DEN: Linear algebra numerical view (GEM: Gauss elimination method for reducing a full rank matrix to upper-triangular
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1 form) Given: matrix C = (c i,j ) n,m i,j=1 ODE and num math: Linear algebra (N) [lectures] c phabala 2016 DEN: Linear algebra numerical view (GEM: Gauss elimination method for reducing a full rank matrix to upper-triangular of real numbers, where m n 0 Set k = 1 1 If c i,k = 0 for all i = k,, n, then rank(a) < n The algorithm fails and stops Otherwise, do the pivoting : Among the rows i = k,, n with non-zero c i,k, choose the row k that has the smallest possible value of max j>k c i,j c i,k If k k, exchange rows k and k The (new) number c k,k is called a pivot Continue with step 2 2 For i = k + 1,, n do the following: Let l i,k = c i,k c k,k, set c i,k = 0 and for j > k compute c i,j = c i,j l i,k c k,j If k < n, increase k by one and go back to step 1 Otherwise the algorithm stops The output is the matrix (c i,j ) n,m i,j=1 Note: The output has the form c 1,1 c 1,2 c 1,n 1 c 1,n c 1,m 0 c 2,2 c 2,n 1 c 2,n c 2,m 0 0 c n 1,n 1 c n 1,n c n 1,m c n,n c n,m The computational complexity of GEM when reducing an n n matrix or an n (n + 1) matrix is 2 3 n3 + O(n 2 ) 1
2 (GJM: Gauss-Jordan elimination method for reducing a full-rank matrix to an extended diagonal matrix) Given: matrix C = (c i,j ) n,m i,j=1 of real numbers, where m n 0 Set k = 1 1 If c i,k = 0 for all i = k,, n, then rank(a) < n The algorithm fails and stops Otherwise, do the pivoting : Among the rows i = k,, n with non-zero c i,k, choose the row k that has the smallest possible value of max j>k c i,j c i,k If k k, exchange rows k and k The (new) number c k,k is called a pivot Continue with step 2 2 Set c k,k = 1, for j > k let c k,j = c k,j c k,k For all i k set c i,k = 0 and for j > k compute c i,j = c i,j c i,k c k,j If k < n, increase k by one and go back to step 1 Otherwise the algorithm stops The output is the matrix (c i,j ) n,m i,j=1 Note: The output has the form c 1,n+1 c 1,m c 2,n+1 c 2,m c n 1,n+1 c n 1,m c n,n+1 c n,m The computational complexity of GJM when reducing an n n matrix or an n (n + 1) matrix is n 3 + O(n 2 ) Any system A x = b whose regular matrix A is upper (resp lower) triangular can be solved by back (resp forward) substitution with computational complexity n 2 2
3 (Gauss elimination for reducing matrix to row-echelon form) Given: matrix C = (c i,j ) n,m i,j=1 of real numbers, where m n 0 Set k = 1, l = 1 1 If c i,k = 0 for all i = k,, n, then go to step 3 Otherwise: Among the rows i = k,, n with non-zero c i,l, choose the row k that has the smallest possible value of max j>l c i,j c i,l If k k, exchange rows k and k Continue with step 2 2 For i = k + 1,, n do the following: Let l i,k = c i,l c k,l, set c i,l = 0 and for j > l compute c i,j = c i,j l i,k c k,j Increase k by one and go to step 3 3 If l < n, increase l by one and go back to step 1 Otherwise the algorithm stops The output is the matrix (c i,j ) n,m i,j=1 Consider a real n n matrix A We say that n n matrices L, U are an LU decomposition or LU factorization of A if U is an upper-triangular matrix, L is a lower-triangular matrix whose diagonal entries are all 1, and A = LU Theorem Let A be a real n n matrix If its rank is k and its first k leading principal minors are non-zero, then it has an LU decomposition 3
4 If GEM without pivoting applied to A yields a triangle matrix U, then A has an LU decomposition and U is the right matrix for it (LU decomposition of a matrix) Given: an n n matrix A = (a i,j ) n i,j=1 of real numbers 0 Let U = A, set k = 1, l = 1 1 If u k,l 0 then continue with step 2 Otherwise: If there is i > k such that u i,l 0, then the given matrix does not have any LU decomposition The algorithm fails and stops If u k,l = 0 for all i k, then go to step 3 2 For i = k + 1,, n do the following: Let l i,k = u i,l u k,l, set u i,l = 0 and for j > l compute u i,j = u i,j l i,k u k,j Increase k by one and go to step 3 3 If l < n, increase l by one and go back to step 1 Otherwise the algorithm stops The output is matrices L = U = l 2, l n 1,1 l n 1,2 1 0 l n,1 l n,2 l n,n 1 1 u 1,1 u 1,2 u 1,n 1 u 1,n 0 u 2,2 u 2,n 1 u 2,n, 0 0 u n 1,n 1 u n 1,n u n,n 4
5 For every square matrix A there exists a permutation matrix P, a lower-triangular matrix L and an upper-triangular matrix U such that P A = LU (LUP decomposition of a matrix) Given: an n n matrix A = (a i,j ) n i,j=1 of real numbers 0 Let U = A and L = P = E n (unit matrix) Set k = 1, l = 1 1 If u k,l = 0 for all i k, then go to step 3 Otherwise: Among the rows i = k,, n with non-zero u i,l, choose the row k that has the smallest possible value of max j>l u i,j u i,l If k > k, then exchange rows k and k in matrices U and P ; in the matrix L exchange the first k 1 entries of rows k and k Continue with step 2 2 For i = k + 1,, n do the following: Let l i,k = u i,l u k,l, set u i,l = 0 and for j > l compute u i,j = u i,j l i,k u k,j Increase k by one and continue with step 3 3 If l < n, increase l by one and go back to step 1 Otherwise the algorithm stops The output is the matrices P, L, U Iterative improvement of solution: 1 Find a solution x of the system A x = b, 2 Determine the residuum r = b A x Solve the system A E x = r If the error E x is not sufficiently small, do the correction x := x + E x 5
6 Traditional norms on IR n : x = n x k 2 (Euclidean norm), k=1 x = max x k (max norm), k=1,,n x 1 = x k (sum norm) k=1 Let V be a vector space A mapping : V IR is called a norm if it has the following properties: x 0 for all x IR n ; x = 0 if and only if x = 0; c x = c x for all x IR n and c IR; x + y x + y for all x, y IR n (triangle inequality) Let V be a vector space of matrices A mapping : V IR is called a matrix norm if it is a norm and also satisfies AB A B for all A, B M n n We define the spectral radius of A as ϱ(a) = max( λ j ), where the maximum runs through all eigenvalues λ j of A (including possible complex j ones) Traditional matrix norms: A = A R = A 1 = A C = max i=1,,n j=1 max j=1,,n A F = a i,j (row-sum norm), a i,j (column-sum norm), i=1 i=1 j=1 a i,j 2 (Frobenius norm) 6
7 Consider a norm x for vectors from IR n and a norm A M for matrices from M n n We say that these norms are compatible if A x A M x for all A M n n and x IR n Theorem (i) Let be a vector norm The number defined for A M n n by the formula { A x } A M = sup x ; x IRn \ { 0} = sup{ A x ; x IR n x 1} determines a matrix norm compatible with We call it the matrix norm induced by (ii) Let M be a matrix norm The number defined for x IR n by the formula x x x = x n 0 0 is a norm on IR n compatible with M It is called the norm induced by M M For an n n matrix A we define its condition number as cond(a) = A A 1 Theorem Assume that vectors x 0, x a b 0, b are related by the formulas A x 0 = b 0 and A x = b Denote E x = x 0 x and E b = b 0 b Then That is, E x x cond(a) E b b ε x cond(a)ε b If a vector b is reliable to k significant digits, then the solution x of an equation A x = b is reliable to k log 10 (cond(a)) significant digits Theorem Assume that matrixec A 0, A and vectors x 0, x and b 0, b are related by the formulas A 0 x 0 = b 0 and A x = b Denote E A = A 0 A, E x = x 0 x and E b = b 0 b Then ( ε x cond(a) ε b + ε A x ) x 0 7
8 (JIM: Jacobi iteration method) Given: a system A x = b of linear equations, tolerance ε, arbitrary initial vector x 0 0 Set k = 0 1 Compute ( x k+1 ) i = 1 (i 1 a i,j ( x k ) j + a i,i j=1 If x k+1 x k, increase k by one and go back to step 1 j=i+1 a i,j ( x k ) j ) + b i a i,i (GSM: Gauss-Seidel iteration) Given: a system A x = b of linear equations, tolerance ε, arbitrary initial vector x 0 0 Set k = 0 1 Compute ( x k+1 ) i = 1 (i 1 a i,j ( x k+1 ) j + a i,i j=1 If x k+1 x k, increase k by one and go back to step 1 j=j+1 a i,j ( x k ) j ) + b i a i,i Theorem If a matrix B satisfies B M < 1 for some compatible matrix norm, then the corresponding iterative method x k+1 = B x k + c converges to x f for arbitrary choice of x 0 and we have x f x k+1 B M 1 B M x k+1 x k 8
9 Theorem An iterative method x k+1 = B x k + c converges if and only if ρ(b) < 1 Consider an n n matrix B The inequality ρ(b) B is true for all induced matrix norms Conversely, for every ε > 0 there is an induced matrix norm such that B ε < ρ(b) Consider an n n matrix A We say that A is strictly diagonally dominant if a i,i > j i a i,j for all i = 1,, n We say that A is positive definite if x T A x > 0 for all non-zero vectors x IR n 9
10 Theorem If A is strictly diagonally dominant, then both JIM and GSM converge for arbitrary choice of initial vector If A is symmetric and positive definite, then GSM converges for arbitrary choice of initial vector (SOR, Successive OverRelaxation method) Given: a system A x = b of linear equations, tolerance ε, parameter of relaxation λ, arbitrary initial vector x 0 0 Set k = 0 1 Compute ( x k+1 ) i = (1 λ)( x k ) i λ (i 1 a i,j ( x k+1 ) j + a i,i j=1 If x k+1 x k, increase k by one and go back to step 1 j=j+1 a i,j ( x k ) j ) + λb i a i,i Theorem (Ostrovsky) Let A be a symmetric n n matrix with positive diagonal entries Then ρ(b λ ) < 1 if and only if A is positive definite and 0 < λ < 2 10
11 Consider an n n matrix A A number λ is called an eigenvalue of A if there is a non-zero vector x such that A x = λ x Vectors x with this property are then called eigenvectors associated with λ If a real matrix is symmetric, then its eigenvalues are real If a real matrix is diagonizable, then there exists a basis of IR n composed of its eigenvectors (power method for finding the largest eigenvalue and an associated eigenvector) Given: an n n matrix A, tolerance ε > 0, arbitrary initial vector x 0 If complex eigenvalues are needed, choose a vector with non-zero imaginary part in each component 0 Set k = 0 1 Compute x k+1 = A x k, A x k If x k+1 x k ε, increase k by one and go back to step 1 For an n n matrix A and a vector x IR n we define their Rayleigh quotient by the formula If x is an eigenvector of a matrix A, then xt A x x T x x T A x x T x is equal to the associated eigenvalue 11
12 ODE and num math: Linear algebra (N) [lectures] c phabala 2016 (power method for finding the largest eigenvalue and an associated eigenvector) Given: an n n matrix A, tolerance ε > 0, arbitrary initial vector x 0 If complex eigenvalues are needed, choose a vector with non-zero imaginary part in each component 0 Set k = 0, let l 0 = x 0 A x 0 x 0 x 0 1 Compute x k+1 = l k A x k, l k+1 = x k+1 A x k+1 l k A x k x k+1 x k+1 If x k+1 x k ε, increase k by one and go back to step 1 Alternative: Use Euclidean norm, that is, x k+1 = λ k A x k λ k A x k 2, l k+1 = x k+1a x k+1 Theorem Let A be an n n matrix with eigenvalues λ j, where λ 1 > λ j for j 2 Let M = max λ j j 2 For x 0 let {l k }, { x k } be sequences generated by the power method Then {l k } converges to λ 1 and { x k } converges to some eigenvector v associated with λ 1 Moreover, λ 1 l k = O ([ M λ 1 ] 2k ) and v1 x k = O ([ M λ 1 ] k ) 12
13 ODE and num math: Linear algebra (N) [lectures] c phabala 2016 (deflation methoda for finding eigenvalue and an associated eigenvector) Given: an n n matrix A and its eigenvalues λ 1 through λ N with eigenvectors v 1 through v N satisfying v i = 1, tolerance ε > 0, arbitrary initial vector x 0 If complex eigenvalues are needed, choose a vector with non-zero imaginary part in each component 0 Set k = 0 1 Let y = x k N j=1 λ j( x k v j ) v j Compute x k+1 = A y A y, l k+1 = x k+1 A x k+1 If x k+1 x k ε, increase k by one and go back to step 1 Then λ k λ N+1, x k v N+1 (inverse power method for finding eigenvalue and an associated eigenvector) Given: an n n matrix A, arbitrary real number µ that is not an eigenvalue of A, tolerance ε > 0, arbitrary non-zero initiation vector x 0 If complex eigenvalues are needed, choose a vector with non-zero imaginary part in each component 1 Determine B = (A µe n ) 1 2 Using power iteration, find the largest eigenvalue λ B of the matrix B and an associated eigenvector v Output: The number µ + 1 λ B is the eigenvalue of A closest to µ with eigenvector v 13
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