# Proof from denition aa u v u v u v v u 2 2 u

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ¥ ¥¡ ¥ C 4 ¦¥%# d¦¥ ¥  ! ¤ ¦  F  0  F ¦  F¡   §0 §¤ &31 A I   ¦¥# §¤  ¡%8Q§% ( h¥' ¦ 4    0 0  I    !¡  ¦¡ ¡ ¦ §£ ¤ \$&' !¥£ \$ Q\$§F \$¡ ©  ! )\$¡ # Q¥ )  #  ¥  ¥  " ¡ A I ¦%) ¦¥# Q¤  00  ¡0 4 ¦  4  ¦¥§S ¢¢  ¡   ©  ¢¢   £# ¥§¦ ¦\$EB¥§ Q0 4&§F%   ¡C ¡ ¡  %& § A !   ¨        '¥§%U!¥§\$¡ § ©  0 Q  QF    ¨¨ Let A ∈ IRm×n h0     8 Full and reduced SVD ¤ Uniqueness First note that σ1 and v1 can be uniquely determined by A 2 Suppose in addition to v1 , there is another linearly independent vector w with w 2 = 1 and Aw 2 = σ1 Deﬁne a unit vector v2 , orthogonal to v1 as a linear combination of v1 and w w − (v1 w)v1 v2 = w − (v1 w)v1 2 Since A 2 = σ1 , Av2 2 ≤ σ1 , but this must be an equality, for otherwise w = c v1 + s v2 for some constants c and s with |c |2 + |s |2 = 1, we would have Aw < σ1 v2 is a second right singular vector of A...
View Full Document

## This document was uploaded on 02/10/2014.

Ask a homework question - tutors are online