Thanks @Steve, but I understand the part about forming all possible combinations of basis vectors. What is \\(\otimes\\) for example in \\(e_1 \otimes f_1\\)? If the answer is "tensor product", then all these sounds a bit circular - we define tensor product of vector spaces in terms of tensor products of basis vectors, while never specifying what \\(\otimes\\) actually does.

Or, if we take for example \\(v_1 = a\cdot e_1 + b\cdot e_2 = (a, b)\\) and \\(v_2 = a'\cdot f_1 + b'\cdot f_2 = (a', b')\\), then doing a tensor product we get a new vector \\(v_3 = v_1 \otimes v_2 = (a'a, a'b, b'a, b'b)\\) in a new 4-dimensional vector space?

There are quite a lot of linear maps, candidates for \\(\cup\\), which send such a vector \\(v_3\\) to the underlying field \\(k\\), for example \\(v_4 = (1, 0, 0, 1)^T\\) will send the tensor product above to the inner product of \\(v_1\\) and \\(v_2\\).

**EDIT** Or, as another example, using \\(v_4 = (0, 1, -1, 0)^T\\) one may obtain the value of determinant (or, in general, exterior product). This tensor product has the universal property of being the freest bilinear operation, so all other things may be factored through it.

Or, if we take for example \\(v_1 = a\cdot e_1 + b\cdot e_2 = (a, b)\\) and \\(v_2 = a'\cdot f_1 + b'\cdot f_2 = (a', b')\\), then doing a tensor product we get a new vector \\(v_3 = v_1 \otimes v_2 = (a'a, a'b, b'a, b'b)\\) in a new 4-dimensional vector space?

There are quite a lot of linear maps, candidates for \\(\cup\\), which send such a vector \\(v_3\\) to the underlying field \\(k\\), for example \\(v_4 = (1, 0, 0, 1)^T\\) will send the tensor product above to the inner product of \\(v_1\\) and \\(v_2\\).

**EDIT** Or, as another example, using \\(v_4 = (0, 1, -1, 0)^T\\) one may obtain the value of determinant (or, in general, exterior product). This tensor product has the universal property of being the freest bilinear operation, so all other things may be factored through it.