Linear System Theory

Transcription

1 Linear System Theory Wonhee Kim Lecture 4 Apr. 4, / 40

2 Recap Vector space, linear space, linear vector space Subspace Linearly independence and dependence Dimension, Basis, Change of Basis 2 / 40

3 Linear Maps (Linear Operators) Recall that Function: f : X Y for any x X, f assigns a unique f (x) = y Y X : domain, Y : codomain {f (x) : x X }: range f : function, operator, map, transformation injective, surjective, bijective 3 / 40

4 Linear Maps (Linear Operators) Linear Mapping: Let (V, F) and (W, F) be (finite-dimensional) linear vector spaces on the SAME FIELD!!!. Let A be a map from V to W, i.e., A : V W such that A(v) = w, v V and w W. Then A is said to be a linear mapping (equiv. linear transformation, linear operator, linear map), if and only if A(α 1 v 1 + α 2 v 2 ) = α 1 A(v 1 ) + α 2 A(v 2 ), α 1, α 2 F We will see that if A is linear, then A is a matrix. In other words, we will see that any linear mapping between finite-dimensional linear spaces can be represented as matrix multiplication 4 / 40

5 Linear Maps (Linear Operators) Example: Consider the following mapping on the set of polynomials of degree 2: A : as 2 + bs + c cs 2 + bs + a, where a, b, c R. Is this a linear map? Let = v 1 = a 1 s 2 + b 1 s + c 1 and v 2 = a 2 s 2 + b 2 s + c 2. Then for any α 1, α 2 R A(α 1 v 1 + α 2 v 2 ) = A(α 1 (a 1 s 2 + b 1 s + c 1 ) + α 2 a 2 s 2 + b 2 s + c 2 ) = (α 1 c 1 + α 2 c 2 )s 2 + (α 1 b 1 + α 2 b 2 )s + α 1 a 1 + α 2 a 2 = α 1 A(v 1 ) + α 2 A(v 2 ) 5 / 40

6 Linear Maps (Linear Operators) Example: The mapping A(v) where w = v / 40

7 Linear Maps (Linear Operators) Theorem Let (V, F) have a basis {v 1,..., v n }, and let (W, F) have a basis {w 1,..., w m }. Then w.r.t. these bases, the linear map A is represented by m n matrix. Namely, y = A(x) = Ax, where A is a m n matrix. Proof: For any x V, there exists a unique α 1,..., α n F such that By linearity, x = n α i v j = [v]α j=1 y = A(x) = α 1 A(v 1 ) + + α n A(v n ) = n α j A(v j ) Note that A : V W; hence for each A(v i ) W, it has a unique representation w.r.t. {w 1,..., w m }: j=1 A(v j ) = a 1j w 1 + a mj w m, j = 1, 2,..., n 7 / 40

8 Linear Maps (Linear Operators) Hence, (A1 (v 1 ) A 1 (v 2 ) A n (v n ) ) = ( ) y 1 y 2 y n a 11 a 12 a 1n = ( ) a 21 a 22 a 2n w 1 w 2 w m = [w]a a m1 a m2 a mn Note that A is a m n matrix, and ith column of A is the representation of y i w.r.t basis of W. Now, for any y = Ax W, there exists a unique β 1,..., β m F w.r.t. basis {w 1,..., w m } such that y = β 1 w β m w m = [w]β = α 1 A(v 1 ) + α n A(v n ) = [A(v)]α = [w]aα Hence, β = Aα. Since we have unique α and β, we have the desired result 8 / 40

9 Linear Maps (Linear Operators) We will now use A instead of A for a given basis. i)the matrix A gives the relation between the representations (coordinates) α and β, not x and y. ii) This also means that A depends on bases. So, with different bases, we have the different representation of the same operator A. This means that A is not unique under different bases. The most important case is when V = W, in which case we can use the same basis for the domain and codomain. For a different basis, we may obtain Ā, which is a different representation of A. 9 / 40

10 Similarity Transformation Suppose that v = {v 1,..., v n } and e = {e 1,..., e n } are bases for V. Then for the linear operator A, it can have different representations: A w.r.t. v or Ā w.r.t. e. We now see the relationship between A and Ā. Think about the change of basis, we have shown that given two bases, we can change the coordinate (representation) of a vector by using the transformation matrix. This means that if the domain and the codomain are the same, then we can use the same transformation matrix 10 / 40

11 Similarity Transformation 11 / 40

12 Similarity Transformation We have β = Āᾱ = Pβ = PAα = PAP 1 ᾱ This implies Ā = PAP 1. Similarly, Ā = Q 1 AQ Moreover, A = P 1 ĀP = QĀQ 1 Similar Matrix: A and Ā are similar if there exists a nonsingular matrix P (or Q) satisfying the above transformation. If P (or Q) exists, the above transformation is called a similarity transformation Note: all the matrix representations (w.r.t the different bases) of the same operator are similar 12 / 40

13 Similarity Transformation Example: Second-order differential equation with x 1 = x and x 2 = ẋ = ẋ 1 ( ) ( ẍ + 3ẋ + 2x = u ẋ = x + u 2 3 1) The eigenvalue and eigenvector of A Let z = Qx. Then ż = Qẋ. Hence, det(si A) = 0 s = 1, 2 Q = ( ( ) ) 1 1 e 1 e 1 = 1 2 ( ) Q = P = 1 1 ż = Q 1 AQz + Q 1 u = ( ) 1 0 z ( ) 1.41 u / 40

14 Range and Null Spaces Definition: Range Space Suppose that (V, F) and (W, F) are n and m dimensional vector spaces, respectively. The range of a linear operator A from V to W is the set R(A) defined by R(A) = {y W : y = Ax for some x V} Theorem: The range of a linear operator A is a subspace of (W, F). Proof: 1) We have R(A) W 2) For any y 1, y 2 R(A), and α, β F, we have α 1 y 1 + α 2 y 2 = α 1 Ax 1 + α 2 Ax 2 = A(α 1 x 1 + α 2 x 2 ) (why??). Hence, by the definition of R(A), α 1 y 1 + α 2 y 2 R(A). 14 / 40

15 Range and Null Spaces R(A) W, and R(A) is a linear space Let x = (x 1,..., x n ) T, and A = (A 1,..., A n ), where A i is the ith column of A. Then Ax = y can be written as y = A 1 x 1 + A n x n R(A) is the set of all the possible linear combinations of the columns of A. Note that R(A) is a linear subspace of (W, F). Hence, dim(r(a)) is the maximum number of linearly independent columns of A. 15 / 40

16 Range and Null Spaces Definition: Null Space Suppose that (V, F) and (W, F) are n and m dimensional vector spaces, respectively. The null space of a linear operator A from V to W is the set N(A) defined by N(A) = {x V : Ax = 0} The null space of a linear operator A is a subspace of (V, F). The null space is a subspace of domain, whereas the range space is a subspace of (W, F) 16 / 40

17 Range and Null Spaces Rank and Nullity The dimension of R(A) is denoted by rank(a) The dimension of N(A) is denoted by nullity(a) Theorem: Dimension Theorem or Rank-Nullity Theorem Suppose that (V, F) and (W, F) are n and m dimensional vector spaces, respectively. Then for a linear operator A from V to W, we have rank(a) + nullity(a) = dim(v) = n 17 / 40

18 Range and Null Spaces R(A T ) = {x V : x = A T y, for some y W} (row space) N(A T ) = {y W : A T y = 0} rank(a T ) + nullity(a T ) = m rank(a) = number of linearly independent columns of A rank(a T ) = number of linearly independent raws of A rank(a) = rank(a T ) A square matrix matrix A in nonsingular if and only if all the rows and columns of A are linearly independent 18 / 40

19 Range and Null Spaces Example: A 1 = ( ) 1 2 3, A = We can check ( ) ( ) ( ) ( x 1 + x x 8 3 =, x = (1, 1, 1) 12 0) T, (2, 0, 1) T v v 2 6 = 0, v = (3, 1) T Hence, nullity(a 1 ) = 2, rank(a 1 ) = 1, nullity(a 2 ) = 1, rank(a 2 ) = 1 19 / 40

20 Invariant Subspace Invariant subspace Let A be a linear operator of (V, F) into itself. A subspace Y of V is said to be an invariant subspace of V under A, or an A-invariant subspace of V if A(Y) Y Ay Y, y Y Trivial Example: A : R n R n : R n is an invariant subspace 20 / 40

21 Eigenvalues and Eigenvectors Definition: Eigenvalues and Eigenvectors Let A be a linear operator that maps (C n, C) into itself. Then a scalar λ C is called an eigenvalue of A if there exists a nonzero x C n such that Ax = λx Any nonzero x satisfying Ax = λx is called an eigenvector of A associated with the eigenvalue λ (A λi )x = 0 or (λi A)x = 0 For a nonzero x C n, λ is an eigenvalue of A if and only if det(λi A) = 0 or det(a λi ) = 0 det(λi A) is called the characteristic polynomial of A The degree of the characteristic polynomial of A is n. Hence, the n n matrix A has n eigenvalues 21 / 40

22 Eigenvalues and Eigenvectors Question: For a n n square matrix A, are all the eigenvalues of A distinct? Answer: Not necessarily. There might be a repetition of some eigenvalues. Namely, it can be λ i = λ j for some i j i, j = 1, 2,.., n Case I: We consider the case when all the eigenvalues of A are distinct Theorem: Let λ 1,..., λ n be distinct eigenvalues of A, and v 1,..., v n are associated eigenvectors. Then the set {v 1,..., v n } is linearly independent, and for a n n matrix V = [v] = (v 1,..., v n ), rank(v ) = n. 22 / 40

23 Eigenvalues and Eigenvectors We prove the first part. Assume that the set {v 1,..., v n } is linearly dependent. Then there exist nonzero α 1,..., α n C such that α 1 v α n v n = 0 Without loss of generality, assume that α 1 0. Note that n (A λ 2 I )(A λ 3 I ) (A λ n I ) α i v i = 0 We have (A λ j I )v i = (λ i λ j )v i, j i (A λ i I )v i = 0 i=1 α 1 (λ 1 λ 2 )(λ 1 λ 3 ) (λ 1 λ n )v 1 = 0 Since all the eigenvalues are distinct, we have n (λ 1 λ i )v 1 = 0 α 1 i=2 which implies α 1 = 0. This is a contradiction. Hence, {v 1,..., v n } is linearly independent 23 / 40

24 Eigenvalues and Eigenvectors Theorem: A is diagonalizable if and only if it has n linearly independent eigenvectors. A consequence of diagonalization A = Q 1 ΛQ det(λi A) = det(λi Q 1 ΛQ) = det(q 1 (λi Λ)Q) = det(λi Λ) Example: Second-order differential equation with x 1 = x and x 2 = ẋ = ẋ 1 ( ) ( ) ẍ + 3ẋ + 2x = u ẋ = x + u The eigenvalue and eigenvector of A det(si A) = 0 s = 1, 2 Q = ( ( ) ( ) ) 1 1 e 1 e 1 =, Q = P = Let z = Qx. Then ż = Qẋ. Hence, ż = Q 1 AQz + Q 1 u = ( ) 1 0 z ( ) 1.41 u / 40

25 Eigenvalues and Eigenvectors Case II: We consider the case when eigenvalues of A are not distinct Example I: The case when A has repeated eigenvalues, but has linearly independent eigenvectors A is diagonalizable (theorem) [v 1 v 2 v 3 ] forms a basis A = 0 1 0, det(λi A) = 0, λ = 1, 1, v 1 = (1, 1, 0) T, v 2 = (0, 1, 0) T, v 3 = ( 1, 0, 1) T Example II: The case when A has repeated eigenvalues, and has linear dependent eigenvectors A = 0 1 3, det(λi A) = 0, λ = 1, 1, v 1 = (1, 0, 0) T, v 2 = (1, 0, 0) T, v 3 = (5, 3, 1) T [v 1 v 2 v 3 ] is not sufficient to form a basis 25 / 40

26 Jordan Form As we have seen, there is repeated eigenvalues of A, A can be diagonalizable or not depending on the corresponding eigenvectors. Assume that A has an eigenvalue of λ i with multiplicity of m i. Then the number of linearly independent eigenvectors associated with an eigenvalue λ i is equal to the dimension of N(λ i I A). Hence, we have q i = n rank(λ i I A) = nullity(λ i I A) Previous examples: λ i = 1 with m i = (I A) 0 0 0, / 40

27 Jordan Form We want to diagonalize a matrix A when A has repeated eigenvalues We want to fine some other eigenvectors of A Generalized eigenvectors: A vector v is said to be a generalized eigenvector of grade k 1 if and only if (A λi ) k v = 0, (A λi ) k 1 v 0 k = 1: v is an eigenvector of A Example II A = 0 1 3, det(λi A) = 0, λ = 1, 1, v 1 = (1, 0, 0) T = v 2, v 3 = (5, 3, 1) T We compute the generalized eigenvector of A associated with λ = / 40

28 Jordan Form B := (A I ) 2 = 0 0 3, Bv 2 = 0, v 2 = (0, 1, 0) T Note that (A I )v 0. Then Q = (v 1, v 2, v 3 ) = 0 1 3, J = Q 1 AQ = J: Jordan matrix N := J diag{1, 1, 2} = N: Nilpotent 28 / 40

29 Jordan Form Assume that A is 5 5 matrix with λ 1 with multiplicity of 4, and with λ 2 with multiplicity of 1 (simple). Then we may have the following Jordan forms λ λ λ A 1 = 0 0 λ λ 1 0, A 0 λ = 0 0 λ λ λ λ 2 λ λ λ A 3 = 0 0 λ λ 1 0, A 0 λ = 0 0 λ λ λ λ 2 A 5 = diag{λ 1, λ 1, λ 1 λ 1, λ 2 } J 1 : order 4, J 2 : order 3 J 3 : two order 2, J 4 : two order 1, order 2 29 / 40

30 Jordan Form Theorem: Suppose that A C n n. Then there exists a nonsingular matrix T C n n and an integer 1 p n such that J 1 0 T 1 J 2 AT = J =..., 0 J p where J k are Jordan matrix (or Jordan block) with order n k 30 / 40

31 Caley-Hamilton Theorem Caley-Hamilon Theorem Let us define the characteristic polynomial of A Then (λ) = det(λi A) = λ n + α 1 λ n 1 + α 2 λ n α n 1 λ + α n (A) = A n + α 1 A n α n 1 A + α n I = 0 Simple proof: when A is diagonalizable (general proof: HW) Similarity transformation implies that Consider (A) = Q Λ = Q 1 AQ A = QΛQ 1 det(a) = det(q) det(λ) det(q) 1 = det(λ) A 2 = QΛ 2 Q 1,..., A k = QΛ k Q 1 [ ] Λ n + α 1 Λ n α n 1 Λ + α n I Q 1 = 0 31 / 40

32 Caley-Hamilton Theorem Note that A n = α 1 A n 1 α n I It is a linear combination of {A n 1,..., I }!!! Similarly A n+1 = α 1 A n α n A It is a linear combination of {A n 1,..., I }!!!, which is also a linear combination of {A n 1,..., I } Given any polynomial function f (λ), f (A) can be expressed as follows f (A) = β 0 I + β 1 A + + β n 1 A n 1 We can extend f (λ) to any function, that is, f (λ) is not necessarily a polynomial. 32 / 40

33 Caley-Hamilton Theorem For A with n n, let λ 1,..., λ m be distinct eigenvalues with multiplicity n 1,..., n m, respectively. (note that m i=1 n i = n) Theorem (Theorem 3.5 of the textbook) Given f (λ) and n n matrix A with Define (A) = Π m i=1(λ λ i ) n i h(λ) = β 0 + β 1 λ + + β n 1 λ n 1 The coefficients {β 0,..., β n 1 } are to be solved from the following set of n equations: Then df l (λ) dλ λ=λ i = dhl (λ) dλ λ=λ i, l = 0, 1,..., n i 1, i = 1, 2,..., m f (A) = h(a) 33 / 40

34 Caley-Hamilton Theorem Example: ( ) 0 1 A = 1 2 We want to compute A 100. det(λi A) = λ 2 + 2λ + 1 = (λ + 1) 2. We let h(λ) = β 0 + β 1 λ, and f (λ) = λ 100. We have f ( 1) = h( 1) ( 1) 100 = β 0 β 1 f ( 1) = h ( 1) 100( 1) 99 = β 1 β 1 = 100, β 0 = 1 + β 1 = 99, h(λ) = λ ( ) f (A) = A = h(a) = 99I 100A = / 40

35 Caley-Hamilton Theorem Example: A = We want to compute e At. det(λi A) = (λ 1) 2 (λ 2). Let h(λ) = β 0 + β 1 λ + β 2 λ 2, and f (λ) = e λt. Then We have f (1) = h(1) e t = β 0 + β 1 + β 2 f (1) = h (1) te t = β 1 + 2β 2 f (2) = h(2) e 2t = β 0 + 2β 1 + 4β 2 β 0 = 2te t + e 2t, β 1 = 3te t + 2e t 2e t β 2 = e 2t e t + te t f (A) = e At = h(a) = β 0 I + β 1 A + β 2 A 2 35 / 40

36 Positive (semi)-definite Matrix A matrix A is symmetric if A = A T For any n n matrix A, and x R n, x T Ax is called a quadratic form Definition: A symmetric matrix A is said to be positive definite if x T Ax > 0 for all x 0 positive semi-definite if x T Ax 0 for all x 0 Theorem: A symmetric matrix A is positive definite (positive semi-definite) if and only if all eigenvalues of A are positive (nonnegative) A symmetric positive definite matrix is invertible For a symmetric positive definite matrix A, det(a) > 0 36 / 40

37 Norm and Inner Product Normed vector space (Normed linear space): Length of the vector Let (V, F) be a n-dimensional vector space. A function x : V R, where x V, is said to be a norm if the following properties hold x 0 and x = 0 if and only if x = 0 (separate points) αx = α x (absolute homogeneity) x + y x + y (triangular inequality) Example: Let (R n, R). Then the norm can be chosen as x 1 := n ( n x i, x 2 := x i 2) 1/2, x := max x i i i=1 Example: signal norm for the real-valued continuous function f (t) where 1 p < i=1 ( t f p = 0 ) 1/p f (t) p dt 37 / 40

38 Norm and Inner Product Inner Product: measure angle of two vectors An inner product between two vectors, x, y, on the vector space (V, C) is a function that maps from V V to C such that the following properties hold x, y = y, x complex conjugate x, α 1 y 1 + α 2 y 2 = α 1 x, y 1 + α 2 x, y 2, x, y 1, y 2 V, α 1, α 2 C x, x 0 and x, x = 0 if and only if x = 0 Example: Let (R n, R). Then the inner product is x 2 2 = x, x = n x i 2, x, y = x T y = i=1 n x i y i Example: signal inner product for the real-valued continuous function f (t) f 2 2 = t 0 f (t) 2 dt, f, g = t 0 i=1 f (t)g(t)dt where 1 p < 38 / 40

39 Orthogonal and Orthonormal Vectors Two vectors x and y are said to be orthogonal if and only if x, y = 0. If x, y = x T y, then x and y are orthogonal if and only if x T y = 0. Note that if x T y = 0, then the angle between two vectors is 90 Fact: Let A be a symmetric matrix, that is, A = A T. If A has distinct eigenvalues, then the corresponding eigenvectors are orthogonal. Example: A 1 = ( ) det(λi A 1 ) = 0, λ = 1, 3, (v 1, v 2 ) = v T 1 v 2 = 0, orthogonal ( ) / 40

40 Orthogonal and Orthonormal Vectors A set of vectors, x 1, x 2,..., x m is said to be orthonormal if { xi T 0 if i j x j = 1 if i = j 40 / 40