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Eigen Values and Vectors Matrices and Eigen Vectors EIGEN VALUES AND VECTORS MATRICES AND EIGEN VECTORS 2 3 1 11 × = [2 1] [3] [ 5 ] 2 3 3 12 3 × = = 4 × [2 1] [2] [ 8 ] [2] • Scale 3 6 2 × = [2] [4] 2 3 6 24 6 × = = 4 × [2 1] [4] [16] [4] 2 EIGEN VECTOR - PROPERTIES • Eigen vectors can only be found for square matrices • Not every square matrix has eigen vectors. • Given an n x n matrix that does have eigenvectors, there are n of them for example, given a 3 x 3 matrix, there are 3 eigenvectors. • Even if we scale the vector by some amount, we still get the same multiple 3 EIGEN VECTOR - PROPERTIES • Even if we scale the vector by some amount, we still get the same multiple • Because all you’re doing is making it longer, not changing its direction. • All the eigenvectors of a matrix are perpendicular or orthogonal. • This means you can express the data in terms of these perpendicular eigenvectors. • Also, when we find eigenvectors we usually normalize them to length one. 4 EIGEN VALUES - PROPERTIES • Eigenvalues are closely related to eigenvectors. • These scale the eigenvectors • eigenvalues and eigenvectors always come in pairs. 2 3 6 24 6 × = = 4 × [2 1] [4] [16] [4] 5 SPECTRAL THEOREM Theorem: If A ∈ ℝm×n is symmetric matrix (meaning AT = A), then, there exist real numbers (the eigenvalues) λ1, …, λn and orthogonal, non-zero real vectors ϕ1, ϕ2, …, ϕn (the eigenvectors) such that for each i = 1,2,…, n : Aϕi = λiϕi 6 EXAMPLE 30 28 A = [28 30] From spectral theorem: Aϕ = λϕ 7 EXAMPLE 30 28 A = [28 30] From spectral theorem: Aϕ = λϕ ⟹ Aϕ − λIϕ = 0 (A − λI)ϕ = 0 30 − λ 28 = 0 ⟹ λ = 58 and λ = 2 [ 28 30 − λ] 8 EXAMPLE 30 28 A = [28 30] From spectral theorem: Aϕ = λϕ 1 30 28 ϕ11 ϕ11 2 = 58 ⟹ ϕ1 = 1 [28 30] [ϕ12] [ϕ12] 2 9 EXAMPLE 30 28 A = [28 30] From spectral theorem: Aϕ = λϕ 1 30 28 ϕ11 ϕ11 2 = 58 ⟹ ϕ1 = 1 [28 30] [ϕ12] [ϕ12] 2 1 30 28 ϕ21 ϕ21 2 = 2 ⟹ ϕ2 = −1 [28 30] [ϕ22] [ϕ22] 2 10 EXAMPLE 30 28 A = [28 30] From spectral theorem: Aϕ = λϕ 1 1 2 2 λ = 58 λ = 2 ϕ1 = 1 1 ϕ2 = −1 2 2 2 1 1 2 2 ϕ = 1 −1 2 2 11 SINGULAR VALUE DECOMPOSITION • Every matrix, A, can be written as A = UΣVT where U and V are orthonormal matrices and Σ is a diagonal matrix. • This is known as Singular Value Decomposition (SVD) of matrix A. • The values in the diagonal of Σ are called the singular values of the matrix A 12 SINGULAR VALUE DECOMPOSITION • Thus, Ax = U(Σ(VT x)) • Applying A to a matrix x amounts to rotating using VT, scaling using Σ and rotating again using U • The rank of the matrix A is the number of non-zero singular values. • Given a matrix A = UΣVT of rank r, if we want the closest matrix A′ of rank r − k, then one can simply zero out the k smallest singular values (smallest in absolute value) in Σ to produce Σ′. A′ T is then UΣ′V . If is the -th column of , is the -th column of and is the • ui i U vi i V σi i -th diagonal element of , then T Σ A = ∑ σiuivi i 13 SINGULAR VALUE DECOMPOSITION Theorem T : Anm = UnnΣnmVmm A - Rectangular matrix, n × m Columns of U are orthonormal eigenvectors of AAT Columns of V are orthonormal eigenvectors of AT A Σ is a diagonal matrix containing the square roots of eigenvalues from U or V in descending order called singular values and T Avi = σiui A ui = σivi Where σ is the singular value value 14 SVD - EXAMPLE 1 1 A = 0 1 [1 0] T A3×2 = U3×3Σ3×2V2×2 Columns of U are orthonormal eigenvectors of AAT 6 0 − 1 3 3 6 2 U = − 1 6 2 3 6 2 1 6 2 3 15 SVD - EXAMPLE 1 1 A = 0 1 [1 0] T A3×2 = U3×3Σ3×2V2×2 Columns of V are orthonormal eigenvectors of AT A 2 2 VT = 2 2 2 2 2 − 2 16 SVD - EXAMPLE 1 1 A = 0 1 [1 0] T A3×2 = U3×3Σ3×2V2×2 Σ is a diagonal matrix containing the square roots of eigenvalues from U or V in descending order 3 0 Σ = 0 1 0 0 17 SVD - EXAMPLE 1 1 A = 0 1 [1 0] T A3×2 = U3×3Σ3×2V2×2 6 0 − 1 3 3 2 2 3 0 1 1 6 2 1 2 2 A = − = 0 1 6 2 3 0 1 2 2 [1 0] 0 0 2 − 2 6 2 1 6 2 3 18 SVD - EXAMPLE U, S, VT = numpy . linalg . svd(img) 19 SVD - EXAMPLE full rank 600 300 100 50 20 10 U[: , k]S[: k]VT[: k, :] 20.
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