scipy.linalg.solve_discrete_lyapunov#

scipy.linalg.solve_discrete_lyapunov(a, q, method=None)[source]#

Solves the discrete Lyapunov equation \(AXA^H - X + Q = 0\).

Parameters:

a, q(M, M) array_like

Square matrices corresponding to A and Q in the equation above respectively. Must have the same shape.

method{‘direct’, ‘bilinear’}, optional

Type of solver.

If not given, chosen to be direct if M is less than 10 and bilinear otherwise.

Returns:

xndarray: Solution to the discrete Lyapunov equation

See also

solve_continuous_lyapunov: computes the solution to the continuous-time Lyapunov equation

Notes

This section describes the available solvers that can be selected by the ‘method’ parameter. The default method is direct if M is less than 10 and bilinear otherwise.

Method direct uses a direct analytical solution to the discrete Lyapunov equation. The algorithm is given in, for example, [1]. However, it requires the linear solution of a system with dimension \(M^2\) so that performance degrades rapidly for even moderately sized matrices.

Method bilinear uses a bilinear transformation to convert the discrete Lyapunov equation to a continuous Lyapunov equation \((BX+XB'=-C)\) where \(B=(A-I)(A+I)^{-1}\) and \(C=2(A' + I)^{-1} Q (A + I)^{-1}\). The continuous equation can be efficiently solved since it is a special case of a Sylvester equation. The transformation algorithm is from Popov (1964) as described in [2].

Added in version 0.11.0.

References

[1]

Hamilton, James D. Time Series Analysis, Princeton: Princeton University Press, 1994. 265. Print. http://doc1.lbfl.li/aca/FLMF037168.pdf

[2]

Gajic, Z., and M.T.J. Qureshi. 2008. Lyapunov Matrix Equation in System Stability and Control. Dover Books on Engineering Series. Dover Publications.

Examples

Given a and q solve for x:

>>> import numpy as np
>>> from scipy import linalg
>>> a = np.array([[0.2, 0.5],[0.7, -0.9]])
>>> q = np.eye(2)
>>> x = linalg.solve_discrete_lyapunov(a, q)
>>> x
array([[ 0.70872893,  1.43518822],
       [ 1.43518822, -2.4266315 ]])
>>> np.allclose(a.dot(x).dot(a.T)-x, -q)
True