MIT6_241JS11_lec04

# MIT6_241JS11_lec04 - 6.241 Dynamic Systems and Control...

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Unformatted text preview: 6.241 Dynamic Systems and Control Lecture 4: Singular Values Readings: DDV, Chapter 4 Emilio Frazzoli Aeronautics and Astronautics Massachusetts Institute of Technology February 14, 2011 E. Frazzoli (MIT) Lecture 4: Singular Values Feb 14, 2011 1/9 Outline 1 Singular Values 2 Norm computations through singular values E. Frazzoli (MIT) Lecture 4: Singular Values Feb 14, 2011 2/9 Unitary Matrices A square matrix U ∈ Cn×n is unitary if U � U = UU � = I . A square matrix U ∈ Rn×n is orthogonal if U T U = UU T = I . Properties: If U is a unitary matrix, then �Ux �2 = �x �2 , for all x ∈ Cn . If S = S � is a Hermitian matrix, then there exists a unitary matrix U such that U � SU is a diagonal matrix. 1 For any matrix A ∈ Rm×n , both A� A ∈ Rn×n , AA� ∈ Rm×m are Hermitian ⇒ can be diagonalized by unitary matrices. For any matrix A, the eigenvalues of A� A and AA� are always real 2 and non-negative 3 (in other words, A� A and AA� are positive deﬁnite). = S � ⇔ �Sx , y � = �x , Sy �. Let v1 be an eigenvector of S , and let M1 = R(v1 )⊥ . If u ∈ M1 , then so is Su : �Su , v1 � = �u , Sv1 � = �u , λ1 v1 � = 0. All other eigenvectors must b e in M1 . Finite induction gets the result. 2 Assuming �v , v � = 1, λ = �Sv , v � = �v , Sv � = �Sv , v �� = λ� 11 1 11 1 1 11 1 3 0 < �Av , Av � = v � A� Av = λ v � v . 1 1 1 111 1 1S E. Frazzoli (MIT) Lecture 4: Singular Values Feb 14, 2011 3/9 Singular Value Decomposition Theorem (SVD) Any matrix A ∈ Cm×n can be decomposed as A = U ΣV , where U ∈ Cm×m and V ∈ Cn×n are unitary matrices. The matrix Σ ∈ Rm×n is “diagonal,” with non-negative elements on the main diagonal. The non-zero elements of Σ are √ called the singular values of A, and satisfy σi = i-th eigenvalue of A� A. Proof (assuming rank(A) = m): Since AA� is Hermitian, there exist a diagonal matrix Λ = diag(λ1 , λ2 , . . . , λm ) > 0 such that U ΛU � = AA� . 2 2 2 Write Λ = Σ2 = diag(σ1 , σ2 , . . . , σm ) 1 � � Deﬁne V1 := Σ−1 U � A ∈ Rm×n . Clearly, V1 V1 = Σ−1 U � AA� U Σ−1 = I m×m . 1 1 1 Construct V = [V1 , V2 ] ∈ Cn×n by choosing the columns in V2 so that V is unitary, and Σ = [Σ1 , 0] ∈ Rn×n , by padding with zeroes. � � Hence, ΣV � = Σ1 V1 + 0V2 = U � A, i.e., A = U ΣV � . E. Frazzoli (MIT) Lecture 4: Singular Values Feb 14, 2011 4/9 Singular Vectors If U and V are written as sequences of column vectors, i.e., � � � � U = u1 , u2 , . . . , um and V = v1 , v2 , . . . , vn , then A = U ΣV � = r � σi ui vi� i =1 The columns of U are called the left singular vectors, and the columns of V are called the right singular vectors. Note: Ax can be written as the weighted sum of the left singular vectors, where the weights are given by the projections of x onto the right singular vectors: Ax = r � σi ui (vi� x ), i =1 The range of A is given by the span of the ﬁrst r vectors in U The rank of A is given by r ; The nullspace of A is given the span of the last (n − r ) vectors in V . E. Frazzoli (MIT) Lecture 4: Singular Values Feb 14, 2011 5/9 Outline 1 Singular Values 2 Norm computations through singular values E. Frazzoli (MIT) Lecture 4: Singular Values Feb 14, 2011 6/9 Induced 2-norm computation Theorem (Induced 2-norm) �A�2 = sup x �=0 �Ax �2 = σmax (A). �x �2 Proof: sup x =0 � �Ax �2 �U ΣV � x �2 �ΣV � x �2 = sup = sup = �x �2 �x �2 �x �2 x �=0 x �=0 ��n � 2 2 1/2 �Σy �2 �Σy �2 i = σ | yi | sup = sup = sup �� 1 i �1/2 ≤ σmax (A). n y =0 �y �2 y =0 y �=0 �Vy �2 � � |yi |2 i =1 Assuming σmax = σ1 , the supremum is attained for y = (1, 0, . . . , 0). This corresponds to x = v1 , and Av1 = σ u1 E. Frazzoli (MIT) Lecture 4: Singular Values Feb 14, 2011 7/9 Minimal ampliﬁcation Theorem Given A ∈ Cm×n , with rank(A) = n, �Ax �2 = σn (A). x �=0 �x �2 inf Proof: �Ax �2 �U ΣV � x �2 �ΣV � x �2 = inf = inf = x �=0 �x �2 x �=0 x �=0 �x �2 �x �2 ��n � 2 2 1/2 �Σy �2 �Σy �2 i =1 σi |yi | inf = inf = inf �� �1/2 ≥ σmin (A). n y =0 �y �2 y �=0 y �=0 �Vy �2 � |yi |2 inf i =1 Assuming σmin = σn , the supremum is attained for y = (0, . . . , 0, 1). This corresponds to x = vn , and Avn = σ un E. Frazzoli (MIT) Lecture 4: Singular Values Feb 14, 2011 8/9 Frobenius norm computation Theorem �A�F = �r � �1/2 2 σi (A) i =1 Proof: ⎞1/2 ⎛ n m �� 1/2 1/2 |aij |2 ⎠ = (Trace(A� A)) = (Trace(V Σ� U � U ΣV � )) = �A�F = ⎝ j =1 i =1 � �1/2 � �1/2 Trace(V � V Σ2 ) = Trace(Σ2 ) = �r � �1/2 σi2 i =1 E. Frazzoli (MIT) Lecture 4: Singular Values Feb 14, 2011 9/9 MIT OpenCourseWare http://ocw.mit.edu 6.241J / 16.338J Dynamic Systems and Control Spring 2011 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms . ...
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