Question

I have two large square sparse matrices, A & B, and need to compute the following: A * B^-1 in the most efficient way. I have a feeling that the answer involves using scipy.sparse, but can’t for the life of me figure it out.

After extensive searching, I have run across the following thread: Efficient numpy / lapack routine for product of inverse and sparse matrix? but can’t figure out what the most efficient way would be.

Someone suggested using LU decomposition which is built into the sparse module of scipy, but when I try and do LU on sample matrix is says the result is singular (although when I just do a * B^-1 i get an answer). I have also heard someone suggest using linalg.spsolve(), but i can’t figure out how to implement this as it requires a vector as the second argument.

If it helps, once I have the solution s.t. A * B^-1 = C, i only need to know the value for one row of the matrix C. The matrices will be roughly 1000×1000 to 1500×1500.

Answer 1

Actually 1000×1000 matrices are not that large. You can compute the inverse of such a matrix using numpy.linalg.inv(B) in less than 1 second on a modern desktop computer.

But you can be much more efficient if you rewrite your problem taking into account the fact that you only need one row of C (this is actually very often the case).

Let us write d_i = [0 0 0 … 0 1 0 … 0 ], a vector with only one one on the i-th element.
You can write, if ^t denotes the transpose :

AB^-1 = C <=> A = CB <=> A^t = B^t C^t

For the i-th row :

A^t d_i = B^t C^t d_i <=> a_i = B^t c_i

So you have a linear inverse problem which can be solved using numpy.linalg.solve

ci = np.linalg.solve(B.T, a[i])