Kronecker product in Python and Matlab

Question

I was trying to reproduce a result in Python from MATLAB. However, I can't seem to get it right. This is the correct MATLAB code:

nx = 5;
ny = 7;

x = linspace(0, 1, nx); dx = x(2) - x(1);
y = linspace(0, 1, ny); dy = y(2) - y(1);

onex = ones(nx, 1);
oney = ones(ny, 1);

Dx = spdiags([onex -2*onex onex], [-1 0 1], nx, nx);
Dy = spdiags([oney -2*oney oney], [-1 0 1], ny, ny);

Ix = eye(nx); Iy = eye(ny);
L = kron(Iy, Dx);

size(L) % 35   35

Now, this is the Python code:

nx = 5
ny = 7
x = linspace(0, 1, nx); dx = x[1] - x[0]
y = linspace(0, 1, ny); dy = y[1] - y[0]

onex = ones(nx)
oney = ones(ny)
Dx = sparse.dia_matrix( ([onex, -2*onex, onex], [-1,0,1] ), shape=(nx,nx))
Dy = sparse.dia_matrix( ([oney, -2*oney, oney], [-1,0,1] ), shape=(ny,ny))

Ix = eye(nx)
Iy = eye(ny)

L = kron(Iy, Dx)

L.shape # (7, 7)

As far I have been able to verify, everything is correct until the definition of L. According to MATLAB kron(Iy, Dx) (which is supposed to be the kronecker product) should produce a 35X35 matrix, but Python thinks it should be a 7X7 matrix. In simpler calculations, both give the correct answer:

Python:

kron(array(([1,2],[2,3])), [1,2])

array([[1, 2, 2, 4],
       [2, 4, 3, 6]])

MATLAB

kron([1 2; 2 3], [1 2]) 

ans = 1   2   2   4
      2   4   3   6

Why do I get different results?

Thanks!

Answer 1

Use sparse.kron for the Kronecker product of sparse matrices.

numpy.kron does not handle sparse matrices. When given sparse matrices, it might not generate an error, but the value it returns will not be correct.