Numpy efficient matrix self-multiplication (gram matrix)

Question

I want to multiply B = A @ A.T in numpy. Obviously, the answer would be a symmetric matrix (i.e. B[i, j] == B[j, i]).

However, it is not cl

Answer 1

As noted in @PaulPanzer's link, dot can detect this case. Here's the timing proof:

In [355]: A = np.random.rand(1000,1000)
In [356]: timeit A.dot(A.T)
57.4 ms ± 960 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
In [357]: B = A.T.copy()
In [358]: timeit A.dot(B)
98.6 ms ± 805 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

Numpy dot too clever about symmetric multiplications