assignment3

If a has full row rank we have shown in chapter 3

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Unformatted text preview: s the fact that for a unitary matrix U , U � U = U U � = I , we have AA + � Σ−1 0 =U VV U� 0 0 � � � � −1 Σ0 Σ 0 =U I U� 00 0 0 � � Ir×r 0 =U U �, 0 0 � Σ0 00 � 4 � � which is symmetric. Similarly, � Σ−1 0 0 0 � Σ−1 0 � + Ir×r 0 AA = V =V =V � Σ0 V� UU 00 � �� 0 Σ0 V� I 0 00 � 0 V� 0 � � � which is again symmetric. The facts derived above can be used to show the other two. + AA A = = = = � Ir×r (AA )A = U 0 � � � Ir×r 0 � UU U 0 0 � � Σ0 U V� 00 A. + 0 0 � U �A � Σ0 V� 00 Also, + + A AA � Ir×r 0 = (A A)A = V V � A+ 0 0 � � � � −1 Ir×r 0 Σ 0 � =V U� VV 0 0 0 0 � −1 � Σ 0 =V U� 0 0 + � + = A+ . b) We have to show that when A has full column rank then A+ = (A� A)−1 A� , and that when A has full row rank then A+ = A� (AA� )−1 . If A has full column rank, then we know that m ≥ n, rank (A) = n, and � � Σn×n A=U V �. 0 Also, as shown in chapter 2, when A has full column rank, (A� A)−1 exists. Hence � � �−1 � � � �� �� Σ � −1 � (A A) A = V Σ 0 UU V� V Σ� 0 U � 0 � � �� � � � −1 V Σ 0 U� = V Σ ΣV � � = V (Σ� Σ)−1 V � V Σ� 0 U � � � = V (Σ� Σ)−1 Σ� 0 U � = V ( Σ−1 0 )U � = A+ . 5 Similarly, if A has full row rank, then n ≥ m, rank (A) = m, and � � A = U Σm×m 0 V � . It can be proved that when A has full row rank, (A� A)−1 exists. Hence, � � −1 A (AA ) � � � � �−1 � �� Σ =V U U Σ 0 VV U� 0 � �� � �− 1 Σ =V U � U ΣΣ� U � 0 � �� Σ =V U � U (ΣΣ−1 )U � 0 � −1 � Σ =V U� 0 � Σ� 0 � � = A+ . c) Show that, of all x that minimize �y − Ax�2 , the one with t...
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This document was uploaded on 03/19/2014 for the course EECS 6.241J at MIT.

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