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Unit 6: Matrix Decomposition

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Unit 6: Matrix Decomposition Unit 6: Matrix decomposition Juan Luis Melero and Eduardo Eyras October 2018 1 Contents 1 Matrix decomposition3 1.1 Definitions.............................3 2 Singular Value Decomposition6 3 Exercises 13 4 R practical 15 4.1 SVD................................ 15 2 1 Matrix decomposition 1.1 Definitions Matrix decomposition consists in transforming a matrix into a product of other matrices. Matrix diagonalization is a special case of decomposition and is also called diagonal (eigen) decomposition of a matrix. Definition: there is a diagonal decomposition of a square matrix A if we can write A = UDU −1 Where: • D is a diagonal matrix • The diagonal elements of D are the eigenvalues of A • Vector columns of U are the eigenvectors of A. For symmetric matrices there is a special decomposition: Definition: given a symmetric matrix A (i.e. AT = A), there is a unique decomposition of the form A = UDU T Where • U is an orthonormal matrix • Vector columns of U are the unit-norm orthogonal eigenvectors of A • D is a diagonal matrix formed by the eigenvalues of A This special decomposition is known as spectral decomposition. Definition: An orthonormal matrix is a square matrix whose columns and row vectors are orthogonal unit vectors (orthonormal vectors). Orthonormal matrices have the property that their transposed matrix is the inverse matrix. 3 Proposition: if a matrix Q is orthonormal, then QT = Q−1 Proof: consider the row vectors in a matrix Q: 0 1 u1 B . C T u j ::: ju Q = @ . A and Q = 1 n un 0 1 0 1 0 1 u1 hu1; u1i ::: hu1; uni 1 ::: 0 T B . C u j ::: ju B . .. C B. .. .C QQ = @ . A 1 n = @ . A = @. .A un hun; u1i ::: hun; uni 0 ::: 1 Remember that: Orthogonal vectors hui; uji = 0; 8i 6= j Unit vectors hui; uii = 1; 8i In addition, the determinant of Q is ±1. Theorem: if Q is an orthonormal matrix ! det(Q) = ±1. Proof: p 2 T T det(Q) = det(Q)det(Q ) = det(QQ ) = det(In) = 1 ! det(Q) = 1 = ±1 Proposition: The scalar product of two vectors is invariant under the trans- formation by a an orthonormal matrix: hQu; Qvi = hu; vi Proof: hQu; Qvi = (Qu)T Qv = uT QT Qv = uT v = hu; vi Corollary: the column (or row) vectors from an orthonormal matrix form an (orthonormal) basis of Rn 2 1 Example: Perform the spectral decomposition of the matrix A = . 1 2 4 We calculate the eigenvalues and the eigenvectors: 2 − λ 1 det(A − λI ) = det = λ2 − 4λ + 3 2 1 2 − λ 2 det(A − λI2) = 0 ! λ − 4λ + 3 = (λ − 1)(λ − 3) = 0 ! λ = 1; 3 2 1 x x 2x + y = 3x x = 3 ! ! x = y ! Eigenvectors of 3 1 2 y y x + 2y = 3y x 2 1 x x 2x + y = x x = 1 ! ! −x = y ! Eigenvectors of 1 1 2 y y x + 2y = y −x We have the following eigenspaces: a E(3) = Ker(A − 3I ) = ; a 2 ⊂ 2 2 a R R b E(1) = Ker(A − I ) = ; b 2 ⊂ 2 2 −b R R We have to choose two orthogonal unit vectors. The eigenvectors are already orthogonal to each other: b a a = ab − ba = 0 −b We normalize both vectors to 1: a 1 a a = 1 ! a = p a 2 b 1 b −b = 1 ! b = p −b 2 p p 1= 2 1= 2 So the basis B = p ; p is an orthonormal basis of 2 1= 2 −1= 2 R With these orthonormal vectors we build the orthonormal matrix that diag- onalize A: p p p p 1= 2 1= 2 1= 2 −1= 2 U = p p U −1 = U T = p p −1= 2 1= 2 1= 2 1= 2 5 The matrix U provides a unique diagonal spectral decomposition: p p p p 1= 2 1= 2 1 0 1= 2 −1= 2 A = UDU T = p p p p −1= 2 1= 2 0 3 1= 2 1= 2 Notice that the order of the eigenvectors in the orthonormal matrix deter- mines the order of the eigenvalues in the diagonal matrix. If the matrix A is non-singular (det(A) 6= 0) A 2 Mn×n(R), we can define the inverse of A using the spectral decomposition A = UDU T as: A−1 = (UDU T )−1 = (U T )−1D−1U −1 = UD−1U T where we have used that U is an orthonormal matrix (built with the eigen- values of A) for which the inverse is the transpose. Here D is the diagonal matrix with th eigenvalues of A, hence D−1 is a diagonal matrix containing the inverse of the eigenvalues. We confirm that this is the inverse of A: −1 T −1 T −1 T T AA = UDU UD U = UDD U = UU = In Example: consider the same matrix as before p p p p 1= 2 1= 2 1 0 1= 2 −1= 2 A = UDU T = p p p p −1= 2 1= 2 0 3 1= 2 1= 2 p p p p 1= 2 1= 2 1 0 1= 2 −1= 2 A−1 = p p p p −1= 2 1= 2 0 1=3 1= 2 1= 2 −1 You can easily check that AA = I2. 2 Singular Value Decomposition Singular Value Decomposition (SVD) is a form of dimensional reduction. SVD allows to identify the dimensions that are most relevant. Once these dimensions are identified, we can find the best approximation to the original data using fewer dimensions. SVD generalizes the symmetrical orthonormal decomposition to non-symmetric matrices and to rectangular (non-square) matrices. 6 A is the input matrix of dimensions m × n, U is an orthonormal matrix of dimensions m × m, Σ contains a diagonal submatrix of size min(m; n) and the rest of the elements are 0, V is an orthonormal matrix n × n. Definition of SVD: Given A a rectangular matrix A 2 Mm×n(R), we can find a decomposition: A = UΣV T Where: • U and V are orthonormal matrices T • The columns of U are orthonormal eigenvectors of AA 2 Mm×m(R). It is the orthonormal matrix that diagonalizes AAT . These are called the left singular vectors of A T • The columns of V are orthonormal eigenvectors of A A 2 Mn×n(R). It is the orthonormal matrix that diagonalizes AT A. These are called the right singular vectors of A • Σ is a rectangular block-diagonal matrix containing the square-roots of the eigenvalues of AT A (which are the same as AAT ). These are called the singular values Let's try to understand where does A = UΣV T come from and how to do the decomposition. The idea is to build symmetric matrices because they have the property that the eigenvectors matrix is orthonormal. To build symmetric matrices we multiply the matrix by its transposed: T T AA 2 Mm×m(R) A A 2 Mn×n(R) 7 These matrices are square and symmetric: (AAT )T = (AT )T AT = AAT (AT A)T = AT (AT )T = AT A AT A is a square symmetric matrix, so it has a set orthonormal eigenvectors. V is the matrix built from these eigenvectors that diagonalizes AT A: AT A = (UΣV T )T UΣV T = V ΣT U T UΣV T = V ΣT ΣV T = V Σ2V T So Σ contains the square root of the eigenvalues of AT A: (AT A)V = V Σ2 Similar for U: AAT is a square symmetric matrix, so it has also a set of orthogonal eigen- vectors. AAT = UΣV T (UΣV T )T = UΣV T V ΣT U T = UΣ2U T (AAT )U = UΣ2 U is the matrix that diagonalizes AAT and Σ contains the square root of the eigenvalues of AAT . AAT and AT A have non-negative eigenvalues (Σ matrix contains squared numbers) and they coincide: Proposition: AT A and AAT share min(m; n) eigenvalues. 8 Proof: consider v an eigenvector of AT A: AT A = V Σ2V T (AT A)v = λv ! A(AT A)v = λAv ! (AAT )Av = λAv So, if v is an eigenvalue of AT A (in V ) with eigenvalue λ, then Av is an eigenvector of AAT with eigenvalue λ. v is a column vector of V , hence hv; vi = 1. However, for Av: hAv; Avi = (Av)T Av = vT (AT A)v = vT λv = λ hv; vi = λ The eigenvectors of AAT in U corresponding to the same eigenvalues can thus be built as: 1 u = p Av λ This definition ensures that hu; ui = hAv; Avi = 1. In addition, it establishes a relationship between the eigenvectors. As the spectral decomposition, we can use SVD to define an inverse matrix. Given a non-singular non-symmetrical matrix A (det(A) 6= 0) A 2 Mn×n(R), we can write: A = UDV T where U and V are the matrix built as before, where: • U is the orthonormal matrix that diagonalizes AAT • V is the orthonormal matrix that diagonalizes AT A • D−1 is a diagonal matrix containing the reciprocal of the eigenvalues 1 ( λ ) of A 9 We can define the inverse of a matrix with: A−1 = VD−1U T −1 T −1 T −1 T T AA = UDV VD U = UDD U = UU = In 4 4 Example of SVD: calculate SVD for the matrix (A = UΣV T −3 3 We calculate V , the orthonormal matrix that diagonalizes AT A 4 −3 4 4 25 7 AT A = = 4 3 −3 3 7 25 We can see numerically here that AT A is symmetric.
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