1 Some Facts on Symmetric Matrices

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1 1 Some Facts on Symmetric Matrices Definition: Matrix A is symmetric if A = A T. Theorem: Any symmetric matrix 1) has only real eigenvalues; 2) is always iagonalizable; 3) has orthogonal eigenvectors. Corollary: If matrix A then there exists Q T Q = I such that A = Q T ΛQ. Proof: 1) Let λ C be an eigenvalue of the symmetric matrix A. Then Av = λv, v, an v Av = λv v, v = v T. But since A is symmetric Therefore, λ must be equal to λ! λv v = v Av = (v Av) = λv v. 2) If the symmetric matrix A is not iagonalizable then it must have generalize eigenvalues of orer 2 or higher. That is, for some repeate eigenvalue λ i there exists v such that But note that (A λ i I) 2 v =, (A λ i I)v = v (A λ i I) 2 v = v (A λ i I)(A λ i I), which is contraiction. Therefore, as there exists no generalize eigenvectors of orer 2 or higher, A must be iagonalizable. 3) As A must have no generalize eigenvector of orer 2 or higher AT = A [ v 1 v n ] = [ v1 v n ] Λ = T Λ, T. That is A = T 1 ΛT. But since A is symmetric or T 1 ΛT = A = A T = (T 1 ΛT ) T = T T ΛT T T T = T 1 T T T = I v T i v i = 1, v T i v j =, i j. MAE 28A 1 Maurício e Oliveira

2 1.1 Positive efinite matrices Definition: The symmetric matrix A is sai positive efinite (A > ) if all its eigenvalues are positive. Definition: The symmetric matrix A is sai positive semiefinite (A ) if all its eigenvalues are non negative. Theorem: If A is positive efinite (semiefinite) there exists a matrix A 1/2 > (A 1/2 ) such that A 1/2 A 1/2 = A. Proof: As A is positive efinite (semiefinite) A = Q T ΛQ, Q T Q = QQ T = I = Q T Λ 1/2 Λ 1/2 Q, Λ 1/2 ii = Q T Λ 1/2 Q }{{} QT Λ 1/2 Q }{{}, A 1/2 A 1/2 = λ i Theorem: A is positive efinite if an only if x T Ax >, x. Proof: Assume there is x such that x T Ax an A is positive efinite. Then there exists Q T Q = I such that A = Q T ΛQ with Λ ii = λ i >. Then for y such that x = Q T y x T Ax = y T QAQy = y T QQ T ΛQQ T y = y T Λy = which is a contraiction. n λ i yi 2 > i=1 MAE 28A 2 Maurício e Oliveira

3 2 Controllability Gramian LTI system in state space ẋ(t) = Ax(t) + Bu(t), y(t) = Cx(t) Problem: Given x() = an any x, compute u(t) such that x( t) = x for some t >. Solution: We know that x = x( t) = t e A( t τ) Bu(τ)τ. If we limit our search for solutions u in the form u(t) = B T e AT ( t t) z we have an t x = e A( t τ) BB T e ( t τ) zτ, AT ( ) t = e A( t τ) BB T e AT ( t τ) τ z, ( ) t = e Aξ BB T e AT ξ ξ z, ξ = t τ z = ( 1 t e Aξ BB T e ξ) AT ξ x, The symmetric matrix X(t) := u(t) = B T e AT ( t t) is known as the Controllability Gramian. e Aξ BB T e AT ξ ξ ( 1 t e Aξ BB T e ξ) AT ξ x MAE 28A 3 Maurício e Oliveira

4 2.1 Properties of the Controllability Gramian Theorem: The Controllability Gramian X(t) = is the solution to the ifferential equation If X = lim t X(t) exists then Proof: For the first part, compute e Aξ BB T e AT ξ ξ, t X(t) = AX(t) + X(t)AT + BB T. AX + XA T + BB T =. t X(t) = e Aξ BB T e AT ξ ξ = e A(t τ) BB T e AT (t τ) τ, t t = t ea(t τ) BB T e AT (t τ) + e A(t τ) BB T e AT (t τ), τ=t ( ) = A e A(t τ) BB T e AT (t τ) τ ( ) + e A(t τ) BB T e AT (t τ) τ A T + BB T, = AX(t) + X(t)A T + BB T. For the secon part, use the fact that X(t) is smooth an therefore lim X(t) = X lim X(t) =. t t t MAE 28A 4 Maurício e Oliveira

5 2.2 Summary on Controllability Theorem: The following are equivalent 1) The pair (A, B) is controllable; 2) The Controllability Matrix C(A, B) has full-row rank; 3) There exists no z such that z A = λz, z B = ; 4) The Controllability Gramian X(t) is positive efinite for some t. Proof: Everything has alreay been prove except the equivalence of 4). Sufficiency: Immeiate from the construction of u(t). Necessity: First part: X(t) = e Aξ BB T e AT ξ ξ by construction. We have to prove that when (A, B) is controllable then X(t) >. To prove this assume that (A, B) is controllable but X(t) is not positive efinite. So there exists z such that z e Aτ B, τ t. But this implies i τ i(i! z e Aτ B) = z A i e Aτ B τ= = z A i B =, i =,..., n 1 τ= which implies C(A, B) oes not have full-row rank (see proof of the Popov-Belevitch-Hautus Test). MAE 28A 5 Maurício e Oliveira

6 3 Observability Gramian LTI system in state space ẋ(t) = Ax(t) + Bu(t), y(t) = Cx(t) Problem: Given u(t) = an y(t) compute x(). Solution: We know that y(t) = Ce At x(). Multiplying on the left by e AT t C T an integrating from to t we have ( ) e AT ξ C T y(ξ)ξ = e AT ξ C T Ce Aξ ξ x() from which ( x() = The symmetric matrix ) 1 e AT ξ C T Ce Aξ ξ e AT ξ C T y(ξ)ξ. Y (t) := is known as the Observability Gramian. e AT ξ C T Ce Aξ ξ MAE 28A 6 Maurício e Oliveira

7 3.1 Properties of the Observability Gramian Theorem: The Observability Gramian Y (t) = is the solution to the ifferential equation If Y = lim t X(t) exists then 3.2 Summary on Observability Theorem: The following are equivalent 1) The pair (A, C) is observable; e AT ξ C T Ce Aξ ξ, t Y (t) = AT Y (t) + Y (t)a + C T C. A T Y + Y A + C T C =. 2) The Observability Matrix O(A, C) has full-column rank; 3) There exists no x such that Ax = λx, Cx = ; 4) The Observability Gramian Y = Y (t) is positive efinite for some t. MAE 28A 7 Maurício e Oliveira

8 4 Controllability, Observability an Duality Primal LTI system in state space ẋ(t) = Ax(t) + Bu(t), y(t) = Cx(t) Dual LTI system in state space ẋ(t) = A T x(t) + C T u(t), y(t) = B T x(t). The primal system is observable if an only if the ual system in controllable. The primal system is controllable if an only if the ual system in observable. MAE 28A 8 Maurício e Oliveira