Standard decompositions and linear solving#
Returns the Cholesky decomposition of the symmetric matrix matA \(=A = PLU\), without pivoting. Here matA is an instance of one of the above classes.
See also Eigen [238], Wikipedia [1521], Wikipedia [1555].
Cholesky decomposition#
- ctx.cholesky(A, tol=None)#
where
ctxisfpm,mpm,ipm,dec,gmporapm.Cholesky decomposition of a symmetric positive-definite matrix \(A\). Returns a lower triangular matrix \(L\) such that \(A = L \times L^T\). More generally, for a complex Hermitian positive-definite matrix, a Cholesky decomposition satisfying \(A = L \times L^H\) is returned.
The Cholesky decomposition can be used to solve linear equation systems twice as efficiently as LU decomposition, or to test whether \(A\) is positive-definite.
The optional parameter
toldetermines the tolerance for verifying positive-definiteness.Examples
Cholesky decomposition of a positive-definite symmetric matrix:
>>> from mpmath import * >>> mp.dps = 25; mp.pretty = True >>> A = eye(3) + hilbert(3) >>> nprint(A) [ 2.0 0.5 0.333333] [ 0.5 1.33333 0.25] [0.333333 0.25 1.2] >>> L = cholesky(A) >>> nprint(L) [ 1.41421 0.0 0.0] [0.353553 1.09924 0.0] [0.235702 0.15162 1.05899] >>> chop(A - L*L.T) [0.0 0.0 0.0] [0.0 0.0 0.0] [0.0 0.0 0.0]
Cholesky decomposition of a Hermitian matrix:
>>> A = eye(3) + matrix([[0,0.25j,-0.5j],[-0.25j,0,0],[0.5j,0,0]]) >>> L = cholesky(A) >>> nprint(L) [ 1.0 0.0 0.0] [(0.0 - 0.25j) (0.968246 + 0.0j) 0.0] [ (0.0 + 0.5j) (0.129099 + 0.0j) (0.856349 + 0.0j)] >>> chop(A - L*L.H) [0.0 0.0 0.0] [0.0 0.0 0.0] [0.0 0.0 0.0]
Attempted Cholesky decomposition of a matrix that is not positive definite:
>>> A = -eye(3) + hilbert(3) >>> L = cholesky(A) Traceback (most recent call last): ... ValueError: matrix is not positive-definite
References
Cholesky decomposition, solve#
- ctx.cholesky_solve(A, b, **kwargs)#
where
ctxisfpm,mpm,ipm,dec,gmporapm.Solves a symmetric positive-definite linear equation system. This is twice as efficient as lu_solve.
Returns the LU decomposition of the general square matrix matA \(= A = PLU\), with partial pivoting.
Matrix LU factorization#
- ctx.lu(A)#
where
ctxisfpm,mpm,ipm,dec,gmporapm.A -> P, L, U
LU factorisation of a square matrix A. L is the lower, U the upper part. P is the permutation matrix indicating the row swaps.
P*A = L*U
If you need efficiency, use the low-level method LU_decomp instead, it’s much more memory efficient.
The function
lucomputes an explicit LU factorization of a matrix:>>> P, L, U = lu(matrix([[0,2,3],[4,5,6],[7,8,9]])) >>> print(P) [0.0 0.0 1.0] [1.0 0.0 0.0] [0.0 1.0 0.0] >>> print(L) [ 1.0 0.0 0.0] [ 0.0 1.0 0.0] [0.571428571428571 0.214285714285714 1.0] >>> print(U) [7.0 8.0 9.0] [0.0 2.0 3.0] [0.0 0.0 0.214285714285714] >>> print(P.T*L*U) [0.0 2.0 3.0] [4.0 5.0 6.0] [7.0 8.0 9.0]
Determinant of a matrix, using LU decomposition#
- ctx.det(A)#
where
ctxisfpm,mpm,ipm,dec,gmporapm.Calculates the determinant of a matrix, using the LU factorization.
Inverse of a matrix, using the LU factorization#
- ctx.inverse(A, **kwargs)#
where
ctxisfpm,mpm,ipm,dec,gmporapm.Calculates the inverse of a matrix, using the LU factorization.
If you want to solve an equation system Ax = b, it’s recommended to use solve(A, b) instead, it’s about 3 times more efficient.
Linear equations: LU solve#
- ctx.lu_solve(A, b)#
where
ctxisfpm,mpm,ipm,dec,gmporapm.Basic linear algebra is implemented; you can for example solve the linear equation system:
x + 2*y = -10 3*x + 4*y = 10
using
lu_solve:>>> from mpmath import * >>> mp.pretty = False >>> A = matrix([[1, 2], [3, 4]]) >>> b = matrix([-10, 10]) >>> x = lu_solve(A, b) >>> x matrix( [['30.0'], ['-20.0']])
Linear equations: residual of LU solve#
- ctx.residual(A, x, b, **kwargs)#
where
ctxisfpm,mpm,ipm,dec,gmporapm.Calculate the residual of a solution to a linear equation system.
r = A*x - b for A*x = b
If you don’t trust the result, use
residualto calculate the residual ||A*x-b||:>>> residual(A, x, b) matrix( [['3.46944695195361e-18'], ['3.46944695195361e-18']]) >>> str(eps) '2.22044604925031e-16'
As you can see, the solution is quite accurate. The error is caused by the inaccuracy of the internal floating point arithmetic. Though, it’s even smaller than the current machine epsilon, which basically means you can trust the result.
lu_solveaccepts overdetermined systems. It is usually not possible to solve such systems, so the residual is minimized instead. Internally this is done using Cholesky decomposition to compute a least squares approximation. This means thatlu_solvewill square the errors. If you can’t afford this, useqr_solveinstead. It is twice as slow but more accurate, and it calculates the residual automatically.
??? LU improve solution#
- improve_solution(ctx, A, x, b, maxsteps=1)#
Improve a solution to a linear equation system iteratively.
This re-uses the LU decomposition and is thus cheap. Usually 3 up to 4 iterations are giving the maximal improvement.
mpmath: LU condition number#
- ctx.cond(A, norm=None)#
where
ctxisfpm,mpm,ipm,dec,gmporapm.Calculates the condition number of a matrix using a specified matrix norm.
The condition number estimates the sensitivity of a matrix to errors. Example: small input errors for ill-conditioned coefficient matrices alter the solution of the system dramatically.
For ill-conditioned matrices it’s recommended to use qr_solve() instead of lu_solve(). This does not help with input errors however, it just avoids to add additional errors.
Definition: cond(A) = ||A|| * ||A**-1||
Returns the QR decomposition of the symmetric matrix matA \(=A = QR\), without pivoting.
QR factorization#
- ctx.qr(A, mode='full', edps=10)#
where
ctxisfpm,mpm,ipm,dec,gmporapm.Compute a QR factorization $A = QR$ where A is an m x n matrix of real or complex numbers where m >= n
mode has following meanings: (1) mode = ‘raw’ returns two matrixes (A, tau) in the internal format used by LAPACK (2) mode = ‘skinny’ returns the leading n columns of Q and n rows of R (3) Any other value returns the leading m columns of Q and m rows of R
edps is the increase in mp precision used for calculations
Examples
>>> from mpmath import * >>> mp.dps = 15 >>> mp.pretty = True >>> A = matrix([[1, 2], [3, 4], [1, 1]]) >>> Q, R = qr(A) >>> Q [-0.301511344577764 0.861640436855329 0.408248290463863] [-0.904534033733291 -0.123091490979333 -0.408248290463863] [-0.301511344577764 -0.492365963917331 0.816496580927726] >>> R [-3.3166247903554 -4.52267016866645] [ 0.0 0.738548945875996] [ 0.0 0.0] >>> Q * R [1.0 2.0] [3.0 4.0] [1.0 1.0] >>> chop(Q.T * Q) [1.0 0.0 0.0] [0.0 1.0 0.0] [0.0 0.0 1.0] >>> B = matrix([[1+0j, 2-3j], [3+j, 4+5j]]) >>> Q, R = qr(B) >>> nprint(Q) [ (-0.301511 + 0.0j) (0.0695795 - 0.95092j)] [(-0.904534 - 0.301511j) (-0.115966 + 0.278318j)] >>> nprint(R) [(-3.31662 + 0.0j) (-5.72872 - 2.41209j)] [ 0.0 (3.91965 + 0.0j)] >>> Q * R [(1.0 + 0.0j) (2.0 - 3.0j)] [(3.0 + 1.0j) (4.0 + 5.0j)] >>> chop(Q.T * Q.conjugate()) [1.0 0.0] [0.0 1.0]
QR solve#
- ctx.qr_solve(A, b, norm=None, **kwargs)#
where
ctxisfpm,mpm,ipm,dec,gmporapm.Ax = b => x, ||Ax - b||
Solve a determined or overdetermined linear equations system and calculate the norm of the residual (error). QR decomposition using Householder factorization is applied, which gives very accurate results even for ill-conditioned matrices. lu_solve is twice as efficient.