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### problem3_sol

Course: EE 221A, Fall 2011
School: Berkeley
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Word Count: 776

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Problem EE221A Set 3 Solutions - Fall 2011 Problem 1. a) A w.r.t. the standard basis is, by inspection, 00 0 4 . 02 2 AE = 1 0 b) Now consider the diagram from LN3, p.8. We are dealing with exactly this situation; we have one matrix representation, and two bases, but we are using them in both the domain and the codomain so we have all the ingredients. So the matrices P and Q for the similarity transform in...

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Problem EE221A Set 3 Solutions - Fall 2011 Problem 1. a) A w.r.t. the standard basis is, by inspection, 00 0 4 . 02 2 AE = 1 0 b) Now consider the diagram from LN3, p.8. We are dealing with exactly this situation; we have one matrix representation, and two bases, but we are using them in both the domain and the codomain so we have all the ingredients. So the matrices P and Q for the similarity transform in this case are, P= e1 e2 e3 = b1 b2 b3 1 b1 b2 b3 , since the matrix formed from the E basis vectors is just the identity; and b1 Q= b2 b3 1 e1 e2 e3 = b1 b2 b3 1 = P 1 . Let AB be the matrix representation of A w.r.t. B . From the diagram, we have AB = QAE P = P 1 AE P b1 = b2 12 = 0 0 21 16 1 7 = 15 21 1 b3 AE b1 b2 b3 1 0 2 00 1 5 1 0 4 0 1 0 02 2 4 12 32 6 6 12 0 5 1 2 0 1 Problem 2. Representation of a linear map. This is straightforward from the denition of matrix representation, 0 0 . .. . 1 . . .. A= 0 1 .0 . .. . . 0 . 0 1 Problem 3. Norms. n 2 n n 2 2 Proof. 1st inequality: Consider the Cauchy-Schwarz inequality, ( i=1 xi yi ) i=1 xi i=1 yi . Now, let 2 2 y = 1 (vector of all ones). Then we have x 1 n x 2 which is equivalent to the rst inequality. 2 2 2nd inequality: Note that x 2 x 1 x 2 x 1 . Consider that x 2 2 2 2 = |x1 | + + |xn | , while x 2 1 2 = (|x1 | + + |xn |) 2 2 2 = |x1 | + |x1 | |x2 | + + |x1 | |xn | + |x2 | + |x2 | |x1 | + + |xn | |xn1 | + |xn | =x 2 2 + (cross terms), showing the second inequality. 2 Problem 4. Proof. First note that the problem implies that A Fmn . By denition, A Consider Au 1 n j =1 = respectively. Then Au 1 Aj uj 1 n j =1 1,i = sup uU Au 1 . u1 , where Aj and uj represent the j -th column of A and the j -th component of u Aj 1 |uj |. Let Amax be the column of A with the maximum 1-norm; that is, m Amax = Then Au 1 n j =1 Amax |uj | = Amax n j =1 |aij | . max j {1,...,n} i=1 |uj | = Amax u 1 . So we have that Au 1 Amax . u1 T Now, it remains to nd a u such that equality holds. Chose u = (0, . . . , 1, . . . 0) , where the 1 is in the k -th component that such Au pulls out a column of A having the maximum 1-norm. Note that u 1 = 1, and we see then that Au 1 = Amax . u1 Thus in this case the supremum is achieved and we have the desired result. Problem 5. Proof. Straightforward; we simply use properties of the inner product at each step: x+y 2 + xy 2 = x + y, x + y + x y, x y = x + y, x + x + y, y + x y, x + x y, y = ( x, x + y + y , x + y + x, x y + y, x y ) = ( x, x + x, y + y , x + y , y + x, x + x, y + y, x + y, y ) =2 x 2 +2 y 2 + x, y + y , x x, y + x, y =2 x 2 +2 y 2 + x, y + x, y =2 x 2 +2 y 2 + ( x, y + x, y ) =2 x 2 +2 y 2 + x, y y =2 x 2 +2 y 2 Problem 6. We will show that the adjoint map A : Cn Cn is identical to matrix multiplication by the complex conjugate transpose of A. Initially we will use the notation Aa for the matrix representation of the adjoint of A and reserve the notation v for the complex conjugate transpose of v . First, we know that we can represent A (w.r.t. the standard basis of Cn ) by a matrix in Cnn ; call this matrix A. Then we can use the dening property of the adjoint to write, Au, v = u, Aa v u A v = u Aa v Now, this must hold for all u, v Cn . Choose u = ei , v = ej (where ek is a vector that is all zeros except for 1 in the k -th entry). This will give, a = aa , ij ij for all i, j {1, . . . , n}. Thus Aa = A ; it is no accident that we use the notation for both adjoints and complex conjugate transpose. 3 Problem 7. Continuity and Linearity. Proof. Let A : (U, F ) (V, F ) with dim U = n and dim V = m be a linear map. Let x, y U , x = y , and z = x y . Since A is a linear map between nite dimensional vector spaces we can represent it by a matrix A. Now, the induced norm, Az A i := sup z z U,z =0 = Given some Az A i z. > 0, let = A i So x y = z < = Az < A i = A i A = Az < i and we have continuity. Alternatively, we can also use the induced matrix norm to show Lipschitz continuity, x, y U, Ax Ay < K x y , where K > A i , which shows that the map is Lipschitz continuous, and thus is continuous (LC = C , note that the reverse implication is not true!).
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Berkeley - EE - 221A
EE221A Linear System TheoryProblem Set 4Professor C. TomlinDepartment of Electrical Engineering and Computer Sciences, UC BerkeleyFall 2011Issued 9/30; Due 10/7Problem 1: Existence and uniqueness of solutions to dierential equations.Consider the fo
Berkeley - EE - 221A
EE221A Problem Set 4 Solutions - Fall 2011Problem 1. Existence and uniqueness of solutions to dierential equations.TTCall the rst system f (x, t) = x1 x2and the second one g (x) = x1 x2.a) Construct the Jacobians:D1 f (x, t) =1 et sin (x1 x2 )et
Berkeley - EE - 221A
EE221A Linear System TheoryProblem Set 5Professor C. TomlinDepartment of Electrical Engineering and Computer Sciences, UC BerkeleyFall 2011Issued 10/18; Due 10/27Problem 1: Dynamical systems, time invariance.Suppose that the output of a system is r
Berkeley - EE - 221A
EE221A Problem Set 5 Solutions - Fall 2011Problem 1. Dynamical systems, time invariance.i) To show that this is a dynamical system we have to identify all the ingredients:First we need a dierential equation of the form x = f (x, u, t): Let x(t) = y (t)
Berkeley - EE - 221A
EE221A Linear System TheoryProblem Set 6Professor C. TomlinDepartment of Electrical Engineering and Computer Sciences, UC BerkeleyFall 2011Issued 10/27; Due 11/4Problem 1: Linear systems. Using the denitions of linear and time-invariance discussed i
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EE221A Problem Set 6 Solutions - Fall 2011Problem 1. Linear systems.a) Call this dynamical system L = (U , , Y , s, r), where U = Rni , = Rn , Y = Rno . So clearly U , , Y are alllinear spaces over the same eld (R). We also have the response map(t, t0
Berkeley - EE - 221A
EE221A Linear System TheoryProblem Set 7Professor C. TomlinDepartment of Electrical Engineering and Computer Sciences, UC BerkeleyFall 2011Issued 11/3; Due 11/10Problem 1.A has characteristic polynomial (s 1 )5 (s 2 )3 , it has four linearly indepe
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EE221A Problem Set 7 Solutions - Fall 2011Problem 1.With the given information, we can determine the Jordan form J = T AT 1 of A to be,10J =1110111200120.012Thus,cos e10=cos eJe1 sin e1cos e1and cos eA = T 1 cos eJcos e01
Berkeley - EE - 221A
EE221A Linear System TheoryProblem Set 8Professor C. TomlinDepartment of Electrical Engineering and Computer Sciences, UC BerkeleyFall 2011Issued 11/10; Due 11/18Problem 1: BIBO Stability.fH , THiTHTCfC , TCVHVCiFigure 1: A simple heat ex
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EE221A Problem Set 8 Solutions - Fall 2011Problem 1. BIBO Stability.a) First write this LTI system in state space form,x = Ax + Bu=( +fC )VCVH=0.3 0.20.2 0.3y = Cx =10VC( +fH )VH01fCVCx+0.10x+0fHVH000.1u,uxwhere x := (TC
Berkeley - EE - 221A
EE221A Linear System TheoryProblem Set 9Professor C. TomlinDepartment of Electrical Engineering and Computer Sciences, UC BerkeleyFall 2011Issued 11/21; Due 12/1Problem 1: Lyapunov Equation.(a) Consider the linear map L : Rnn Rnn dened by L(P ) = A
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EE221A Linear System TheoryProblem Set 10Professor C. TomlinDepartment of Electrical Engineering and Computer Sciences, UC BerkeleyFall 2011Issued 12/2; Due 12/9Problem 1: Feedback control design by eigenvalue placement. Consider the dynamic system:
Berkeley - EE - 221A
8/26/11EE221A Section 11Administrivia1.1Section and Oce Hours SchedulingIf you have a conict, please contact Prof. Tomlin and/or Pat by email and let us know about it.Well try to accomodate everyone but it may not be possible.1.2Homework collabor
Berkeley - EE - 221A
9/2/11EE221A Section 21Fields1. Show that the set cfw_0, 1, with multiplication dened as binary AND and additiondened as binary XOR, is a eld. (AND)01+ (XOR) 0 10011100100012. Show that F, 0 = 0 = 0.2Vector Spaces1. Does C form a vecto
Berkeley - EE - 221A
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Berkeley - EE - 221A
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