[Python] Is there really an @ operator in Python to calculate dot product? Why doesn't it work...

Is this answer correct?

In Python 3.5, there is a new operator for the dot product, so you can write a = A @ B instead of a = numpy.dot(A, B)
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— Henri Andre, on "What is the pythonic way to calculate dot product?"

It does not seem to work for me.

$ python3
Python 3.6.1 (default, Apr 4 2017, 09:40:21)
[GCC 4.2.1 Compatible Apple LLVM 8.1.0 (clang-802.0.38)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> a = [1, 2, 3]
>>> b = [4, 5, 6]
>>> a @ b
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: unsupported operand type(s) for @: 'list' and 'list'


But the linked answer has received 6 upvotes, so I must be missing something. Can you provide a complete example that shows how to use the @ operator to calculate a dot product?

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