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Lecture 13 Operations in Graphics and Geometric Modeling I: Projection, Rotation, and Reflection Shang-Hua Teng

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Projection Projection onto an axis (a,b) x axis is a vector subspace

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Projection onto an Arbitrary Line Passing through 0 (a,b)

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Projection on to a Plane

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Projection on to a Line b a p

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Projection Matrix: on to a Line b a p What matrix P has the property p = Pb

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Properties of Projection on to a Line b a p p is the points in Span(a) that is the closest to b

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Projection onto a Subspace Input: 1. Given a vector subspace V in R m 2.A vector b in R m … Desirable Output: –A vector in p in V that is closest to b –The projection p of b in V –A vector p in V such that (b-p) is orthogonal to V

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How to Describe a Vector Subspace V in R m If dim(V) = n, then V has n basis vectors –a 1, a 2, …, a n –They are independent V = C(A) where A = [a 1, a 2, …, a n ]

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Projection onto a Subspace Input: 1. Given n independent vectors a 1, a 2, …, a n in R m 2.A vector b in R m … Desirable Output: –A vector in p in C([a 1, a 2, …, a n ]) that is closest to b –The projection p of b in C([a 1, a 2, …, a n ]) –A vector p in V such that (b-p) is orthogonal to C([a 1, a 2, …, a n ])

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Using Orthogonal Condition

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Think about this Picture C(A T ) N(A) RnRn RmRm C(A) N(A T ) xnxn xrxr b dim r dim n- r dim m- r p b-p

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Connection to Least Square Approximation

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Rotation

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Properties of The Rotation Matrix

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Q is an orthonormal matrix: Q T Q = I

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Rotation Matrix in High Dimensions Q is an orthonormal matrix: Q T Q = I

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Rotation Matrix in High Dimensions Q is an orthonormal matrix: Q T Q = I

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Reflection u b mirror

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Reflection u b

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u b mirror

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Reflection is Symmetric and Orthonormal u b mirror

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Orthonormal Vectors are orthonormal if

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Orthonormal Matrices Q is orthonormal if Q T Q = I The columns of Q are orthonormal vectors Theorem: For any vectors x and y,

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Products of Orthonormal Matrices Theorem: If Q and P are both orthonormal matrices then QP is also an orthonormal matrix. Proof:

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Orthonormal Basis and Gram-Schmidt Input: an m by n matrix A Desirable output: Q such that –C(A) = C(Q), and –Q is orthonormal

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Basic Idea Suppose A = [a 1 … a n ] If n = 1, then Q = [a 1 /|| a 1 ||] If n = 2, –q 1 = a 1 /|| a 1 || –Start with a 2 and subtract its projection along a 1 –Normalize

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Gram-Schmidt Suppose A = [a 1 … a n ] –q 1 = a 1 /|| a 1 || –For i = 2 to n What is the complexity? O( mn 2 )

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Theorem: QR-Decomposition Suppose A = [a 1 … a n ] –There exist an upper triangular matrix R such that –A = QR

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Using QR to Find Least Square Approximation Can be solved by back substitution

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