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I've repeatedly read things that reference tensors, and despite reading the wiki page and other answers here on stackexchange I still don't know what a tensor is. I'm fine with hearing things in the framework on linear algebra, abstract algebra, and physics-but wiki just says it's a geometric object and in physics my professors all ways say it builds spaces out of other spaces. Can somebody clarify things?

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Try this video which i found helpful.… – neofoxmulder Jan 30 '14 at 17:57
"A tensor is something that transforms like a tesnor" --A. Zee – Dror Feb 6 '14 at 18:59
up vote 7 down vote accepted

A $k$-tensor is a multilinear function from $V\times V\times\dots\times V$ to the reals, where $V$ is a vector space and $k$ is the number of the $V$'s in the above Cartesian product. (Calculus on Manifolds, Michael Spivak, 1965, page 75).

This is the best definition I can find. I am with you, and thank you for asking the question, because I hate definitions that lack a noun. So and so is a __ (noun), please!

I don't know if a k-tensor is the most general type of tensor or not.

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Welcome to MSE! For me as well (all those years ago), Spivak's definition of a tensor was the first definition that made sense. Not sure why you got downvoted.... – Andrew D. Hwang Apr 5 '14 at 16:56
Thank you! As I am new, one thing I don't get is that there is another question on MSE almost identical (what is a tensor, really?) Can these be combined or linked easily? – user140651 Apr 5 '14 at 23:00
The site does try to avoid duplicate questions (by linking to answered versions and "closing" the newer duplicate), but it tends to happen more for sharply-focused or homework-type questions than for "philosophical" questions such as this one. As for how to link, clicking "help" (below the "add comment" button) pops up a micro-reference on Markdown. See also the site help center page. – Andrew D. Hwang Apr 5 '14 at 23:15

At the lowest level of understanding a tensor $T$ of rank $r$ is an $r$-dimensional array (think of a spreadsheet) whose "side-lengths" are all equal to a given $n\geq1$. Therefore $T$ has $n^r$ entries, which we assume to be real numbers in the following.

When we are setting up such a tensor we have some application in mind, say in geometry or physics. That's where the difficulties come in. The tensor is meant to be "applied" to one or several (variable) vectors, and the result will be a number or a vector of interest in the context at hand. E.g., the value $T(x,y)$ could be the scalar product of $x$ and $y$, or the area of the parallelogram spanned by $x$ and $y$, or the image of $x$ under $T$ when $T$ is considered as a linear map, or the retaliatory force felt when moving in direction $x$, and on and on. For the computation of actual values we need the coordinates of $x$ and $y$. Now these depend on the choice of basis in the ground space ${\mathbb R}^n$, and when we change the basis the coordinate values of the points $x$ change. But the scalar product or some induced force, being "well defined" geometric or physical quantities, should not change. This in turn implies that the entries in our tensor (spreadsheet) $T$ will have to change, albeit in a characteristic way, called "contravariant" or "covariant", depending on the case at hand.

(As an aside: When a certain Excel-spreadsheet is meant to be a price list for various fabrics, then its entries will change as well in a characteristic way when the currency or the units of measurement are changed.)

But we have a definite feeling that there is some hidden "robust identity" incorporated in $T$ that is independent of the more or less accidental values appearing in the spreadsheet. It is only in the second half of the last century that mathematics has found a universal (and abstract!) way to express and to deal with this "hidden identity" of $T$. The field of mathematics concerned with this is called multilinear algebra. Only in this realm it then makes sense to talk about the tensor product. But I won't go into this here.

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This was REALLY HELPFULL!!!!!! – John Aug 6 '14 at 5:54

The simplest case is a tensor product of two vector spaces. If $V$ is a vector space with basis $\{v_i\}$ and $W$ is a vector space with basis $\{w_i\}$ then $V \otimes W$ is a vector space with basis $\{v_i \otimes w_i\}$.

There is more theory behind it than that. I'm sure you've read stuff about it being universal with respect to bilinear maps and such. But in terms of "building spaces out of other spaces" it's not that complicated. A tensor product of an $n$-dimensional vector space with an $m$-dimensional vector space is just an $nm$-dimensional vector space.

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But what is that space filled with? The vectors of V and W, or some hyrbidization? – user82004 Jan 31 '14 at 0:09
Any pair of vectors $v \in V$ and $w \in W$ gives a vector $v \otimes w \in V \otimes W$, but unfortunately there are more vectors in $V \otimes W$ then just what you can get from such pairs. So there is no easy description of the elements of $V \otimes W$ in terms of elements of $V$ and elements of $W$. The elements of $V \otimes W$ are just linear combinations of the symbols $v_i \otimes w_j$. – Jim Jan 31 '14 at 0:21
How do you read that symbol between v and w? – user82004 Jan 31 '14 at 0:22
$V \otimes W$ is read as "V tensor W" – Jim Jan 31 '14 at 0:23
And building a tensor product of two vectors- does this just give a plane? Are you saying then that this plane can only be described via pairs of these vectors? – user82004 Jan 31 '14 at 0:23

Every tensor is associated with a linear map that produces a scalar.

For instance, a vector can be identified with a map that takes in another vector (in the presence of an inner product) and produces a scalar. If I have a vector $v$ and some input vector $a$, then I define the map $\underline v(a) \equiv v \cdot a$.

A matrix is just a representation of a map that takes in two vectors. Usually we say matrices take in vectors and produce vectors. $T(a) \mapsto a'$ for instance. But you can instead use the inner product and say there is a map $\underline T(a, b)$ which produces a scalar by $\underline T(a, b) = T(a) \cdot b$.

Tensors obey certain transformation laws. A change of basis for a matrix is a similarity transformation; for tensors the rule is just a generalization of this idea. This gives a way to compute the components of a tensor in a new basis, but the underlying map can be thought of as unchanging, the same way a vector expressed in a new basis is geometrically considered the same as it was before.

Some tensors correspond to geometric objects or primitives. As I said, vectors can be thought of as very simple tensors. Some other tensors correspond to planes, volumes, and so on, formed directly from 2, 3, or more vectors. Clifford algebra is a part of the tensor algebra, dealing directly with such geometrically significant objects. Not all tensors are so easily to visualize or imagine, though; the rest can only be thought of abstractly as maps.

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Think of a tensor as a matrix transforming one vector into another (or one-form into another one form). For example in linear algebra we learned that given a vector $\nu$ and the matrix $A$, we can get a new vector by multiplying the matrix by the vector $\nu$, i.e. $$ \nu\cdot A = \omega $$ In linear algebra $A$ is just a matrix, in tensor analysis $A$ would be a dyad tensor (tensor of order two), in linear algebra we have $A_{i,j}$ in tensor analysis we have $A_{ij}$. In other words a tensor is a matrix.

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