Tensors are mathematical objects that are needed in physics to define certain quantities. I have a couple of questions regarding them that need to be clarified:

  1. Are matrices and second rank tensors the same thing?

  2. If the answer to 1 is yes, then can we think of a 3rd rank tensor as an ordered set of numbers in 3D lattice (just in the same way as we can think of a matrix as an ordered set of numbers in 2D lattice)?


Matrices are often first introduced to students to represent linear transformations taking vectors from $\mathbb{R}^n$ and mapping them to vectors in $\mathbb{R}^m$. A given linear transformation may be represented by infinitely many different matrices depending on the basis vectors chosen for $\mathbb{R}^n$ and $\mathbb{R}^m$, and a well-defined transformation law allows one to rewrite the linear operation for each choice of basis vectors.

Second rank tensors are quite similar, but there is one important difference that comes up for applications in which non-Euclidean (non-flat) distance metrics are considered, such as general relativity. 2nd rank tensors may map not just $\mathbb{R}^n$ to $\mathbb{R}^m$, but may also map between the dual spaces of either $\mathbb{R}^n$ or $\mathbb{R}^m$. The transformation law for tensors is similar to the one first learned for linear operators, but allows for the added flexibility of allowing the tensor to switch between acting on dual spaces or not.

Note that for Euclidean distance metrics, the dual space and the original vector space are the same, so this distinction doesn't matter in that case.

Moreover, 2nd rank tensors can act not just as maps from one vector space to another. The operation of tensor "contraction" (a generalization of the dot product for vectors) allows 2nd rank tensors to act on other second rank tensors to produce a scalar. This contraction process is generalizable for higher dimensional tensors, allowing for contractions between tensors of varying ranks to produce products of varying ranks.

To echo another answer posted here, a 2nd rank tensor at any time can indeed be represented by a matrix, which simply means rows and columns of numbers on a page. What I'm trying to do is offer a distinction between matrices as they are first introduced to represent linear operators from vector spaces, and matrices that represent the slightly more flexible objects I've described

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    $\begingroup$ Is there a reference where this difference is discussed with examples? $\endgroup$ – Revo Feb 2 '12 at 20:50
  • $\begingroup$ Most introductory textbooks on general relativity offer great discussions of tensors and their relations to linear operators and dual spaces. One example would be Sean Carrol's book "Spacetime and Geometry," although different people have their own favorites $\endgroup$ – kleingordon Feb 2 '12 at 20:58
  • $\begingroup$ @Revo: The relation beteen tensors and matrices is explained in the entry ''How are matrices and tensors related?'' in Chapter B8: Quantum gravity of my theoretical physics FAQ at mat.univie.ac.at/~neum/physfaq/physics-faq.html $\endgroup$ – Arnold Neumaier May 14 '12 at 15:36
  • $\begingroup$ @kleingordon For future reference we have MathJax active on the site which allows you to write neatly marked up mathematical notation using LaTeX-alike markup. I've done this one for you. $\endgroup$ – dmckee --- ex-moderator kitten Jul 14 '12 at 18:37

A second-order tensor can be represented by a matrix, just as a first-order tensor can be represented by an array. But there is more to the tensor than just its arrangement of components; we also need to include how the array transforms upon a change of basis. So tensor is an n-dimensional array satisfying a particular transformation law.

So, yes, a third-order tensor can be represented as a 3-dimensional array of numbers -- in conjunction with an associated transformation law.

  • $\begingroup$ Nice answer. As a simple example, we could have a matrix that came up in economics, and was a linear map from a space of economic conditions to a space of economic outputs. There is no way this would be a tensor, because it wouldn't transform properly. $\endgroup$ – Ben Crowell Sep 23 '14 at 0:06
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    $\begingroup$ This is an old comment, and I am not an economist, but if it is a linear map, then it IS a tensor, if the spaces you mentioned are finite-dimensional. If we denote the vector space of economic conditions as $\mathbb{EC}$ and the space of economic outputs as $\mathbb{EO}$, then that tensor would be an element of the space $\mathbb{EO}\otimes\mathbb{EC}^{*}$, where the star denotes algebraic dualspace. $\endgroup$ – Bence Racskó Jan 13 '15 at 12:19

A matrix is a special case of a second rank tensor with 1 index up and 1 index down. It takes vectors to vectors, (by contracting the upper index of the vector with the lower index of the tensor), covectors to covectors (by contracting the lower index of the covector with the upper index of the tensor), and in general, it can take an m upper/n-lower tensor to either m-upper/n-lower by acting on one of the up indices, to m-upper/n-lower by acting on one of the lower indices, or to m-1-upper/n-1-lower by contracting with one upper and one lower indices.

There is no benefit to matrix notation if you know tensors, it's a special case where the operation of tensor product plus one contraction produces an object of the same type. The tensor notation generalizes the calculus of vectors and linear algebra properly to make the right mathematical objects.


Strictly speaking matrices and rank 2 tensors are not quite the same thing, but there is a close correspondence that works for most practical purposes that physicists encounter.

A matrix is a two dimensional array of numbers (or values from some field or ring). A 2-rank tensor is a linear map from two vector spaces, over some field such as the real numbers, to that field. If the vector spaces are finite dimensional then you can select a basis for each one and form a matrix of components. This correspondence between matrices and rank-2 tensors is one-to-one so you can regard them as the same thing, but strictly speaking they are just equivalent.

You can make up cases of infinite dimensional vector spaces where no meaningful representation in terms of matrices for the corresponding tensors is possible even when the field is the real numbers and the matrices can have an infinity of components. Some of these examples are relevant to physics, e.g. when the vector spaces are functionals whose dimension is (in lose terms) uncountably infinite. For this reason it is a good idea to bear in mind the distinction between what tensors and matrices of arrays actually are, even if you are just a physicist.

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    $\begingroup$ A matrix is a two dimensional array of numbers (or values from some field or ring). No, a matrix has more structure than that. $\endgroup$ – Ben Crowell Sep 23 '14 at 0:04
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    $\begingroup$ No it doesn't. You can add more structure and define all kinds of operations but a matrix is just a 2d array of numbers. $\endgroup$ – Philip Gibbs - inactive Sep 23 '14 at 7:06

This is a pet peeve of mine. Having in the early part of my career been a geometer. Much of the discussion before is correct. A tensor of various ranks are linear transformations. However, a tensor is an invariant under coordinate systems selected.

Easiest way to think of it is a vector is a magnitude and direction and only can be expressed as an array once a coordinate system is chosen. Likewise a rank 2 tensor can only be expressed as a matrix when a coordinate system is chosen.

That is why it is used in physics like the stress energy tensor or the refractive index tensor of anistropic crystals. It's this coordinate invariance that makes it useful for describing physical properties.


No. A matrix can mean any number of things, a list of numbers, symbols or a name of a movie. But it can never be a tensor. Matrices can only be used as certain representations of tensors, but as such, they obscure all the geometric properties of tensors which are simply multilinear functions on vectors.

  1. All scalars are not tensors, although all tensors of rank 0 are scalars (see below).
  2. All vectors are not tensors, although all tensors of rank 1 are vectors (see below).
  3. All matrices are not tensors, although all tensors of rank 2 are matrices.

Example for 3: Matrix M (m11=x , m12=-y , m21=x^2 , m22=-y^2) .This matrix is not tensor rank 2. Test matrix M to rotation matrix.

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    $\begingroup$ All scalars are indeed tensors(of rank 0), there is no flaw in it. So is true for vectors, or matrices. Tensor is much bigger class(or abstract object), which contains this. Scalars, vectors, or matrices are just representation in some basis. $\endgroup$ – L.K. Mar 21 '17 at 12:14
  • $\begingroup$ I said to example for it. It is for it. you search it and will see many vectors is not tensor rank1. or many scalers are not tensors. In up example you solve it and will see is not a tensor but it is a matrix. please see: An Introduction to Tensors for Students of Physics and Engineering by Kolecki. from NASA $\endgroup$ – Farshad Ashkbous Mar 31 '17 at 15:02
  • $\begingroup$ Nowhere it is written that vectors are not tensor rank $1$. I have checked it. Even the concept is wrong. Means see this your referred book, see page number 4. $\endgroup$ – L.K. Mar 31 '17 at 15:39
  • $\begingroup$ See p.7 in this ref...In elementary books about tensors say scalers are tensor rank0 and vector are rank1 but in advanced books and some articles they aren't. scalers is difference to tensors rank 0 and vector is difference to rank1 and matrix difference to rank2. If you say 3(no unit) then it isn't mean to tensor rasnk0 but it is scalar . Or spinor is not vector or scaler.In my example matrix : m11=x , m12=-y , m21=x^2 , m22=-y^2 is not tensor otherwise please prove it!!!In reverse is vector but is not tensor. or pseudovector is not tensor rank1 etc... $\endgroup$ – Farshad Ashkbous Mar 31 '17 at 17:09
  • $\begingroup$ But concept is wrong. You said but in advanced books and some articles they aren't.. Let me know what are those references. $\endgroup$ – L.K. Apr 2 '17 at 19:02

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