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It V T V = L In mm LV = I with L = V T LEFT INVERSE COMPSCI 527 - PowerPoint PPT Presentation

Orthogonal Matrices The Left Inverse of an Orthogonal Matrix M E m q i = v T i p (Finding coefficients q i is easy!) O_O Rewrite v T i v j = ij in matrix form V = [ v 1 , . . . , v n ] 2 R m n It V T V = L In mm LV = I with L =


  1. Orthogonal Matrices The Left Inverse of an Orthogonal Matrix M E m q i = v T i p (Finding coefficients q i is easy!) O_O • Rewrite v T i v j = δ ij in matrix form V = [ v 1 , . . . , v n ] 2 R m × n It V T V = L In mm • LV = I with L = V T LEFT INVERSE COMPSCI 527 — Computer Vision The Singular Value Decomposition 5 / 21

  2. Orthogonal Matrices II Left and Right Inverse of an Orthogonal Matrix nem t.nl rank R E n mxm • LV = I with L = V T • Can we have R such that VR = I ? rank L E n • That would be the right inverse I m m • What if m = n ? vis m m LV If all square VR Imo El left inv left A anyinvertible matrix sight inv R LAR L R I LA L If invertible LALR I AR R L COMPSCI 527 — Computer Vision The Singular Value Decomposition 6 / 21

  3. Orthogonal Matrices Orthogonal Transformations Preserve Norm ( m � n ) t d s y = V x : R n ! R m m ETV'VE E HEIR jTy k y k 2 = HEH 1154 IIII.AT Yuere rotation COMPSCI 527 — Computer Vision The Singular Value Decomposition 7 / 21

  4. Orthogonal Projection If m Projection Onto a Subspace ( m  n ) • Projection of b 2 R m onto subspace S ✓ R m is the point p 2 S closest to b • Let V 2 R m × n an orthonormal basis for S (That is, V is an orthogonal matrix ) • b � p ? v i for i = 1 , . . . , n that is, V T ( b � p ) = 0 • Projection of b 2 R m onto S is p = VV T b Es i (optional proofs in an Appendix) t VE E WT F COMPSCI 527 — Computer Vision The Singular Value Decomposition 8 / 21

  5. The Singular Value Decomposition rowspeaCA Linear Mappings µ EIR E for men √ √   3 3 0 1 b = A x : R n → R m . Example: A = − 3 3 0 ( m = n = 3 ) √   2 will A 1 1 0 f d'm range ( A ) ↔ rowspace ( A ) ELLIPSOID EffortzelR f b3 HYPER TIFIELSPHERE I 5 x raranklA in e 2 Fx Rfp cE b be COMPSCI 527 — Computer Vision The Singular Value Decomposition 9 / 21

  6. The Singular Value Decomposition The Singular Value Decomposition: Geometry Ari TE √ √   3 3 0 1 11511 1 11411 1 b = A x where A = √ − 3 3 0   2 1 1 0 420 Tz VI quiz Ariz pm FIL ate t z Tz Hui 11 1 th Iwaki 03 0 ITE 3 0 O ctfu F REIKO I Tfz NON TRIVIAL 0 Fz spans null A Hies L HFzH 1 a 3 spans left hull A 5 0 T.tv cities COMPSCI 527 — Computer Vision The Singular Value Decomposition 10 / 21

  7. The Singular Value Decomposition SMALL SVD The Singular Value Decomposition: Algebra hi.info 3 Atria A v 1 = σ 1 u 1 A v 2 = σ 2 u 2 z Ati.int ≥ σ 2 > σ 3 = 0 σ 1 u T 1 u 1 = 1 u T 2 u 2 = 1 FARGE SVD 1 I u T u 3 u 3 = 1 u T I 1 u 2 = 0 u T 1 u 3 = 0 iE u T 2 u 3 = 0 v T 1 v 1 = 1 v T 2 v 2 = 1 v T 3 v 3 = 1 v T 1 v 2 = 0 v T E UTAV 1 v 3 = 0 v T 2 v 3 = 0 COMPSCI 527 — Computer Vision The Singular Value Decomposition 11 / 21

  8. The Singular Value Decomposition The Singular Value Decomposition: General For any real m ⇥ n matrix A there exist orthogonal matrices 2 R m × m ⇥ ⇤ U = u 1 · · · u m 2 R n × n ⇥ ⇤ V = v 1 v n · · · SAFE safe such that U T AV = Σ = diag ( σ 1 , . . . , σ p ) 2 R m × n where p = min( m , n ) and σ 1 � . . . � σ p � 0. Equivalently, A = U Σ V T . COMPSCI 527 — Computer Vision The Singular Value Decomposition 12 / 21

  9. The Singular Value Decomposition Rank and the Four Subspaces rank can yr    v 1  σ 1 . ...    .  stows I .     pep 0     v r     A = U Σ V T = [ u 1 , . . . , u r , u r + 1 , . . . , u m ] σ r     0 v r + 1     4     p . ...     . sperm     .     0 v n spar span range lefty COMPSCI 527 — Computer Vision The Singular Value Decomposition 13 / 21

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