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How to solve a circulant matrix equation using Python SciPy?
The linear function named scipy.linalg.solveh_banded is used to solve the banded matrix equation. In the below given example we will be solving the circulant system Cx = b −
Example
from scipy.linalg import solve_circulant, solve, circulant, lstsq import numpy as np c = np.array([2, 2, 4]) b = np.array([1, 2, 3]) solve_circulant(c, b)
Output
array([ 0.75, -0.25, 0.25])
Example
Let’s see a singular example, it will raise an LinAlgError −
from scipy.linalg import solve_circulant, solve, circulant, lstsq import numpy as np c = np.array([1, 1, 0, 0]) b = np.array([1, 2, 3, 4]) solve_circulant(c, b)
Output
--------------------------------------------------------------------------
LinAlgError Traceback (most recent call last)
<ipython-input-6-978604ed0a97> in <module>
----> 1 solve_circulant(c, b)
~\AppData\Roaming\Python\Python37\site-packages\scipy\linalg\basic.py in
solve_circulant(c, b, singular, tol, caxis, baxis, outaxis)
865 if is_near_singular:
866 if singular == 'raise':
--> 867 raise LinAlgError("near singular circulant matrix.")
868 else:
869 # Replace the small values with 1 to avoid errors in the
LinAlgError: near singular circulant matrix.Now, to remove this error we need to use the option singular = ‘lstsq’ as follows −
from scipy.linalg import solve_circulant, solve, circulant, lstsq import numpy as np c = np.array([1, 1, 0, 0]) b = np.array([1, 2, 3, 4]) solve_circulant(c, b, singular='lstsq')
Output
array([0.25, 1.25, 2.25, 1.25])
Updated on: 24-Nov-2021
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