λ 1 0 λ λ λ λ 67/77

Jordan Form (cont.) This result can now be applied to each of the blocks A i in the previous theorem to obtain the complete Jordan decomposition in the general case. It is best to illustrate the procedure with an example. Example Let A be a 15 15 matrix with with λ i being distinct. det(si A) = (s λ 1 ) 8 (s λ 2 ) 4 (s λ 3 ) 3 68/77

Jordan Form (cont.) Example (cont.) We compute (A λ i I ) k, k = 0, 1, 2,, 8, 9 and set h k = dimn (A λ 1 I ) k, k = 0, 1, 2,, 9. Similarly, let i j = dimn (A λ 2 I ) j, j = 0, 1, 2,, 5 and l s = dimn (A λ 3 I ) s, s = 0, 1, 2, 3, 4. 69/77

Jordan Form (cont.) Example (cont.) Suppose, for example, h 0 = 0, h 1 = 2, h 2 = 4, h 3 = 6, h 4 = 7, h 5 = h 6 = h 7 = h 8 = 8 i 0 = 0, i 1 = 3, i 2 = 4, i 3 = i 4 = i 5 = 4 l 0 = 0, l 1 = 1, l 2 = 2, l 3 = l 4 = 3. By Theorem 29, the Jordan form of A has the following structure. A 1 0 0 T 1 AT = 0 A 2 0 0 0 A 3 with A 1 R 8 8, A 2 R 4 4 and A 3 R 3 3. Furthermore, the detailed structure of A 1, by Theorem 30, consists of the following numbers of Jordan blocks. 70/77

Jordan Form (cont.) Example (cont.) a) of size 1 1 = 2h 1 h 0 h 2 = 0 blocks b) of size 2 2 = 2h 2 h 1 h 3 = 0 blocks c) of size 3 3 = 2h 3 h 2 h 4 = 1 blocks d) of size 4 4 = 2h 4 h 3 h 5 = 0 blocks e) of size 5 5 = 2h 5 h 4 h 6 = 1 blocks f) of size 6 6 = 2h 6 h 5 h 7 = 0 blocks g) of size 7 7 = 2h 7 h 6 h 8 = 0 blocks h) of size 8 8 = 2h 8 h 7 h 9 = 0 blocks 71/77

Jordan Form (cont.) Example (cont.) Similarly, the numbers of Jordan blocks in A 2 are: a) of size 1 1 = 2i 1 i 0 i 2 = 2 blocks b) of size 2 2 = 2i 2 i 1 i 3 = 1 blocks c) of size 3 3 = 2i 3 i 2 i 4 = 0 blocks d) of size 4 4 = 2i 4 i 3 i 5 = 0 blocks and the numbers of Jordan blocks in A 3 are: a) of size 1 1 = 2l 1 l 0 l 2 = 0 blocks b) of size 2 2 = 2l 2 l 1 l 3 = 0 blocks c) of size 3 3 = 2l 3 l 2 l 4 = 1 blocks 72/77

Jordan Form (cont.) Example (cont.) With this information, we know that the Jordan form of A is given by T 1 AT = λ 1 1 0 0 0 0 λ 1 1 0 0 0 0 λ 1 1 0 0 0 0 λ 1 1 0 0 0 0 λ 1 λ 1 1 0 0 λ 1 1 0 0 λ 1 λ 2 1 0 λ 2 λ 2 2 λ 3 1 0 0 λ 3 1 0 0 λ 3 = A J 73/77

Finding the Transformation Matrix T Once the Jordan form is found, it is relatively easy to find the transformation matrix. We illustrate using the previous example. Write T in terms of its columns. T = t 1 t 2 t 3 t 4 t 5 }{{} block 1 t 6 t 7 t 8 }{{} block 2 t 9 t 10 }{{} block 3 t 11 }{{} block 4 t 12 }{{} block 5 t 13 t 14 t }{{ 15. } block 6 74/77

Finding the Transformation Matrix T (cont.) Then we have as well as At 1 = λ 1 t 1 or (A λ 1 I ) t 1 = 0 At 2 = λ 1 t 2 + t 1 or (A λ 1 I ) t 2 = t 1 At 3 = λ 1 t 3 + t 2 or (A λ 1 I ) t 3 = t 2 (45) At 4 = λ 1 t 4 + t 3 or (A λ 1 I ) t 4 = t 3 At 5 = λ 1 t 5 + t 4 or (A λ 1 I ) t 5 = t 4 At 6 = λ 1 t 6 or (A λ 1 I ) t 6 = 0 At 7 = λ 1 t 7 + t 6 or (A λ 1 I ) t 7 = t 6 (46) At 8 = λ 1 t 8 + t 7 or (A λ 1 I ) t 8 = t 7 75/77

Finding the Transformation Matrix T (cont.) Therefore, we need to find two linearly independent vectors t 1, t 6 in N (A λ 1 I ) such that the chains of vectors t 1, t 2, t 3, t 4, t 5 and t 6, t 7, t 8 are linearly independent. Alternatively, we can attempt to find t 5 such that (A λ i I ) 5 t 5 = 0 and find t 4, t 3, t 2, t 1 from the set in eq. (45). Similarly, we can find t 8, such that (A λ i I ) 3 t 8 = 0 and t 6, t 7, t 8 found from eq. (46) are linearly independent. 76/77

Finding the Transformation Matrix T (cont.) Moving to the blocks associated with λ 2, we see that we need to find t 9, t 11, t 12 so that (A λ 2 I ) t 9 = 0 (A λ 2 I ) t 11 = 0 (A λ 2 I ) t 12 = 0 and (t 9, t 10, t 11, t 12 ) are independent with Likewise, we find t 13, such that At 10 = λ 2 t 10 + t 9. (A λ 3 I ) t 13 = 0 (A λ 3 I ) t 14 = t 13 (A λ 3 I ) t 15 = t 14 with (t 13, t 14, t 15 ) linearly independent. In each of the above cases, the existence of the vectors t ij are guaranteed by the existence of the Jordan forms. 77/77