Chapter 4 General Vector Spaces Face Recognition ( á¹¢å± å½å¬)

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ExampleLA 04_53 v 1 = (1 0 0), v 2 = (2 1 0), v 3 = (3 2 1) (1)(2)u1u2 v v12 (1, 0, 0) proj(2,1,0) (2,1,0) (1,0,0)u 1v2v2(((1,0,0)(1,0,0)vu21uu11)u)1(1,0,0) (2,1,0) (2,0,0) (0,1,0) ( u1)(3') u3 v3 proju 1v3 (0,2,1) ( u1)(Not u2)(3) u3 v3 proju 1v3 proju 2v3 (0,0,1)( u1, u2) 1 4 8 2 0 1A 0 5 5 3 8 6 From Example 4, orthonormal basis {q 1 , q 2 , q 3 }: 1 2 3 2 4 5 2 4 1 , , 0, , , , , , , , 0, 14 14 14 7 7 7 7 21 21QR-Decomposition (1/2)uu u 12212 LA 04_553 Then by Theorem 4.20: u 1 = (u 1 · q 1 ) q 1 + (u 1 · q 2 ) q 2 +(u 1 · q 3 ) q 3 = (u 1 · q 1 ) q 1 = 14q 1 u 2 = (u 2 · q 1 ) q 1 + (u 2 · q 2 ) q 2 + (u 2 · q 3 ) q 3 =2 14 q q172 u 3 = (u 3 · q 1 ) q 1 + (u 3 · q 2 ) q 2 + (u 3 · q 3 ) q 3 =2 14 q q1 7 q2213Example 4: {(1, 2, 0, 3), (4, 0, 5, 8), (8, 1, 5, 6)} linearlyindependent in R 4 a basis for a 3-dim subspace.Construct an orthonormal basis for V.• Solution: Let v 1 = (1, 2, 0, 3), v 2 = (4, 0, 5, 8), v 3 = (8, 1, 5, 6).LA 04_54uuu123vvv123 (1, 2, 0, 3)projproju 1u 1vv23(2,proj 4, 5,u 2v32)(4,1, 0, 2)• The set {(1, 2, 0, 3), (2, – 4, 5, 2), (4, 1, 0, – 2)} orthogonal:Check 1 of 2: (1, 2, 0, 3) · (2, – 4, 5, 2) = 2 – 8 + 0 + 6 = 0(1, 2, 0, 3) 12 22 02 32 14, (2, 4, 5, 2) 7, (4,1, 0, 2) 21• Orthonormal basis 114,214,0,314 , 27,47,57,27 , 421,121,0,221 QR-Decomposition (2/2)LA 04_5614 21 1 4 8 2 4 1 14 2 14 2 14 A 2 0 1 1 2 3 0 5 5 5 0 0 0 0 21 3 8 6 7 3 2 2 14 7 21 uu u 14 7 21 0 7 7 QR1274 Theorem: If A R mn has linearly independent columnvectors, then A = QR, where Q is an mn matrix withorthonormal column vectors and R is an nn invertibleupper triangular matrix. One of top 10 algorithms in the 20th century (computeeigenvalue in chapter 5 by QR algorithm, check courseweb site)ttt
LA 04_57Projection of a <strong>Vector</strong> onto a Subspace• Let W be a subspace of R n .• Let {u 1 , …, u m } be anorthonormal basis for W.• If v is a vector in R n , theprojection of v onto W isdenoted proj W v and isdefined byproj2Wv ( v u1) u1 ( v u2) u ( v um) umExample 5• Consider the vector v = (3, 2, 6) in R 3 .• W : Subspace of R 3 of the form (a, b, b).• Decompose v into the sum of a vector that lies in W and a vectororthogonal to W.SolutionLA 04_59• Need an orthonormal basis for W.• Write an arbitrary vector of W as follows(a, b, b) = a (1, 0,0) + b ( 0, 1, 1)• The set {(1, 0, 0), (0, 1, 1)} spans W and is linearly independent. a basis for W• Orthogonal Normalize () to get an orthonormal basisu 1 1, 0, 0), u(2 0,21,21 LA 04_58Definition: A vector v is orthogonal to a subspace Wif v is orthogonal to every vector u in W.-- Basis: v u = v (c 1 u 1 + … + c n u n )• Theorem 4.24: Let W be asubspace of R n .Every vector v in R n can bewritten uniquely in the formv = w + w where w is in W and w isorthogonal to W.The vectors w and w arew = proj W vw = v –proj W vLA 04_60,1(3,2w projWv ( v u1) u1 ( v u2) u2 ((3, 2, 6) (1, 0, 0))(1, 0, 0) (3, 0, 0) (0, 4, 4) (3, 4, 4) W2,6)0,12,12 0,12wW v proj v (3, 2, 6) (3, 4, 4) (0, 2, 2) Orthogonal to W:(0, 2, 2) · (1, 0, 0) = 0 , (0, 2, 2) · (0, 1, 1) = 0• Thus the decomposition of v is (3, 2, 6) = (3, 4, 4) + (0, 2, 2)• Another example: t = (3, 2, 2) W12,12projWt((3,(3,2,2,2)2)(1,0,0))(1,0,0)(3,2,2)0,12,12 0,
Page 2 and 3: Vector Spaces of Matrices (1/2)•
Page 4 and 5: Section 4.2 to 4.4• Consider R 2
Page 6 and 7: Example 7Show that the function h (
Page 8 and 9: Theorem 4.8Let {v 1 , …, v n } be
Page 10 and 11: Rank• Theorem 4.13: The row space
Page 12 and 13: 4.6 Orthonormal Vectors and Project
Page 16 and 17: Example 6Find the distance of the p
Page 18 and 19: Example 2 (1/2)x x2 x 3 y yy639 12
Page 20 and 21: 4.7 Kernel (), Range (), and theRan
Page 22 and 23: 4.9 Transformations and Systems of
Page 24: Demo Tacoma BridgeLA 04_93

Chapter 4 General Vector Spaces Face Recognition ( á¹¢å ± å½å¬)

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Chapter 4 General Vector Spaces Face Recognition ( á¹¢å± å½å¬)