Use numpy.dot or a.dot(b). See the documentation here.
>>> a = np.array([[ 5, 1 ,3],
[ 1, 1 ,1],
[ 1, 2 ,1]])
>>> b = np.array([1, 2, 3])
>>> print a.dot(b)
array([16, 6, 8])
This occurs because numpy arrays are not matrices, and the standard operations *, +, -, / work element-wise on arrays. Instead, you could try using numpy.matrix, and * will be treated like matrix multiplication.
Also know there are other options:
As noted below, if using python3.5+ the @ operator works as you'd expect:
>>> print(a @ b)
array([16, 6, 8])
If you want overkill, you can use numpy.einsum. The documentation will give you a flavor for how it works, but honestly, I didn't fully understand how to use it until reading this answer and just playing around with it on my own.
>>> np.einsum('ji,i->j', a, b)
array([16, 6, 8])
As of mid 2016 (numpy 1.10.1), you can try the experimental numpy.matmul, which works like numpy.dot with two major exceptions: no scalar multiplication but it works with stacks of matrices.
>>> np.matmul(a, b)
array([16, 6, 8])
numpy.inner functions the same way as numpy.dot for matrix-vector multiplication but behaves differently for matrix-matrix and tensor multiplication (see Wikipedia regarding the differences between the inner product and dot product in general or see this SO answer regarding numpy's implementations).
>>> np.inner(a, b)
array([16, 6, 8])
# Beware using for matrix-matrix multiplication though!
>>> b = a.T
>>> np.dot(a, b)
array([[35, 9, 10],
[ 9, 3, 4],
[10, 4, 6]])
>>> np.inner(a, b)
array([[29, 12, 19],
[ 7, 4, 5],
[ 8, 5, 6]])
If you have tensors (arrays of dimension greater than or equal to one), you can use numpy.tensordot with the optional argument axes=1:
>>> np.tensordot(a, b, axes=1)
array([16, 6, 8])
Don't use numpy.vdot if you have a matrix of complex numbers, as the matrix will be flattened to a 1D array, then it will try to find the complex conjugate dot product between your flattened matrix and vector (which will fail due to a size mismatch n*m vs n).