Suppose I have a symmetric matrix A and a vector b and want to find A^(-1) b. Now, this is well-known to be doable in time O(N^2) (where N is the dimension of the vector\matrix), and I believe that in MATLAB this can be done as b\A. But all I can find in python is numpy.linalg.solve() which will do Gaussian elimination, which is O(N^3). I must not be looking in the right place...
scipy.linalg.solve has an argument to make it assume a symmetric matrix:
x = scipy.linalg.solve(A, b, assume_a="sym")
If you know your matrix is not just symmetric but positive definite you can give this stronger assumption instead, as "pos".
Related
How can I create Matrix P consisting of three eigenvector columns by using a double nested loop.
from sympy.matrices import Matrix, zeros
from sympy import pprint
A = Matrix([[6,2,6], [2,6,6], [6,6,2]])
ew_A = A.eigenvals()
ev_A = A.eigenvects()
pprint(ew_A)
pprint(ev_A)
# Matrix P
(n,m) = A.shape
P = TODO # Initialising
# "filling Matrix P with ...
for i in TODO:
for j in TODO:
P[:,i+j] = TODO
## Calculating Diagonalmatrix
D= P**-1*P*A
Thanks so much in Advance
Finding the eigenvalues of a matrix, or diagonalizing it, is equivalent to finding the zeros of a polynomial with a degree equal to the size of the matrix. So in your case diagonalizing a 3x3 matrix is equivalent to finding the zeros of a 3rd degree polynomial. Maybe there is a simple algorithm for that, but mathematicians always go for the general case.
And in the general case you can show that there is no terminating algorithm for finding the zeros of a 5th-or-higher degree polynomial (that is called Galois theory), so there is also no simple "triple loop" algorithm for matrices of size 5x5 and higher. Eigenvalue software works by an iterative approximation algorithm, so that is a "while" loop around some finite loops.
This means that your question has no answer in the general case. In the 3x3 case maybe, but even that is not going to be terribly simple.
I have a very big sparse matrix A = 7Mi-by-7Mi matrix. I am using Matlab's eigs(A,k) function which can calculate first k eigenvalues and vectors.
I need all of its eigenvector and values. But I can't store all of the eigenvectors because it requires a lot of memory.
Is there any way (in Matlab or Python) by which I can get eigenvectors one by one in a for loop? i.e. in ith iteration, I get the ith eigenvector and value.
If you have a good guess about how large the eigenvalue you are looking for is, say lambda_guess, you can use the Power iteration on
(A - lambda_guess* Id)^-1
This approach is sometimes referred to as the inverse-shift method. Here the method will converge to the eigenvalue closest to lambda_guess (and the better your guess the faster the convergence). Note that you wouldn't store the inverse, but only compute the solution of
x_next_iter = solve(A - lambda_guess*Id, x_iter), possibly itself with an iterative linear solver.
I would combine this with a subspace iteration method with subspace at least size two. This way, on your first iteration, you can find the smallest and second smallest eigenvalues lambda1, lambda2.
Then you can try lambdaguess= lambda2+ epsilon so that The first and second eigenvector outputted correspond to the second and third smallest eigenvalues, respectively.(if the first eigenvalue of this iteration is not the same as the value of lambda2 for your previous iteration, you need to make epsilon smaller and repeat. In practice you would test that their difference is small enough, to account for roundoff error and the fact that iterative methods are never exact). You repeat this until you get the eigenvalue number you are looking for. It's going to be slow, but you will only use two eigenvectors at any time.
NOTE: we assume all eigenvalues are distinct, otherwise this problem will not have a low memory solution with the usual techniques. In general, if the maximal multiplicity of an eigenvalue is m, you would need m vectors in memory for subspace iteration to converge.
I need to solve a set of simultaneous equations of the form Ax = B for x. I've used the numpy.linalg.solve function, inputting A and B, but I get the error 'LinAlgError: Last 2 dimensions of the array must be square'. How do I fix this?
Here's my code:
A = matrix([[v1x, v2x], [v1y, v2y], [v1z, v2z]])
print A
B = [(p2x-p1x-nmag[0]), (p2y-p1y-nmag[1]), (p2z-p1z-nmag[2])]
print B
x = numpy.linalg.solve(A, B)
The values of the matrix/vector are calculated earlier in the code and this works fine, but the values are:
A =
(-0.56666301, -0.52472909)
(0.44034147, 0.46768087)
(0.69641397, 0.71129036)
B =
(-0.38038602567630364, -24.092279373295057, 0.0)
x should have the form (x1,x2,0)
In case you still haven't found an answer, or in case someone in the future has this question.
To solve Ax=b:
numpy.linalg.solve uses LAPACK gesv. As mentioned in the documentation of LAPACK, gesv requires A to be square:
LA_GESV computes the solution to a real or complex linear system of equations AX = B, where A is a square matrix and X and B are rectangular matrices or vectors. Gaussian elimination with row interchanges is used to factor A as A = PL*U , where P is a permutation matrix, L is unit lower triangular, and U is upper triangular. The factored form of A is then used to solve the above system.
If A matrix is not square, it means that you either have more variables than your equations or the other way around. In these situations, you can have the cases of no solution or infinite number of solutions. What determines the solution space is the rank of the matrix compared to the number of columns. Therefore, you first have to check the rank of the matrix.
That being said, you can use another method to solve your system of linear equations. I suggest having a look at factorization methods like LU or QR or even SVD. In LAPACK you can use getrs, in Python you can different things:
first do the factorization like QR and then feed the resulting matrices to a method like scipy.linalg.solve_triangular
solve the least-squares using numpy.linalg.lstsq
Also have a look here where a simple example is formulated and solved.
A square matrix is a matrix with the same number of rows and columns. The matrix you are doing is a 3 by 2. Add a column of zeroes to fix this problem.
So I would like to generate a 50 X 50 covariance matrix for a random variable X given the following conditions:
one variance is 10 times larger than the others
the parameters of X are only slightly correlated
Is there a way of doing this in Python/R etc? Or is there a covariance matrix that you can think of that might satisfy these requirements?
Thank you for your help!
OK, you only need one matrix and randomness isn't important. Here's a way to construct a matrix according to your description. Start with an identity matrix 50 by 50. Assign 10 to the first (upper left) element. Assign a small number (I don't know what's appropriate for your problem, maybe 0.1? 0.01? It's up to you) to all the other elements. Now take that matrix and square it (i.e. compute transpose(X) . X where X is your matrix). Presto! You've squared the eigenvalues so now you have a covariance matrix.
If the small element is small enough, X is already positive definite. But squaring guarantees it (assuming there are no zero eigenvalues, which you can verify by computing the determinant -- if the determinant is nonzero then there are no zero eigenvalues).
I assume you can find Python functions for these operations.
How do I get the inverse of a matrix in python? I've implemented it myself, but it's pure python, and I suspect there are faster modules out there to do it.
You should have a look at numpy if you do matrix manipulation. This is a module mainly written in C, which will be much faster than programming in pure python. Here is an example of how to invert a matrix, and do other matrix manipulation.
from numpy import matrix
from numpy import linalg
A = matrix( [[1,2,3],[11,12,13],[21,22,23]]) # Creates a matrix.
x = matrix( [[1],[2],[3]] ) # Creates a matrix (like a column vector).
y = matrix( [[1,2,3]] ) # Creates a matrix (like a row vector).
print A.T # Transpose of A.
print A*x # Matrix multiplication of A and x.
print A.I # Inverse of A.
print linalg.solve(A, x) # Solve the linear equation system.
You can also have a look at the array module, which is a much more efficient implementation of lists when you have to deal with only one data type.
Make sure you really need to invert the matrix. This is often unnecessary and can be numerically unstable. When most people ask how to invert a matrix, they really want to know how to solve Ax = b where A is a matrix and x and b are vectors. It's more efficient and more accurate to use code that solves the equation Ax = b for x directly than to calculate A inverse then multiply the inverse by B. Even if you need to solve Ax = b for many b values, it's not a good idea to invert A. If you have to solve the system for multiple b values, save the Cholesky factorization of A, but don't invert it.
See Don't invert that matrix.
It is a pity that the chosen matrix, repeated here again, is either singular or badly conditioned:
A = matrix( [[1,2,3],[11,12,13],[21,22,23]])
By definition, the inverse of A when multiplied by the matrix A itself must give a unit matrix. The A chosen in the much praised explanation does not do that. In fact just looking at the inverse gives a clue that the inversion did not work correctly. Look at the magnitude of the individual terms - they are very, very big compared with the terms of the original A matrix...
It is remarkable that the humans when picking an example of a matrix so often manage to pick a singular matrix!
I did have a problem with the solution, so looked into it further. On the ubuntu-kubuntu platform, the debian package numpy does not have the matrix and the linalg sub-packages, so in addition to import of numpy, scipy needs to be imported also.
If the diagonal terms of A are multiplied by a large enough factor, say 2, the matrix will most likely cease to be singular or near singular. So
A = matrix( [[2,2,3],[11,24,13],[21,22,46]])
becomes neither singular nor nearly singular and the example gives meaningful results... When dealing with floating numbers one must be watchful for the effects of inavoidable round off errors.
For those like me, who were looking for a pure Python solution without pandas or numpy involved, check out the following GitHub project: https://github.com/ThomIves/MatrixInverse.
It generously provides a very good explanation of how the process looks like "behind the scenes". The author has nicely described the step-by-step approach and presented some practical examples, all easy to follow.
This is just a little code snippet from there to illustrate the approach very briefly (AM is the source matrix, IM is the identity matrix of the same size):
def invert_matrix(AM, IM):
for fd in range(len(AM)):
fdScaler = 1.0 / AM[fd][fd]
for j in range(len(AM)):
AM[fd][j] *= fdScaler
IM[fd][j] *= fdScaler
for i in list(range(len(AM)))[0:fd] + list(range(len(AM)))[fd+1:]:
crScaler = AM[i][fd]
for j in range(len(AM)):
AM[i][j] = AM[i][j] - crScaler * AM[fd][j]
IM[i][j] = IM[i][j] - crScaler * IM[fd][j]
return IM
But please do follow the entire thing, you'll learn a lot more than just copy-pasting this code! There's a Jupyter notebook as well, btw.
Hope that helps someone, I personally found it extremely useful for my very particular task (Absorbing Markov Chain) where I wasn't able to use any non-standard packages.
You could calculate the determinant of the matrix which is recursive
and then form the adjoined matrix
Here is a short tutorial
I think this only works for square matrices
Another way of computing these involves gram-schmidt orthogonalization and then transposing the matrix, the transpose of an orthogonalized matrix is its inverse!
Numpy will be suitable for most people, but you can also do matrices in Sympy
Try running these commands at http://live.sympy.org/
M = Matrix([[1, 3], [-2, 3]])
M
M**-1
For fun, try M**(1/2)
If you hate numpy, get out RPy and your local copy of R, and use it instead.
(I would also echo to make you you really need to invert the matrix. In R, for example, linalg.solve and the solve() function don't actually do a full inversion, since it is unnecessary.)