Working Matrix Square root

Question

I'm trying to take the square root of a matrix. That is find the matrix B so B*B=A. None of the methods I've found around gives a working result.

First I found this formula on Wikipedia:

Set Y_0 = A and Z_0 = I then the iteration:

Y_{k+1} = .5*(Y_k + Z_k^{-1}),

Z_{k+1} = .5*(Z_k + Y_k^{-1}).

Then Y should converge to B.

However implementing the algorithm in python (using numpy for inverse matrices), gave me rubbish results:

>>> def denbev(Y,Z,n):
    if n == 0: return Y,Z
    return denbev(.5*(Y+Z**-1), .5*(Z+Y**-1), n-1)

>>> denbev(matrix('1,2;3,4'), matrix('1,0;0,1'), 3)[0]**2
matrix([[ 1.31969074,  1.85986159],
        [ 2.78979239,  4.10948313]])

>>> denbev(matrix('1,2;3,4'), matrix('1,0;0,1'), 100)[0]**2
matrix([[ 1.44409972,  1.79685675],
        [ 2.69528512,  4.13938485]])

As you can see, iterating 100 times, gives worse results than iterating three times, and none of the results get within a 40% error margin.

Then I tried the scipy sqrtm method, but that was even worse:

>>> scipy.linalg.sqrtm(matrix('1,2;3,4'))**2
array([[ 0.09090909+0.51425948j,  0.60606061-0.34283965j],
       [ 1.36363636-0.77138922j,  3.09090909+0.51425948j]])

>>> scipy.linalg.sqrtm(matrix('1,2;3,4')**2)
array([[ 1.56669890+0.j,  1.74077656+0.j],
       [ 2.61116484+0.j,  4.17786374+0.j]])

I don't know a lot about matrix square rooting, but I figure there must be algorithms that perform better than the above?

Answer 1

(1) the square root of the matrix [1,2;3,4] should give something complex, as the eigenvalues of that matrix are negative. SO your solution can't be correct to begin with.

(2) linalg.sqrtm returns an array, NOT a matrix. Hence, using * to multiply them is not a good idea. In your case, the solutions is thus correct, but you're not seeing it.

edit try the following, you'll see it's correct:

asmatrix(scipy.linalg.sqrtm(matrix('1,2;3,4')))**2

Answer 2

Your matrix [1 2; 3 4] isn't positive so there is no solution to the problem in the domain of real matrices.