If I have numpy arrays A and B, then I can compute the trace of their matrix product with:
tr = numpy.linalg.trace(A.dot(B))
You can improve on @Bill's solution by reducing intermediate storage to the diagonal elements only:
from numpy.core.umath_tests import inner1d
m, n = 1000, 500
a = np.random.rand(m, n)
b = np.random.rand(n, m)
# They all should give the same result
print np.trace(a.dot(b))
print np.sum(a*b.T)
print np.sum(inner1d(a, b.T))
%timeit np.trace(a.dot(b))
10 loops, best of 3: 34.7 ms per loop
%timeit np.sum(a*b.T)
100 loops, best of 3: 4.85 ms per loop
%timeit np.sum(inner1d(a, b.T))
1000 loops, best of 3: 1.83 ms per loop
Another option is to use np.einsum and have no explicit intermediate storage at all:
# Will print the same as the others:
print np.einsum('ij,ji->', a, b)
On my system it runs slightly slower than using inner1d, but it may not hold for all systems, see this question:
%timeit np.einsum('ij,ji->', a, b)
100 loops, best of 3: 1.91 ms per loop