assignment3

# X2 x2 3 maximum so ax x a na2 or equivalently 1

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Unformatted text preview: dt This is a system of two equations and two unknowns, which we can rewrite in matrix form: � � � T3 T2 � � � y a1 3 2 , = T 2 a2 0 T 2 So, � a1 a2 � � = T3 3 T2 2 T2 2 �−1 � T y 0 � . Exercise 4.1 Note that for any v ∈ C m , (show this!) �v �∞ ≤ �v �2 ≤ √ m�v �∞ . (1) Therefore, for A ∈ C m×n with x ∈ Cn �Ax�2 ≤ √ √ Ax �∞ m�Ax�∞ → for x = 0, ��x��2 ≤ m �Ax�2 . � �x 2 But, from equation (1), we also know that �Ax�2 ≤ �x�2 1 �x�∞ ≥ 1 �x�2 . Thus, √ √ m�Ax�∞ m�Ax�∞ √ ≤ m�A�∞ , ≤ �x�2 �x�∞ � Equation (2) must hold for all x = 0, therefore 3 (2) maxx=0 � To prove the lower bound �Ax�∞ 1 √ �A�∞ n �Ax�2 �x�2 = �A�2 ≤ √ m�A�∞ . ≤ �A�2 , reconsider equation (1): �Ax�∞ �Ax�2 ≤ �Ax�2 → for x �= 0, ≤ ≤ �A�2 → �x�2 �x�2 But, from equation (1) for x ∈ C n , √ n �x�2 �Ax�∞ �x�∞ for all x = 0 including x that makes � ≥ �x1 ∞ . So, � √ n�Ax�∞ √ ≤ ≤ n�A�2 �x�2 �Ax�∞ �x�∞ maxx=0 � √ √ n�Ax�∞ n�Ax�2 √ ≤ ≤ n�A�2 . �x�2 �x�2 (3) maximum, so, �Ax�∞ �x�∞ = �A�∞ ≤ √ n�A�2 , or equivalently, 1 √ �A�∞ ≤ �A�2 . n Exercise 4.5 Any m × n matrix A, it can be expressed as � � Σ0 A=U V �, 00 where U and V are unitary matrices. The “Moore-Penrose inverse”, or pseudo-inverse of A, denoted by A+ , is then deﬁned as the n × m matrix � −1 � Σ 0 A+ = V U �. 0 0 a) Now we have to show that A+ A and AA+ are symmetric, and that AA+ A = A and A+ AA+ = A+ . Suppose that Σ is a diagonal invertible matrix with the dimension of r × r. Using the given deﬁnitions as well a...
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## This document was uploaded on 03/19/2014 for the course EECS 6.241J at MIT.

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