So based on the inputs here, I’m marking my original code block with the explicit test as the solution: if linalg.cond(x) < 1/sys.float_info.epsilon: i = linalg.inv(x) else: #handle it Surprisingly, the numpy.linalg.inv function doesn’t perform this test. I checked the code and found it goes through all it’s machinations, then just calls the lapack routine … Read more
This could be what you are looking for if len(a) < len(b): c = b.copy() c[:len(a)] += a else: c = a.copy() c[:len(b)] += b basically you copy the longer one and then add in-place the shorter one
Sympy makes this straightforward. >>> from sympy import Matrix >>> A = [[2, 3, 5], [-4, 2, 3], [0, 0, 0]] >>> A = Matrix(A) >>> A * A.nullspace()[0] Matrix([ [0], [0], [0]]) >>> A.nullspace() [Matrix([ [-1/16], [-13/8], [ 1]])]
Since this question has garnered a lot of views, I thought I’d post code for the final answer that I found, in part, by posting to the Eigen forums. The code uses Boost for the univariate normal and Eigen for matrix handling. It feels rather unorthodox, since it involves using the “internal” namespace, but it … Read more
np.linalg.solve(A, b) does not compute the inverse of A. Instead it calls one of the gesv LAPACK routines, which first factorizes A using LU decomposition, then solves for x using forward and backward substitution (see here). np.linalg.inv uses the same method to compute the inverse of A by solving for A-1 in A·A-1 = I … Read more
You should be able to exploit the fact that the matrix is affine to speed things up over a full inverse. Namely, if your matrix looks like this A = [ M b ] [ 0 1 ] where A is 4×4, M is 3×3, b is 3×1, and the bottom row is (0,0,0,1), then … Read more
Assuming you have two lines of the form Ax + By = C, you can find it pretty easily: float delta = A1 * B2 – A2 * B1; if (delta == 0) throw new ArgumentException(“Lines are parallel”); float x = (B2 * C1 – B1 * C2) / delta; float y = (A1 * … Read more
**if your matrix is sparse, then instantiate your matrix using a constructor from scipy.sparse then use the analogous eigenvector/eigenvalue methods in spicy.sparse.linalg. From a performance point of view, this has two advantages: your matrix, built from the spicy.sparse constructor, will be smaller in proportion to how sparse it is. the eigenvalue/eigenvector methods for sparse matrices … Read more
you can use the transpose operation to do this: Example: In [2]: a = np.array([[1,2], [3,4], [5,6]]) In [5]: a.shape Out[5]: (3, 2) In [6]: a_trans = a.T #or: np.transpose(a), a.transpose() In [8]: a_trans.shape Out[8]: (2, 3) In [7]: a_trans Out[7]: array([[1, 3, 5], [2, 4, 6]]) Note that the original array a will still … Read more