# V v f f v v mv v v f t singular value decomposition

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v v 0 F F ( ( v v 0 ) ) Mv v v F t = ) (

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Singular Value Decomposition Proof (Step 1) : Let F ( v ) be the function on the unit sphere (|| v ||=1) defined by: Then F must have a maximum at some point v 0 . Then F ( v 0 )= λ v 0 . Mv v v F t = ) ( v v 0 F F ( ( v v 0 ) )
Singular Value Decomposition If F has a maximum at some point v 0 then F ( v 0 )= λ v 0 . If w 0 is on the sphere, next to v 0 , then w 0 - v 0 is nearly perpendicular to v 0 . And for any small vector w 1 perpendicular to v 0 , v 0 + w 1 is nearly on the sphere v v 0 w w 0 v v 0 w w 0 -v -v 0 w w 0

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Singular Value Decomposition If F has a maximum at some point v 0 then F ( v 0 )= λ v 0 . For small values of w 0 close to v 0 , we have: For v 0 to be a maximum, we must have: for all w 0 near v 0 . Thus, F ( v 0 ) must be perpendicular to all vectors that are perpendicular to v 0 , and hence must itself be a multiple of v 0 . 0 0 0 0 0 ), ( ) ( ) ( v w v F v F w F - + 0 ), ( 0 0 0 = - v w v F
Singular Value Decomposition Proof (Step 1) : Let F ( v ) be the function on the unit sphere (|| v ||=1) defined by: Then F must have a maximum at some point v 0 . Then F ( v 0 )= λ v 0 . Mv v v F t = ) ( v v 0 F F ( ( v v 0 ) )

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Singular Value Decomposition Proof (Step 1) : Let F ( v ) be the function on the unit sphere (|| v ||=1) defined by: Then F must have a maximum at some point v 0 . Then F ( v 0 )= λ v 0 . But F ( v )=2 Mv Mv v v F t = ) ( v v 0 F F ( ( v v 0 ) ) v 0 is an eigenvector of M .
Singular Value Decomposition Proof : 1. Every symmetric matrix has at least one eigenvector v . 2. If v is an eigenvector and w is perpendicular to v then Mw is also perpendicular to v.

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Singular Value Decomposition Proof (Step 2) : If w is perpendicular to v , then v,w =0. Since M is symmetric: so that Mw is also perpendicular to v . 0 , , , = = = w v w Mv Mw v λ
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• Spring '10
• unknown
• Variance, Singular value decomposition, Euclidean space, varp

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