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

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Unformatted text preview: ¥ ¥¡ ¥ C 4 ¦¥%# d¦¥ ¥  ! ¤ ¦  F  0  F ¦ ’ F¡   §0 §¤ &31 A I   ¦¥# §¤  ¡%8Q§% ( 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...
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