Are matrices rank 2 tensors?

Question

I know that this is sometimes the case, but that some matrices are not tensors. So what is the intuitive and specific demands of a matrix to also be a tensor? Does it need to be quadratic, singular or something else?

Some sources I read seem to suggest that all rank 2 matrices are tensors while other just claims that "some" matrices are rank 2 tensors.

What's the connection between tensors and matrices?

http://math.stackexchange.com/questions/2400/is-it-misleading-to-think-of-rank-2-tensors-as-matrices?rq=1 — HipsterMathematician, Sep 07 '12 at 15:11
Matrices (as such in itself) are never tensors. However, a matrix can represent a tensor once a set of coordinate systems (basis vectors) is fixed. (Before asking whether matrices are rank 2 tensors, first ask yourself whether you understand the difference between a rank 2 matrix, a linear operator on a vector space $V$, and a bilinear form on a vector space $V$.) — Willie Wong, Sep 07 '12 at 15:13
Matrices are $(1,1)$-tensors. A matrix $A = a^{i}{}{j}$ maps any vector (or $(0,1)$-tensor) $v = v^i$ to another vector $w = Av = a^{i}{}{j}v^{j}$. In other cases, bilinear forms $B = b_{ij}$ and some physical tensors, such as angular momentum tensor $I = I_{ij}$, are in fact $(0,2)$-tensors. They yield a number when two vectors are acted. We make little distinction in Euclidean space $E = \Bbb{R}^n$ via the identification $E^{\ast} = E$, but they behave differently under the coordinate transform. MeAndMath's link deals with a more detailed discussion. — Sangchul Lee, Sep 07 '12 at 15:17
@WillieWong: I'd say that matrices are always tensors; they can be canonically identified with the members of $K^n\otimes K^n$. (One might argue about whether one or both of the factors should be duals, but $K^n$ is canonically self-dual, so that doesn't matter). If you want them to represent things about other vector spaces than $K^n$, then you need to choose a basis, or equivalently a particular isomorphism with $K^n$. — hmakholm left over Monica, Sep 07 '12 at 16:32
@Henning: mathematically, fair enough. But if the OP is thinking from a physics background, then your explanation is not satisfactory compared to the historic genesis of ¨tensors¨ in that field as ¨higher dimensional matrices with associated ´change of variables´ formulae¨ (the $K^n\to K^n$ automorphisms are built-in). — Willie Wong, Sep 10 '12 at 08:53

score 26 · Accepted Answer · answered Sep 07 '12 at 17:09

This question doesn't have a single good answer, because there isn't a universally agreed upon definition of "tensor" in mathematics. In particular:

Tensors are sometimes defined as multidimensional arrays, in the same way that a matrix is a two-dimensional array. From this point of view, a matrix is certainly a special case of a tensor.
In differential geometry and physics, "tensor" refers to a certain kind of object that can be described at a point on a manifold (though the word "tensor" is often used to refer to a tensor field, in which one tensor is chosen for every point). From this point of view, a matrix can be used to describe a rank-two tensor in local coordinates, but a rank-two tensor is not itself a matrix.
In linear algebra, "tensor" sometimes refers to an element of a tensor product, and sometimes refers to a certain kind of multilinear map. Again, neither of these is a generalization of "matrix", though you can get a matrix from a rank-two tensor if you choose a basis for your vector space.

You run into the same problem if you ask a question like "Is a vector just a tuple of numbers?" Sometimes a vector is defined as a tuple of numbers, in which case the answer is yes. However, in differential geometry and physics, the word "vector" refers to an element of the tangent space to a manifold, while in linear algebra, a "vector" may be any element of a vector space.

On a basic level, the statement "a vector is a rank 1 tensor, and a matrix is a rank 2 tensor" is roughly correct. This is certainly the simplest way of thinking about tensors, and is reflected in the Einstein notation. However, it is important to appreciate the subtleties of this identification, and to realize that "tensor" often means something slightly different and more abstract than a multidimensional array.

I suspected this was the problem. I am currently taking a Quantum Field Theory course and I am having trouble figuring out what the connections between the Tensors and matrix representations of them are.
My professor defines tensors as anything that transforms covariantly (or contra variantly) with eg. Lorentz transformations, but at the same time says that not all matrices fulfill this? Given this definition of a tensor what does it take for a matrix to be a tensor? Is it enough for it to be a square matrix? — HansHarhoff, Sep 08 '12 at 13:20
One definition of a tensor is ¨matrix + transformation laws¨. Note that the ¨transformation law¨ is not built in to the definition of a matrix (which is just a bunch of numbers arranged in a particular fashion). You have to specify the transformation law. Think about the matrix $\begin{pmatrix} 1 & 0 \ 0 & 1\end{pmatrix}$. What is it under the change of variable $u = y$, $v = x+y$? You can´t say, because I haven´t told you what the matrix represents! If treated as a linear operator, in the new basis it is still the identity; if treated as a bilinear form, in the new basis it won´t be. — Willie Wong, Sep 10 '12 at 08:59

score 6 · Answer 2 · answered Sep 07 '12 at 15:35

The connection is this: a matrix consists of the coefficients of a (1,1) tensor, but it is not a tensor itself.

Suppose we are talking about a linear transformation $T$ on an $n$ dimensional vector space $V$.

Now $T$ is certainly a tensor (tensors are, after all, multilinear maps on copies of $V$ and $V^\ast$, and a linear transformation can be interpreted as a multilinar function from $V\times V^\ast$ to $\mathbb{F}$.)

Once a basis for $V$ is fixed, then you can talk about the matrix $A$ for $T$ which is written in terms of the basis. The same can be said for general multilinear functions on copies of $V$ and $V^\ast$, that after you have fixed a basis, you have a big array holding its coefficients.

It's important to remember not to confuse the array for the tensor. The tensor is a basis independent entity: it's a kind of function. The components are just one particular representation of that function, and the components depend upon a choice of basis.

Sometimes, however, a matrix consists of the coefficients of a (0,2) -- or even (2,0) -- tensor. And there is also an argument to be made that a matrix is itself a tensor, more precisely an element of the particular tensor product $(K^n)^\otimes K^n$ where $K$ is the scalar field. (Then one can use $(K^n)^\otimes K^n$ to represent other tensor products by choosing appropriate isomorphisms to/from $K^n$). — hmakholm left over Monica, Sep 07 '12 at 16:19

Farshad Ashkbous · Answer 3 · 2017-04-05T19:00:10.800

All matrices are not tensors, although all tensors of rank 2 are matrices.

Example: Matrix T (T11=x , T12=-y , T21=x^2 , T22=-y^2) .This matrix is not tensor rank 2. We test matrix T to rotation matrix A: a11=cos(theta),a12=sin(theta),a21=-sin(theta),a22=cos(theta). Now Expand tensor equation rank2,for example T'11=Σ(a1ia1jTij) (1)

Now calculate T'11=x'=xcos(theta)+ysin(theta) (2)

You see (1) unequal to (2) then conclude Matrix T isn't a tensor rank2!!

Tensor must follow the conversion(transformation) rules, but matrices generally are not.

Are matrices rank 2 tensors?

3 Answers3