Chapter 3 Linear Algebra In this Chapter we provide a review of some basic concepts from Linear Algebra which will be required in order to compute solutions of LTI systems in state space form, discuss controllability and observability, design state feedback controllers, and state observers. The main topics are Vector Spaces. These spaces are a useful concept when solving linear algebraic equations and in particular eigenvectors. A vector space gives a proper foundation for the notion of a state space with a choice of vector space basis being analogous to a change of state coordinates as seen in Section 2.3. Linear Algebraic Systems of Equations. Solving these systems is important when finding eigenvectors and the vector spaces which arise when discussing controllability and observability. Eigenvalues, Eigenvectors. Eigenvalues and eigenvector will be fundamentally related to the nature of the solutions of state space systems. 3.1 Vector Spaces See [Bay99, Sec. 2.2.1, 2.2.2, 2.2.3, 2.2.4, 2.2.5]. A vector has a clear geometric interpretation as a directed line segment in either 2- or 3-dimensional space. In that setting notions of magnitude, direction, and dimension have an intuitive meaning. We have already informally used the term vector when referring to the input, output, and state vectors of a system in state space form. The state vectors of the systems we have considered could have components which represent different physical quantities such as displacement, velocity, voltage, and current. As well, there can be more than three components in a vector. In this case, there is no obvious geometrical interpretation for a vector available. In fact, vectors are more generally defined as an abstract concept which does not require a geometric interpretation. For example, the set of continuous functions defined on some interval is a vector space. In this case, there is no immediate interpretation for a magnitude or direction of a vector. The abstract definition of a vector space is a collection of elements called vectors which can be added (vector addition) and multiplied by scalars (scalar multiplication) which belong to a 51
52 CHAPTER 3. LINEAR ALGEBRA field. The operations of vector addition and scalar multiplication must satisfy certain axioms such as commutativity, associativity of vector addition (see [Bay99, Pg. 50]). Primarily we will be concerned with the vector space of n-tuples of real numbers R n defined over the field R. Linear dependence A set of r vectors {v 1,...,v r } belonging to a vector space X is said to be linearly dependent if there exists scalars a i, 1 i r such that not all a i = 0 and r a i v i = a 1 v 1 + + a r v r = 0 (3.1) i=1 Note here we use subscript in a different way than in Chapter 2. That is, the subscript k in v k denotes the kth vector and not a component of a vector as in Chapter 2. We refer to the sum a 1 v 1 + + a r v r in (3.1) as a linear combination of vectors v 1,...,v r. The a i are called scalar coefficients. Linear independence A set of vectors {v 1,...,v r } belonging to a vector space X is said linearly independent (LI) if it is not linearly dependent. That is, if r a i v i = a 1 v 1 + + a r v r = 0 a i = 0, 1 i r i=1 Suppose r = 2, linear independence means that v 2 is not a scalar multiple of v 1. That is, the two vectors point in different directions. In general, if a set of r vectors {v 1,...,v r } is linearly dependent, there exists at least one a i nonzero such that a 1 v 1 + + a r v r = 0 Assuming a 1 0, we can express v 1 in terms of the remaining vectors v 1 = 1 a 1 (a 2 v 2 + + a r v n ) Hence, if vectors {v 1,..., v r } are linearly dependent, we can write at least one vector as a linear combination of the others. Note that a zero vector is linearly dependent on any vector. Example 14 Suppose the vector space is X = R 3 and we have v 1 = (1, 0, 1) T, v 2 = (2, 1, 2) T, v 3 = (3, 2, 0) T Is {v 1, v 2, v 3 } LI? To show linear dependence we attempt to solve [ ] a 1 v1 v 2 v 3 a 2 = 0 a 3
3.1. VECTOR SPACES 53 for some a i 0. We know that a 1 = a 2 = a 3 = 0 if and only if det( [ ] v 1 v 2 v 3 ) 0. In this case we have 1 2 3 det 0 1 2 = 11 1 2 0 Hence, {v 1, v 2, v 3 } is LI. The determinant provides and easy check for linear independence when the number of components in the vectors equals the number of vectors in the set. When this is not the case, we can reduce the matrix [ v 1 v 2 v r ] to row echelon form in order to determine whether the vectors are LI. Span, dimension, and basis Span Given a set of vectors {v 1,...,v r } belonging to some vector space X, the set of all linear combinations of these vectors is called their span and is denoted span{v 1,...,v r }. For example, suppose X = R 3 and v 1, v 2 X: span{v 1, v 2 } = {sv 1 + tv 2 : s, t R} The set V = span{v 1,...,v r } is a subspace of X and we write V U. That is, V is itself a vector space using the vector addition and scalar multiplication of X and V is closed under the vector addition and scalar multiplication of X (i.e.,, any linear combination of elements in V belong to V ). Example 15 Defining the vectors e 1 = (1, 0,..., 0) T, e 2 = (0, 1, 0,..., 0) T,..., e n = (0, 0, 0,..., 1) T belonging to X = R n. We have R n = span{e 1, e 2,...,e n } The vectors e k are called the standard basis vectors of R n. Example 16 The vectors v 1 = (1, 1, 1) T, v 2 = (1, 1, 0) T, v 3 = (0, 1, 1) T span R 3, i.e., R 3 = span{v 1, v 2, v 3 } Basis A set of vectors {v 1,...,v r } belonging to a vector space X is called a basis if 1. {v 1,...,v r } is LI 2. X = span{v 1,...,v r } This definition implies that any vector in X can be expressed as a unique linear combination of v 1,..., v n. Evidently a set of basis vectors is the minimum number of vectors required to span a vector space. It is important to note that for a given vector space, the choice of basis is not unique. However, the number of elements in a basis is unique and leads to the following definition.
54 CHAPTER 3. LINEAR ALGEBRA Dimension If {v 1,...,v r } is a basis of a vector space X, the dimension of X is r and we write dim(x) = r. Example 17 Since det( [ e 1 e 2... e n ] ) = 1, we know {e1, e 2,...,e n } is LI and since R n = span{e 1, e 2,...,e n } we conclude {e 1, e 2,...,e n } is a basis of R n and dim(r n ) = n. Coordinates or representations of vectors In geometry we label points in a three dimensional space with an ordered triple of numbers called the coordinates of the point. These coordinates are determined from a definition of coordinate axes. Sometimes it is more convenient to choose a different coordinate system in which to represent a vector. For example, spherical coordinates may be more convenient than cartesian. A choice of new coordinates can be viewed as a choice of new basis for R 3. We now look at how the representation or coordinates of a vector changes as we vary the basis. Representation of a vector in a vector space Given the ordered basis {v 1,..., v r } of a vector space X, the representation or coordinates of any v X in the basis {v 1,...,v r } is the unique vector of coefficients (a 1,...,a r ) T such that v = r a i v i i=1 Example 18 Find the representation of v = (2, 1, 3) T in the basis {v 1, v 2, v 3 } of R 3, where We solve v 1 = (1, 1, 0) T, v 2 = (1, 0, 1) T, v 3 = (0, 1, 1) T 2 1 1 0 a 1 1 = 1 0 1 a 2 3 0 1 1 a 3 The unique solution of the above system of equation yields the coordinates (a 1, a 2, a 3 ) T = (0, 2, 1) T such that v = 3 i=1 a iv i. We have already used the notion of change of basis in Section 2.3 when considering linear changes of state coordinates. Although previously we introduced the notion of change of state coordinates without any concept of vector space or basis. This relation can be seen by considering the LTI system ẋ = Ax + Bu y = Cx + Du
3.2. LINEAR ALGEBRAIC EQUATIONS 55 We can expand x R n in terms of another basis {v 1,...,v n }, v i R n. The coordinates of x in the basis {v 1,...,v n }, v i R n are denoted ξ i : x = ξ 1 v 1 +...ξ n v n = V ξ where V = [ v 1 v n ], ξ = [ ξ1 ξ n ] T Taking the time derivative of x gives ẋ = V ξ = AV ξ + Bu (3.2) which implies and we have ξ = V 1 AV ξ + V 1 Bu y = CV ξ + Du Hence we observe that a change of basis is equivalent to a linear change of state variables as discussed in Section 2.3 (where no notion of basis or vector space was formally used). 3.2 Linear algebraic equations See [Bay99, Sec. 3.3]. We consider the systems of m linear equations in n variables y = Ax, where A R m n, x R n, y R m (3.3) A number of fundamental questions arise regarding (3.3). When do solutions to (3.3) exist? And if they exist, how many solutions exist? In order to answer these questions we need to define two vector spaces: the range space of A and the null space of A. Range space of A The range space of A, denoted R(A), is the set of all possible linear combinations of the columns of A, i.e., R(A) = {y R m : y = Ax for some x R n } = span{a 1,...,a n }, where a k R m denotes the k column of A. R(A) is sometimes called the column space of A. The matrix A can be thought of a linear mapping from R n to R m, and R(A) is a subspace of R m. Rank of A The rank of A is denoted r(a) and is defined as r(a) = dim(r(a)). Null space of A The null space of A, denoted N(A), is the set of all linear combinations of vectors x satisfying Ax = 0, i.e., N(A) = {x R n : Ax = 0}. Evidently, N(A) is a subspace of R n.
56 CHAPTER 3. LINEAR ALGEBRA Nullity of A The nullity of A is denoted q(a), it is defined as q(a) = dim(n(a)). We have the following useful facts r(a) min(n, m) q(a) = n r(a) r(a) = r(a T ) Dimension Theorem dimension of column space equals the dimension of row space r(a) = r(ua) = r(av ), for any nonsingular U R m m, V R n n Solutions to y = Ax exist if and only if y R(A). This is because, Ax is a linear combination of the columns of A. That is, if we let a i R m denote the columns of A we have Ax = [ a 1 a n ] x = a1 x 1 + + a n x n Evidently, for a solution of (3.3) to exist, y must be expressible as a linear combination of the columns of A. The statement y R(A) is equivalent to r(a) = r( [ A y ] ) or y is LI of the columns of A. When y / R(A) the equations (3.3) are said inconsistent. When y R(A), the equations (3.3) are said consistent and at least one solution exists. This solution is unique iff q(a) = 0. When a solution exists the two cases are Case I r(a) = n and the columns of A are LI and form a basis for R n. In this case, the solution to (3.3) is unique. The solution x is the representation of y in the basis formed by the columns of A. Case II r(a) < n or since q(a) = n r(a) we have q(a) > 0. In this case, an infinite number of solutions exist since Ax 0 = 0 has a nontrivial solution and we can take any scalar multiple of x 0 to generate an infinite number of solutions in the null space of A. Hence, we can add these solutions to x satisfying Ax = y to generate an infinite number of solutions to (3.3). That is, A(x + αx 0 ) = Ax + αax 0 = Ax + 0 = Ax = y, for any α R Solutions to Ax = y are parameterized as x = x p + x h = x p + α 1 n 1 + α 2 n 2 + + α k n k where x p is called the particular solution satisfying Ax p = y. The homogeneous solution is denoted x h = α 1 n 1 + α 2 n 2 + + α k n k and satisfies Ax h = A(α 1 n 1 + + α k n k ) = 0 That is x h N(A) and we have taken q(a) = k, N(A) = span{n 1,...,n k }. Solving Ax = y by hand In order to solve Ax = y we must perform elementary row operations to reduce a matrix to row echelon form (REF) or reduced row echelon form (RREF). Elementary row operations consist of
3.2. LINEAR ALGEBRAIC EQUATIONS 57 Multiplying any row by a nonzero scalar Interchanging any two rows Adding a multiple of one row to another row A matrix is in REF if all zero rows are at the bottom rows of the matrix the first nonzero entry from the left in any nonzero row is a one. This one is called a leading one each leading one is to the right of all leading ones above it A matrix is in RREF if it is in REF and each leading one is the only nonzero entry in its column. Examples of matrices in REF are [ ] 0 1 1, 0 0 1 0 0 1 0 0 0 0 where denotes some nonzero number. Examples of matrices in RREF are [ ] 0 1 0 1 0, 0 0 1 0 0 1 0 0 0 0 Procedure for Computing N(A), R(A), q(a), r(a), x h 1. Using elementary row operations, reduce the augmented matrix [ A 0 ] to REF. Here A R m n and 0 is a m 1 zero vector. 2. The rank of A, r(a) is the number of nonzero rows in the REF computed in previous step. 3. From the Dimension Theorem we have q(a) = n r(a). 4. The columns of A corresponding to the columns of [ A 0 ] in REF which contain leading ones are a basis for R(A). Be careful, the REF matrix tells you which columns of the original A matrix to pick. 5. Assign nonleading or free variables in the REF of [ A 0 ] to parameters. The number of parameters assigned equals q(a) = n r(a). 6. Solve for the leading or pivot variables in terms of the free variables. The number of pivot variables is r(a). 7. The solution to Ax h = 0 you obtain will be in the form x h = α 1 n 1 + α 2 n 2 + + α k n k where q(a) = n r(a) = k and N(A) = span{n 1,..., n k }. The α k are the free variables.
58 CHAPTER 3. LINEAR ALGEBRA Procedure for Computing x p 1. Reduce the augmented matrix [ A y ] to RREF. If the RREF has a leading one in the last column, no solution exists i.e., y / R(A) or the equations are inconsistent. 2. Introduce parameters for nonleading or free variables and solve for leading or pivot variables. The solution you obtain takes the form x = x p + α 1 n 1 + α 2 n 2 + + α k n k where q(a) = k and N(A) = span{n 1,...,n k }. Evidently, if you are required to compute x p, as a byproduct you obtain N(A), R(A), q(a), r(a), x h. Example 19 Given 0 1 1 2 4 A = 1 2 3 4, y = 8 2 0 2 0 0 Find N(A), R(A), q(a), r(a), x h, x p. Note that n = 4, m = 3. Since we require x p we reduce the augmented matrix [ A y ] to RREF. 0 1 1 2 4 1 2 3 4 8,2R 2 R 3 R 3, 0 4 4 8 16 1 2 3 4 8 0 1 1 2 4,R 1 R 2, 4R 1 R 3 R 3 0 0 0 0 0 1 0 1 0 0 0 1 1 2 4,R 1 2R 2 R 1 0 0 0 0 0 The number of nonzero rows in RREF (or REF) is the rank of A. Hence, r(a) = 2. We assign parameters to free variables x 3, x 4. x 3 = t, x 4 = s, t R s R Next we solve for pivot variables x 1, x 2 : x 1 = x 3 = t x 2 = 4 x 3 2x 4 = 4 t 2s Hence the solution can be written 0 1 0 x = x p + tn 1 + sn 2 = 4 0 + t 1 1 + s 2 0 (3.4) 0 0 1
3.2. LINEAR ALGEBRAIC EQUATIONS 59 Hence, the null space is given by 1 0 N(A) = span{n 1, n 2 } = span{ 1 1, 2 0 } 0 1 The dimension of N(A) is q(a) = n 2 = 2. A basis for R(A) is obtained by picking the column of A which correspond to the leading ones in the RREF of [ A y ]. In this case, the leading ones are in column 1 and 2. Hence, 0 1 R(A) = span{b 1, b 2 } = span{ 1, 2 } 2 0 We remark that the third column of A is b 1 + b 2 and the fourth column is 2b 2. As expected, y R(A) and in this case y = 4b 2. Solving Ax = y by MATLAB MATLAB provides a number of useful commands for computing N(A), R(A), q(a), r(a), x h, x p rref(a) computes the RREF of A. rrefmovie(a) shows the sequence of row operations to get RREF. rank(a) computes r(a) using a singular value decomposition. null(a) computes an orthonormalized basis for N(A). null(a, r ) computes a basis for N(A) in rational number form from rref output. orth(a) computes an orthonormalized basis for R(A). MATLAB s linear systems of equations commands y=[-4-8 0] ; A=[0 1 1 2;... 1 2 3 4;... 2 0 2 0]; AM=[A y]; Augmented matrix disp( Reduced row echelon form ); rref(am) disp( Basis for N(A) ); null(a, r ) disp( r(a) ); rank(a) disp( orth(a) ); orth(a) Program Output: >> Reduced row echelon form
60 CHAPTER 3. LINEAR ALGEBRA ans = 1 0 1 0 0 0 1 1 2-4 0 0 0 0 0 Basis for N(A) ans = -1 0-1 -2 1 0 0 1 r(a) ans = 2 Orthonormalized basis for R(A) ans = -0.3782 0.3084-0.8877 0.1468-0.2627-0.9399 3.3 Eigenvalues, eigenvectors See [Bay99, Sec. 4.1, 4.2, 4.3, 4.4.1, 4.4.2, 4.4.5]. For square matrices A R n n, a number λ C is an eigenvalue of A if there exists a nonzero vector v C n such that Av = λv (3.5) In this case v is called an eigenvector of A. Evidently (3.5) implies (A λi)v = 0 (3.6) We know v 0 iff det(a λi) = 0, and we know the number of LI solutions to (3.6) is the nullity of A λi or q(a λi). The set of all eigenvectors corresponding to an eigenvalue λ is a vector space and called the eigenspace of λ. We denote the eigenspace E λ (A) = {v C n : Av = λv}. Note that an eigenspace contains the zero vector and is a subspace of C n. Recall that φ(λ) = det(λi A) (3.7)
3.3. EIGENVALUES, EIGENVECTORS 61 is the characteristic polynomial of A, and the nth degree polynomial equation φ(λ) = det(λi A) = 0 (3.8) is the characteristic equation of A. Evidently the roots of the characteristic equation of A are the eigenvalues of A. [ ] a11 a Example 20 Consider A = 12. The characteristic polynomial of A is a 21 a 22 ([ ]) λ a11 a φ(λ) = det 12 a 21 λ a 22 = (λ a 11 )(λ a 22 ) a 12 a 21 = λ 2 λ(a 11 + a 22 ) + a 11 a 22 a 12 a 21 = λ 2 λ trace(a) + det(a) where trace(a) is the sum of the diagonal entries of A. Eigenvalues are given by the roots of the characteristic equation: λ = 1 2 (a 11 + a 22 ± (a 11 + a 22 ) 2 4(a 11 a 22 a 12 a 21 )) = 1 2 (a 11 + a 22 ± (a 11 a 22 ) 2 + 4a 12 a 21 ) We make the following useful observations about eigenvalues and eigenvectors: Eigenvectors and eigenvalues have a geometric interpretation. If v is an eigenvector of A, the effect of multiplying A by v is simple when the eigenvalue λ corresponding to v is real: it involves a scaling of v by λ. Eigenvectors are not unique. They can be scaled by any number. If A has complex eigenvalues, they appear in complex pairs. If λ C has a corresponding eigenvector x, this implies λ has a corresponding eigenvector x ( denotes complex conjugate). This is because A 1 exists iff A has no zero eigenvalues. Ax = λx (Ax) = (λx) Ax = λ x All eigenvalues of a symmetric matrix (i.e., A T = A) are real. All eigenvalues of a skew-symmetric matrix (i.e., A T = A) are purely imaginary. For A R n n we have trace(a) = n i=1 λ i and det(a) = Π n i=1 λ i.
62 CHAPTER 3. LINEAR ALGEBRA The eigenvalues of an upper or lower triangular matrix can be read off of its diagonal entries. This is because φ(λ) = Π n i=1 (λ a ii) = 0, where a ij denotes the (i, j)th entry of A. The eigenvalues of an upper or lower block triangular matrix is the union of eigenvalues of the submatrices appearing on its block diagonal. the eigenvalue of A and A T are the same. If λ is an eigenvalue of A, then λ m, m 0 is an eigenvalue of A m, and any eigenvector of A is an eigenvector of A m. If λ is an eigenvalue of A, then 1/λ is an eigenvalue of A 1, and any eigenvector of A is an eigenvector of A 1. Numerical methods which compute eigenvalues do not solve for the roots of the characteristic equation as this is not numerically stable. Diagonalizing matrices We suppose A R n n has n LI eigenvectors. Such matrices are said non-defective. We denote this set of eigenvectors as {v 1,...,v n }. We define a modal matrix as the matrix whose columns are the eigenvectors of A: M = [ v 1 v n ] (3.9) If we compute  = M 1 AM (3.10) we find that  is a diagonal matrix. This is because AM = [ Av 1 Av n ] = [ λ1 v 1 λ n v n ] = M diag(λ1,..., λ n ) = M (3.11) where λ 1 0 0 0 λ 2 0 diag(λ 1,...,λ n ) =...... 0 0 λ n Example 21 Given 1 0 1 A = 0 1 0 0 0 2 The eigenvalues of A can be read off the diagonal entries for upper or lower triangular matrices. Hence, in this case we have λ 1 = 2 with algebraic multiplicity m 1 = 1 and λ 2 = 1
3.3. EIGENVALUES, EIGENVECTORS 63 with algebraic multiplicity m 2 = 2. We compute eigenvector associated with λ 1 by solving the homogeneous linear system 1 0 1 (λ 1 I A)x = (2I A)x = 0 1 0 x = 0 0 0 0 Since λ 1 I A is already in RREF we assign x 3 = t, t R and solve for pivot variables which implies For λ 2 we solve x 1 = t x 2 = 0 1 x = t 0, t R 1 0 0 1 (I A)x = 0 0 0 x = 0 0 0 1 The augmented matrix [ I A 0 ] in RREF is 0 0 1 0 0 0 0 0 0 0 0 0 We assign parameters to free variables: x 1 = s, x 2 = t, where s, t R. And solve for the pivot variable to get 1 0 x = 0 s + 1 t, s, t R 0 0 We remark that q(i A) = 2 = m 2 and hence the matrix A has three LI eigenvectors and we can diagonalize it: 1 0 1 1 1 0 1 Â = M 1 AM = 0 1 0 A 0 1 0 = diag(1, 1, 2) 0 0 1 0 0 1 Example 22 Given 1 1 2 A = 0 1 3 0 0 2 As before, we can read the eigenvalues of A off its diagonal entries as A is upper triangular. Hence we have λ 1 = 2 with algebraic multiplicity m 1 = 1 and λ 2 = 1 with algebraic multiplicity m 2 = 2.
64 CHAPTER 3. LINEAR ALGEBRA We compute the eigenvector associated with λ 1 = 2 from 1 1 2 (2I A)x = 0 1 3 x = 0 0 0 0 Since 2I A is in REF we assign free variables x 3 = t, t R and solve for pivot variables x 1, x 2 : x 2 = 3t, x 1 = 3t + 2t = 5t Hence For λ 2 we have 5 x = 3 t, 1 t R 0 1 2 (I A)x = 0 0 3 x = 0 0 0 1 Reducing the augmented system to REF gives 0 1 2 0 0 0 1 0 0 0 0 0 Assigning x 1 = t and solving for x 2 = x 3 = 0 gives 1 x = 0 t, t R 0 In this case we only have two LI eigenvectors and diagonalization is not possible. In general, if for each distinct eigenvalue λ j we have q(λ j I A) = m j, i.e., the nullity of λ j I A equals the algebraic multiplicity of λ j then it is possible to diagonalize A. If on the other hand q(λ j I A) < m j for some j, we cannot diagonalize A. The number q(λ j I A) is called the geometric multiplicity of λ j. In general, the algebraic multiplicity is greater than or equal to the geometric multiplicity, and when they are equal diagonalization is possible. Since q(λ j I A) 1 we observe that if a matrix has distinct eigenvalues, it can always be diagonalized. It turns out that for symmetric matrices (i.e., A = A T ), n LI eigenvectors can always be found and that the eigenvalues of A are real. Similarly skew-symmetric matrices (i.e., A T = A) and unitary matrices (i.e., A T = A 1 ) have n LI eigenvectors and can also always be diagonalized. It should also be noted that although not all matrices can be diagonalized, any matrix can be put into a block diagonal so-called Jordan Form which is almost diagonal : see [Bay99, pg. 172]. The Jordan form of a matrix has the form J = blockdiag{j 1,...,J p }
3.3. EIGENVALUES, EIGENVECTORS 65 where J i R r i r i, 1 i p, p i=1 r i = n, and λ i 1 0 0 0 0 0 λ i 1 0 0 0. 0 0 λ i.. 0 0 0 J i =........... 0 0 0 λ i 1 0 0 0 0 0 λ i 1 0 0 0 0 0 λ i The matrices J i are called Jordan blocks and have an eigenvalue on its diagonal entries and ones on its superdiagonal (except when r i = 1). It can be readily shown that each J i has only one LI eigenvector. Note that in general there are many possible choices of r i such that the above conditions hold. To see this we consider a 5 5 matrix with algebraic multiplicities m 1 = 4 and m 2 = 1. Such a matrix could have one of 5 possible Jordan forms depending on the geometric multiplicity of λ 1. If the geometric multiplicity q(λ 1 I A) = 1 then the Jordan form has 1 Jordan block associated with λ 1 : λ 1 1 0 0 0 0 λ 1 1 0 0 0 0 λ 1 1 0 0 0 0 λ 1 0 0 0 0 0 λ 2 If the geometric multiplicity q(λ 1 I A) = 2 then 2 Jordan blocks are associated with λ 1 and we have two possible cases λ 1 1 0 0 0 0 λ 1 0 0 0 0 0 λ 1 1 0 0 0 0 λ 1 0 0 0 0 0 λ 2 λ 1 1 0 0 0 0 λ 1 1 0 0 0 0 λ 1 0 0 0 0 0 λ 1 0 0 0 0 0 λ 2 When q(λ 1 I A) = 3 then 3 Jordan blocks are associated with λ 1 and we have 1 possible case λ 1 1 0 0 0 0 λ 1 0 0 0 0 0 λ 1 0 0 0 0 0 λ 1 0 0 0 0 0 λ 2
66 CHAPTER 3. LINEAR ALGEBRA And finally when q(λ 1 I A) = m 1 = 4 the matrix is diagonalizable which is a special case of the Jordan form: λ 1 0 0 0 0 0 λ 1 0 0 0 0 0 λ 1 0 0 0 0 0 λ 1 0 0 0 0 0 λ 2 Diagonalizing LTI state space models Just as we can diagonalize matrices, we can also transform the state coordinates of an LTI state space system model by diagonalizing its system matrix. Example 23 1 2 0 1 ẋ = 1 2 0 x + 0 u 2 1 3 0 [ ] 1 0 1 y = x 1 1 0 The eigenvalues of the system matrix are determined from λ + 1 2 0 det(λi A) = det 1 λ 2 0 2 1 λ + 3 Taking an expansion down column 3 gives det(λi A) = (λ + 3)((λ + 1)(λ 2) + 2) = λ(λ + 3)(λ 1) Hence we have distinct eigenvalues λ 1 = 0, λ 2 = 1, λ 3 = 3. Computing eigenvectors 2 λ 1 = 0 x = 1 t, t R 1 4 λ 2 = 1 x = 4 t, t R 1 0 λ 3 = 3 x = 0 t, t R 1 Hence a change of coordinates Mξ = x puts 1 the system into the form ξ = Âξ + ˆBu y = Ĉξ (3.12) 1 Note that compared to Section 2.3 we have changed convention in our definition of new coordinates from ξ = Mx to Mξ = x.
3.3. EIGENVALUES, EIGENVECTORS 67 where the matrices Â, ˆB, Ĉ are given byâ = M 1 AM ˆB = M 1 B Ĉ = CM We have and we obtain 2 4 0 1 1 0 M = 1 4 0 M 1 = 1/4 1/2 0 1 1 1 3/4 1/2 1 Â = diag(0, 1, 3) 1 1 0 1 1 ˆB = 1/4 1/2 0 0 = 1/4 3/4 1/2 1 0 3/4 [ ] 2 4 0 1 0 1 Ĉ = = 1 1 0 1 4 0 1 1 1 [ 1 3 1 3 8 0 When a system has an A-matrix which is diagonal (assuming eigenvalues are real), we sometimes say it is in modal form. Note that since eigenvectors are not uniquely defined, neither are the matrices ˆB, Ĉ. The JCF we saw in Chapter 2 is a generalization of the modal form. MATLAB has the command [CSYS,T]= canon(sys, modal ) transforms a systemsys into modal form. In this case if ξ denote the modal form coordinates, ξ = Tx. ]
Bibliography [Bay99] John S. Bay. Fundamentals of linear state space systems. McGraw-Hill, 1999. [Che99] Chi-Tsong Chen. Linear System Theory and Design. Oxford, New York, NY, 3rd edition, 1999. [Max68] J.C. Maxwell. On governors. Proceedings of the Royal Society of London, 16:270 283, 1868. 123