FFTs in Graphics and Vision The Laplace Operator 1 Outline Math Stuff Symmetric/Hermitian Matrices Lagrange Multipliers Diagonalizing Symmetric Matrices The Laplacian Operator 2 Linear Operators Definition: Given a real inner product space (푉, ⋅,⋅ ) and a linear operator 퐿: 푉 → 푉, the adjoint of the 퐿 is the linear operator 퐿∗, with the property that: 푣, 퐿푤 = 퐿∗푣, 푤 ∀푣, 푤 ∈ 푉 3 Linear Operators Note: If 푉 is the space of 푛-dimensional, real-valued, arrays with the standard inner product: 푛 푣 ⋅ , 푤 ⋅ = 푣 푖 ⋅ 푤 푖 = 푣푡푤 푖=1 then the adjoint of a matrix 푀 is its transpose: 푣, 푀푤 = 푣푡푀푤 = 푀푡푣 푡푤 = 〈푀푡푣, 푤〉 4 Linear Operators Definition: A real linear operator 퐿 is self-adjoint if it is its own adjoint, i.e.: 푣, 퐿푤 = 퐿푣, 푤 ∀푣, 푤 ∈ 푉 5 Linear Operators Note: If 푉 is the space of 푛-dimensional, real-valued, arrays with the standard inner product: 푛 푣 ⋅ , 푤 ⋅ = 푣 푖 ⋅ 푤 푖 = 푣푡푤 푖=1 then a matrix 푀 is self-adjoint if it is symmetric: 푀 = 푀푡 6 Linear Operators Definition: Given a complex inner product space (푉, ⋅,⋅ ) and given a linear operator 퐿: 푉 → 푉, the adjoint of the 퐿 is the linear operator 퐿∗, with the property that: 푣, 퐿푤 = 퐿∗푣, 푤 ∀푣, 푤 ∈ 푉 7 Linear Operators Note: If 푉 is the space of 푛-dimensional, complex- valued, arrays with the standard inner product: 푛 푣 ⋅ , 푤 ⋅ = 푣 푖 ⋅ 푤 푖 = 푣푡푤 푖=1 then the adjoint of a matrix 푀 is just the complex conjugate of the transpose: 푣, 푀푤 = 푣푡푀푤 = 푀 푡푣 푡푤 = 푀 푡푣, 푤 8 Linear Operators Definition: A complex linear operator 퐿 is self-adjoint if it is its own adjoint, i.e.: 푣, 퐿푤 = 퐿푣, 푤 ∀푣, 푤 ∈ 푉 9 Linear Operators Note: If 푉 is the space of 푛-dimensional, complex- valued, arrays with the standard inner product: 푛 푣 ⋅ , 푤 ⋅ = 푣 푖 ⋅ 푤 푖 = 푣푡푤 푖=1 then a matrix 푀 is self-adjoint if it is Hermitian: 푀 = 푀 푡 10 Outline Math Symmetric/Hermitian Matrices Lagrange Multipliers Diagonalizing Symmetric Matrices The Laplacian Operator 11 Implicit Surface Definition: Given a function 푔(푥, 푦, 푧), the implicit surface or iso-surface defined by 푔(푥, 푦, 푧) is the set of points in 3D satisfying the condition: 푔 푥, 푦, 푧 = 0 2 푔 푥, 푦, 푧 = 푥2 + 푦2 + 푧2 − 1 푔 푥, 푦, 푧 = 푥2 + 푦2 + 푧2 − 푅2 + 푟2 − 4푅2 푟2 − 푧2 Lagrange Multipliers Goal: Given an implicit surface defined by a function 푔(푥, 푦, 푧) and given a function 푓(푥, 푦, 푧), we want to find the extrema of 푓 on the surface. This can be done by finding the points on the surface where the 0 gradient of 푓 is parallel to the surface normal. 13 http://www.answers.com Lagrange Multipliers Since the implicit surface is the set of points with: 푔 푥, 푦, 푧 = 0 the normal at a point on the surface is parallel to the gradient of 푔. Finding the extrema amounts to finding the points (푥, 푦, 푧) such that: 푔 푥, 푦, 푧 = 0 (the point is on the surface) 휆∇푓 = ∇푔 (the point is a local extrema) 14 Outline Math Symmetric/Hermitian Matrices Lagrange Multipliers Diagonalizing Symmetric Matrices The Laplacian Operator 15 Diagonalizing Symmetric Matrices Claim: Given the space of 푛-dimensional, real-valued, arrays and given a symmetric matrix 푀: The eigenvectors of 푀 form an orthogonal basis 16 Diagonalizing Symmetric Matrices The Eigenvectors Form an Orthogonal Basis: To show this we will show two things: 1. If 푣 is an eigenvector, then the space of vectors orthogonal to 푣 is fixed by 푀. 2. At least one eigenvector exists. 17 Diagonalizing Symmetric Matrices 1. If 푣 is an eigenvector, then the space of vectors orthogonal to 푣 is fixed by 푀. Suppose that 푣 is an eigenvector and 푤 is some other vector that is orthogonal to 푣: 푣, 푤 = 0 Since 푣 is an eigenvector, this implies that: 푀푣, 푤 = 휆푣, 푤 = 0 Since 푀 is symmetric, we have: 푣, 푀푤 = 푀푣, 푤 = 0 18 Diagonalizing Symmetric Matrices 1. If 푣 is an eigenvector, then the space of vectors orthogonal to 푣 is fixed by 푀. If 푊 is the subspace of vectors orthogonal to 푣: 푊 = 푤 ∈ 푉 푤, 푣 = 0 then we have: 푣, 푀푤 = 0 ∀푤 ∈ 푊 ⇕ 푀 푤 ∈ 푊 ∀푤 ∈ 푊 19 Diagonalizing Symmetric Matrices 1. If 푣 is an eigenvector, then the space of vectors perpendicular to 푣 is fixed by 푀. Implications: If we know that we can find one eigenvector 푣, we can consider the restriction of 푀 to 푊. We know that: 푀 푊 ⊂ 푊 푀푢, 푤 = 푢, 푀푤 ∀푢, 푤 ∈ 푊 So we can repeat to find the next eigenvector. 20 Diagonalizing Symmetric Matrices 2. At least one eigenvector must exist We will show this using Lagrange multipliers: The implicit surface will be the sphere in ℝ푛: 2 2 푔 푥1, ⋯ , 푥푛 = 푥1 + ⋯ 푥푛 − 1 ⇕ 푔 푣 = 푣 2 − 1 The function we optimize will be: 푓 푣 = 〈푣, 푀푣〉 Because the sphere is compact, an extrema must exist. 푔 푥, 푦 = 푥2 + 푦2 − 1 Diagonalizing Symmetric Matrices 2. At least one eigenvector must exist The normal of a point on the sphere is parallel to the gradient of 푔: ∇푔 푥1, ⋯ , 푥푛 = 2 푥1, ⋯ , 푥푛 ⇕ ∇푔 푣 = 2푣 푔 푥, 푦 = 푥2 + 푦2 − 1 Diagonalizing Symmetric Matrices 2. At least one eigenvector must exist Claim: The gradient of 푓 is: ∇푓 푣 = 2푀푣 23 Diagonalizing Symmetric Matrices 2. At least one eigenvector must exist Proof: Let 푒푖 be the vector with zeros everywhere but in the 푖-th entry: 푒푖 = (0, ⋯ , 0,1,0, ⋯ , 0) 푖-th entry 24 Diagonalizing Symmetric Matrices 2. At least one eigenvector must exist Proof: Let 푒푖 be the vector with zeros everywhere but in the 푖-th entry. Then the 푖-th coefficient of the gradient is: 휕 푑 푓 푣 = 푓 푣 + 푠푒푖 휕푥푖 푑푠 푠=0 푑 푡 = 푣 + 푠푒푖 푀 푣 + 푠푒푖 푑푠 푠=0 푑 푡 푡 푡 2 푡 = 푣 푀푣 + 푠푒푖 푀푣 + 푠푣 푀푒푖 + 푠 푒푖 푀푒푖 푑푠 푠=0 25 Diagonalizing Symmetric Matrices 2. At least one eigenvector must exist Proof: Let 푒푖 be the vector with zeros everywhere but in the 푖-th entry. Then the 푖-th coefficient of the gradient is: 휕 푡 푡 푓 푣 = 푒푖 푀푣 + 푣 푀푒푖 휕푥푖 푡 = 푒푖, 푀푣 + 푀 푣, 푒푖 = 2〈푒푖, 푀푣〉 = 2 푀푣 푖 26 Diagonalizing Symmetric Matrices 2. At least one eigenvector must exist Proof: Since, the 푖-th coefficient of the gradient of 푓 at 푣 is twice the 푖-th coefficient of 푀푣, we have: ∇푓 푣 = 2푀푣 27 Diagonalizing Symmetric Matrices 2. At least one eigenvector must exist We know that the normal of the point 푣 on the unit sphere is parallel to the gradient, which is: ∇푔 푣 = 2푣 And we know that the gradient of 푓 is: ∇푓 푣 = 2푀푣 28 Diagonalizing Symmetric Matrices 2. At least one eigenvector must exist Since the function 푔 must have a maximum on the sphere, we know that there exists 푣 s.t.: 휆∇푔 푣 = ∇푓(푣) ⇕ 휆푣 = 푀푣 So at the maximum, we have our eigenvalue. 29 Outline Math Symmetric/Hermitian Matrices Lagrange Multipliers Diagonalizing Symmetric Matrices The Laplacian Operator 30 The Laplacian Operator Recall: The Laplacian of a function 푓 at a point measures how similar the value of 푓 at the point is to the average values of its neighbors. 푓(휃) Δ푓 휃 31 The Laplacian Operator Recall: Formally, for a function in 2D, the Laplacian is the sum of unmixed partial second derivatives: 휕2푓 휕2푓 Δ푓 푥, 푦 = + 휕푥2 휕푦2 32 The Laplacian Operator Observation 1: The Laplacian is a self-adjoint operator. To show this, we need to show that for any two functions 푓 and 푔, we have: 푓, Δ푔 = 〈Δ푓, 푔〉 33 The Laplacian Operator Observation 1: First, we show this in the 1D case, for functions 푓(휃) and 푔(휃): 푓, 푔′′ = 〈푓′′, 푔〉 Writing the dot-product as an integral gives: 2휋 푓, 푔′′ = 푓 휃 ⋅ 푔′′ 휃 푑휃 0 34 The Laplacian Operator Observation 1: Using the product rule for derivatives: 푓 ⋅ 푔 ′ = 푓′ ⋅ 푔 + 푓 ⋅ 푔′ ⇓ 2휋 2휋 2휋 ′ ′ ′ 푓 ⋅ 푔 휃 푑휃 = 푓 휃 ⋅ 푔 휃 푑휃 + 푓 휃 ⋅ 푔 휃 푑휃 0 0 0 Since 푓 and 푔 are functions on a circle, their values at 0 and 2휋 are the same: 2휋 푓 ⋅ 푔 ′ 휃 푑휃 = 푓 ⋅ 푔 2휋 − 푓 ⋅ 푔 0 = 0 0 35 The Laplacian Operator Observation 1: Thus, we have: 2휋 2휋 푓 휃 ⋅ 푔′ 휃 푑휃 = − 푓′ 휃 ⋅ 푔 휃 푑휃 0 0 “Moving” the derivative twice gives: 2휋 2휋 푓′′ 휃 ⋅ 푔 휃 푑휃 = − 푓′ 휃 ⋅ 푔′ 휃 푑휃 0 0 2휋 = −1 2 푓 휃 ⋅ 푔′′ 휃 푑휃 0 ⇕ 푓′′, 푔 = 〈푓, 푔′′〉 36 The Laplacian Operator Observation 1: To generalize this to higher dimensions, we write out the dot-product as: 2휋 2휋 휕2푓 2휋 2휋 휕2푓 Δ푓, 푔 = 2 ⋅ 푔 푑휃 푑휙 + 2 ⋅ 푔 푑휙 푑휃 0 0 휕휃 0 0 휕휙 2휋 2휋 휕2푔 2휋 2휋 휕2푔 = 푓 ⋅ 2 푑휃 푑휙 + 푓 ⋅ 2 푑휙 푑휃 0 0 휕휃 0 0 휕휙 = 푓, Δ푔 37 The Laplacian Operator Observation 2: The Laplacian operator commutes with rotation – i.e. computing the Laplacian and rotating gives the same function as first rotating and then computing the Laplacian: Δ 휌푅 푓 = 휌푅 Δ 푓 38 The Laplacian Operator Implications: Observation 1: Since the Laplacian operator is self-adjoint, it must be diagonalizable.