tensorly.solve

solve(a, b)

Solve a linear matrix equation, or system of linear scalar equations.

Computes the “exact” solution, x, of the well-determined, i.e., full rank, linear matrix equation ax = b.

Parameters:

a(…, M, M) array_like

Coefficient matrix.

b{(M,), (…, M, K)}, array_like

Ordinate or “dependent variable” values.

Returns:

x{(…, M,), (…, M, K)} ndarray

Solution to the system a x = b. Returned shape is (…, M) if b is shape (M,) and (…, M, K) if b is (…, M, K), where the “…” part is broadcasted between a and b.

Raises:

LinAlgError

If a is singular or not square.

See also

scipy.linalg.solve

Similar function in SciPy.

Notes

Added in version 1.8.0.

Broadcasting rules apply, see the numpy.linalg documentation for details.

The solutions are computed using LAPACK routine _gesv.

a must be square and of full-rank, i.e., all rows (or, equivalently, columns) must be linearly independent; if either is not true, use lstsq for the least-squares best “solution” of the system/equation.

Changed in version 2.0: The b array is only treated as a shape (M,) column vector if it is exactly 1-dimensional. In all other instances it is treated as a stack of (M, K) matrices. Previously b would be treated as a stack of (M,) vectors if b.ndim was equal to a.ndim - 1.

References

[1]

G. Strang, Linear Algebra and Its Applications, 2nd Ed., Orlando, FL, Academic Press, Inc., 1980, pg. 22.

Examples

Solve the system of equations: x0 + 2 * x1 = 1 and 3 * x0 + 5 * x1 = 2:

>>> import numpy as np
>>> a = np.array([[1, 2], [3, 5]])
>>> b = np.array([1, 2])
>>> x = np.linalg.solve(a, b)
>>> x
array([-1.,  1.])

Check that the solution is correct:

>>> np.allclose(np.dot(a, x), b)
True