rkjt50r983 - 1 year ago 94

Python Question

When a sparse matrix is generated by

`scipy.sparse.rand`

`"RuntimeError: superlu failure (singular matrix?) at line 100 in file scipy/sparse/linalg/dsolve/SuperLU/SRC/dsnode_bmod.c"`

`dim = 20000`

ratio = 0.000133

A = scipy.sparse.rand(dim,dim,ratio)

inv_sparse = scipy.sparse.linalg.inv(A)

Is there a way to generate non-singular sparse matrix?

What I really want to do is comparing performance (process time) of

`scipy.sparse.linalg.inv`

`np.linalg.inv`

Answer Source

The density `ratio = 0.000133`

of your matrix is very low. It means that about one item out of 7518 is non-null. Hence, the probability of each term to be null is about 7517/7518.

Each row is made of 20000 independent terms. So the probability for a row to be null is (7517/7518)^20000=6.99%. Hence, the probability for a row to be non-null is 1-(7517/7518)^20000=93.0%.

Then, the matrix is made of 20000 rows. The rows can be considered as being independent. Hence, the probability that the matrix does not contain null rows is (1-(7517/7518)^20000)^20000=(93.0%)^20000. This probability is very low.

As the matrix is likely to contain a null row, it is often singular.

Moreover, due to the the limited precision of floating-point numbers, programs often consider ill-conditionned matrices as singular. Indeed, in such cases, the computed inverse would be highly unprecise and meaningless.

Finally, to compare the inverse function, it may be better to use matrices known to be invertible... At least, you could try to increase the density so that the probability of a null row becomes very low.