Tensors have geometric meaning just like vectors (just like vectors are not just single-axis arrays). It would be too long to delve into the geometry (especially if it is abstract "geometry" like the geometrical space spanned by the songs in the Pandora database or the space spanned by the faces in a face-recognition software). Instead I will try to give an example:

You all know what rotating a vector means. All the components change in certain way, but the geometrical meaning of this object is preserved in some sense. This is not true for some arbitrary list of numbers. Well, a tensor is just a more complicated geometrical structure that still has very specific rules governing what its components can be and how they can change.

And a counterexample: the Christoffel Symbol is a multi-dimensional array that is not a tensor.

P.S. It is true that it is a very common abuse of nomenclature to call all multidimensional arrays "tensors".

To me, a tensor is a function that maps a collection of m vectors to a collection of n vectors, such that every output is linear in each input. It's true that if you choose a basis, then you can write down an array of numbers which identifies the function, and that changing the basis causes those numbers to change in a certain way, but that's not what a tensor "is".

Of course, the beauty of mathematics is that there are many different ways of looking at the same object, so physicists can deal with their arrays of numbers, I can play with my multilinear functions, and everyone gets the same answer when we ask the same question.

It is not totally without merit if you consider the tensor bundle as associated to the principal bundle of linear frames. However, that's hardly suitable as an introduction to the topic.

I'm also confused: do you think about a tensor as a vector from vector space? as a object that maps between vector spaces? or something completely different?

I realize there's a LOT of detail here - just trying to flesh out my 500,000,000 mile view a little more.

One weird consequence of this is that if A and B are vectors, the cross product (AxB) is NOT a vector in the tensorial sense. [Aside: Cross products are really weird since they are defined in the familiar way 3 and 7 dimensions only! (see http://math.stackexchange.com/questions/185991/is-the-vector...]

But usually when people say that a vector "is" a list of numbers, they mean that if you choose a coordinate system, then the coordinates of a vector uniquely identify it. This is exactly analogous to locations on the earth being uniquely identified by their latitude and longitude: we first have to make a (somewhat arbitrary) choice of coordinates. You wouldn't say that a point on the earth "is" a pair of numbers, though. The case of R^n is confusing, because if you choose the usual coordinate system, then the coordinates of a vector are the same list of numbers as the vector itself.

I like to think of tensors as functions that map some number of vectors to some other number of vectors. But you can add together two tensors (with the same number of inputs and outputs) or multiply a tensor by a number, so you can also think of them as vectors in some other vector space.

This is similar with tensors. It's even harder to talk about them, though, without the geometric/algebraic bits.

They also disagree about what "vector" and "matrix" mean for the same reason, but they can't really put that horse back in the barn. But "tensors" remain unfamiliar enough that this disagreement can come up whenever anyone says "tensor".

Here's a possible truce:

- Programmers should default to talking about "arrays"

- Geometers and physicists should default to talking about "vectors", "matrices", and "tensors"

- We all generally recognize that some (vectors|matrices|tensors) are arrays, and some arrays are (vectors|matrices|tensors), and it's not the worst thing ever if someone uses one to mean the other

- We all generally recognize that you can sometimes do damn cool things by using an array as a tensor or a tensor as an array, so there isn't that much benefit in shouting "NO THEY'RE TOTALLY DIFFERENT".

No mathematician will bat an eye if you say that a tensor can be represented by an nd-array---but if you equate them, then you're clearly missing something.

edit: that was actually a serious question, because i really don't understand this idea that programmers have a common notion of what a matrix is.

Finally, of course, a matrix is a special case of a tensor. Tensor is such a broader topic that it becomes more and more difficult to talk about it without its geometric underpinnings. This is what I was writing about above.

[0] Is a black-and-white image a matrix? I'd imagine a lot of people would have a bit of a tough time saying yes directly---and for good reason. But there's also a reasonable argument for pretending like it is, and even reasonable arguments for turning it into one!

[1] Numpy distinguishes between nd-arrays and matrices exactly right in that matrices are 2d-arrays only and are imbued with multiplication as composition of maps. In practice, many people I know who are Numpy users just think of matrix as a convenient way to overload *, though.

std::vector was an exceptionally poor choice of name...

is it so hard to just call it pseudotensor like it is? its confusing for people trying to research the subject when they come across all kinds of geometry and calculus stuff which is irrelevant to the usage here.

I think this is really a flimsy use of a technical term which does have a precise meaning in the context of mathematical physics to make the idea of multi-dimensional array seem "deeper" than it really is.

There are enough hard problems in data science - I don't think it needs to burdened with terms like this.

There's a new textbook on classical physics by Kip Thorne and Roger Blandford. It hasn't been published yet, but the lecture notes on which it's based are online here:

http://www.pmaweb.caltech.edu/Courses/ph136/yr2012/

The first chapter presents a pretty simple introduction to tensors.

The other reference I've found helpful is Chapter 31 of Volume II of the Feynman Lectures on Physics. Feynman (as usual) gives a great introduction to the concept.

To echo krastanov, it is the case that all tensors can be represented as multi-dimensional arrays, but not all multi-dimensional arrays are tensors. At least in physics, a tensor is ultimately a thing, or at least some description of a physical thing. It therefore has some existence independent of whatever coordinate system you use and therefore, if you change your coordinate system, the values of the tensor have to change in certain ways. This means that your tensor cannot, in general be some arbitrary multidimensional array.

There's a bit of confusion about nomenclature since in fields outside physics and math a tensor doesn't need to represent some physical thing so the special transformation properties of a tensor are not so important and it's often just used as a shorthand term for "multidimensional array."

Heh, it's not exactly new, and I'll believe it's been published when I have it in my hands. That book had already been in development for years—and was due out any day now—back when I used it in Physics 136 (Blandford himself taught one of the terms). That was 1997.

The material is great, though, including the coverage of tensors, which served me well the next year in General Relativity (Thorne's last time teaching it). If you're looking for a solid intro to tensor algebra & analysis, I definitely recommend it.

    f(x, y) = x' A y

    f(x, y, z, w, q)

Why isn't any old multidimensional array a tensor? Because we want to see them as arising not as just an arbitrary collection of numbers but instead as representatives of these kinds of multilinear mappings. In particular, we want to know that tensors are invariant under changes of bases. This is similar to how we like to think of matrices as representing "some" linear mapping, but there might be many changes of bases in the source and target vector spaces which change the actual numbers inside without changing the ultimate mapping itself.

Ultimately, to get the fullest understanding of this we would need to talk about vector dual spaces and note that expressions like f(x, y, z, w, q) could involve two "types" of vector spaces, covariant and contravariant, which have different kinds of transformation-under-basis-change properties. To truly be a tensor your multi-dimensional matrix must understand and respect all of these kinds of invariances.

Matrices are 2d arrays, and matrix decomposition uses methods inspired by geometry, to analyze data in matrix form. Even though the matrix in question might have no intrinsic geometric meaning (e.g. a matrix of which individual likes which beer), the geometric methods still give useful result.

Similarly, tensor decompositions are inspired by geometry, but are also a simple way to get information from a 3d array. For tensors, the most basic tensor decomposition, where a tensor a_ijk is approximated by sum_l b_li * c_lj * d_lk.

tl;dr don't read too much into the geometric side of tensors. Data science applications don't have a geometric interpretation, but the techniques can still work.

Formally what you need to understand is the "tensor product": Given two vector spaces V,W over some field k (think n-tuples of real numbers), their tensor product is characterized by an universal property: Any bilinear map from the Cartesian product V x W to some other vector space U induces a unique linear map from the tensor product V o W to U. You can then proceed to show that the tensor product of vector spaces is associative (V o W) o U == V o (W o U) and symmetric V o W == W o V and the field k is its unit k o V == V == V o k. Interestingly if you are now given two k-linear maps f : V -> V' and g: W -> W' then you can get a map f o g : V o W -> V' o W'. And in the string diagram notation, you write that as

   V |  W |
     |    |
     f    g
     |    | 
   V'|  W'|

     V1 | ... | Vn
        | ... |
       ~~~~~~~~~
       |   a   | 
       ~~~~~~~~~
      W1| ... | Wm

Differential geometry only enters the picture if you want to study "vector space information", for example the vector space of all possible directions at a point also known as the tangent bundle, attached to points of some geometric space. The simplest manifestation of that concept are vector bundles, since forming the tensor product is an example of a "smooth functor" you can also define a tensor product of vector bundles. What physicists call tensors are sections, that is a "smooth" assignment of a vector in that tensor product at every point, of such vector bundles (typically the tangent bundle/cotangent bundle or some vector bundle associated to a principal bundle of "symmetries")

f(x+dx) = f(x) + f'(x)dx + O(dx^2)

This should look familiar: taking a = f(x) and b = f'(x), this is just the line a + b.dx! In other words, Calculus is just a way transforming questions about (differentiable) functions into questions about lines.

So this is all well and good, but what if x is a vector? Or f(x) is a vector? Or both? Well, now we can approximate

f(x+dx) = f(x) + Df(x).dx + O(|dx|^2)

Here, Df(x) is nothing other than the matrix [(df_i/dx_j)(x)]. In other words, we still get a linear approximation. But now suppose we get greedy and want to take higher order derivatives: what is the derivative of Df(x)? It's a tensor!

There are a few ways to think about this (which gives rise to the different interpretations of a tensor):

1. It's just df_i/(dx_jdx_k). This is a multi-dimensional array.

2. In 1D, we have

f(x+dx) = f(x) + f'(x)dx + (1/2)f''(x)dx^2 + O(dx^3)

In higher dimensions, we get

f(x+dx) = f(x) + Df(x)*dx + dx'.D^2f(x).dx + O(|dx|^3)

Here, D^2f(x) is a multi-linear map: it takes in two copies of the vector dx, and returns a new vector.

More generalizations exist for various reasons, but I think this intuition captures why tensors show up so often.

https://news.ycombinator.com/item?id=8651645