I would like to get a square matrix B from a linear vector A such that B = A * transpose(A). A is a numpy array and np.shape(A) returns (10,). I would like B to be a (10,10) array. I tried B = np.matmut(A, A[np.newaxis]) but I get an error :
shapes (10,) and (1,10) not aligned: 10 (dim 0) != 1 (dim 0)
you can do this using outer:
import numpy as np
vector = np.arange(10)
np.outer(vector, vector)
The solution is a little ugly, but it does what you need.
import numpy as np
vector = np.array([1,2,3,4,5,6,7,8,9,10],)
matrix = np.dot(vector[:,None],vector[None,:])
print(matrix)
You can also do the following:
import numpy as np
vector = np.array([1,2,3,4,5,6,7,8,9,10],)
matrix = vector*vector[:,None]
print(matrix)
The issue comes from the fact that transposing a one dimensional array does not have the effect you might expect.
Variation on outer product:
a = A.reshape(-1, 1) # make sure it's a column vector
B = a @ a.T
