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The tensor product of two 1 dimensional vector spaces is 1 dimensional so it is smaller not bigger than the direct sum. The tensor product tof two 2 dimensional vector spaces is 4 dimensional so this is the the same size as the direct sum not bigger.This is correct but missing the relevant point: that the presentation contains a false statement.
The complete stress tensor, $\sigma$, tells us the total force a surface with unit area facing any direction will experience. Once we fix the direction, we get the traction vector from the stress tensor, or, I do not mean literally though, the stress tensor collapses to the traction vector.
A part of the tensor history must come from tenses (past present future) and how Aristotle defined time as the measure of change / motion / movement. So really descriptions of changes of the state (or unchanging stillness) whether static or dynamic. EDIT: Tensor:= Tense + or (REF-1, includes William Rowan Hamilton algebraic origin)
The components of a rank-2 tensor can be written in a matrix. The tensor is not that matrix, because different types of tensors can correspond to the same matrix. The differences between those tensor types are uncovered by the basis transformations (hence the physicist's definition: "A tensor is what transforms like a tensor").
$\begingroup$ @BokaPeer : It appears that what I should have said is that the set of all tensor products of two vectors is a quotient of the Cartesian product, in that many different members of the Cartesian product of two spaces correspond to just one tensor product of two vectors. $\endgroup$
Consider the following example. Let's say we have a certain vector v that we would like to first transform somehow and then rotate.(Lets say that the transformation we want to apply to v is a simple scaling, which is simpler to describe wrt to orthodox v i.e. before we rotate it) This leads us to having two options for the operations:
One thing I am confused about though, is that in one of the answer to this question there was a point made that we are taking tensor products of a Vector space with itself (and possibly it's dual), however in the quantum mechanics picture (although I didn't rely on it in this example) we often take tensor products between different, often disjoint, subspaces of the enormous Hilbert space that ...
$\begingroup$ You can take A as a 3D tensor and 4D as $\frac{\text{d}A}{\text{d}q}$ where q is an $8 \times 1$ vector $\endgroup$ – Nirvana Commented Aug 22, 2014 at 19:00
This is usually treated using the universal property of the tensor product, as in the other answers, or in Dummit and Foote, or left as an exercise. In my opinion, leaving this as an exercise is mostly about not wanting to write this much detail.
Tensor products are important in areas of abstract algebra, homological algebra, algebraic topology and algebraic geometry. and tensor products of vector spaces are also important in differential geometry and physics. I think it is better to learn about these applications thoroughly than to have someone attempt to summarize them.