8
$\begingroup$

I've recently begun reading about tensors and am trying to understand the second order variety in the context of euclidean $\mathbb{R}^n$ with orthonormal basis {$e_1, e_2,\ldots, e_n$}. This seems like a simple starting point that leaves topics like covariance and contravariance for another day. My question is w.r.t. this limited context.

Concerning the relationship between n$\times n$ matrices and second order tensors, I've read that "not every matrix is a tensor" and I'm trying to find a concise statement of which n$\times n$ matrices are second order tensors. Is it true that every n$\times n$ matrix $M$ is a tensor if and only if:

$$M_{ij}=S_{ij}e_i\otimes e_j$$ and $S_{ij} \in \mathbb{R}$

If "yes", is it true that there aren't any other constraints on $S$ in order for $M$ to be considered a tensor?

Improve this question
$\endgroup$
  • 3
    $\begingroup$ I can't make any sense of that statement. Every square matrix is a $(1, 1)$-tensor by definition. What's the source? $\endgroup$ – Qiaochu Yuan Apr 17 '14 at 4:55
  • $\begingroup$ @QiaochuYuan I think they mean "elementary tensor", i.e. one of the form $v\otimes w$ for vectors $v,w$ (and abusing notation by identifying $w$ with $w^*$). $\endgroup$ – Alex Becker Apr 17 '14 at 5:09
  • $\begingroup$ As in "All that is gold does not glitter, Not all those who wander are lost;" it means that things are not always as they appear. $\endgroup$ – Will Jagy Apr 17 '14 at 6:21
4
$\begingroup$

In physics, the terms "vector" and more generally "tensor", refers to an object that transforms like a vector/tensor.

First of all, a matrix is just a representation of a tensor in a certain coordinate frame, the tensor itself is an abstract notion, independent of notation and choice of coordinates. Secondly, a tensor in physics is an object, that, when expressed in another coordinate frame, takes the form

$$T'_{ij}=P_{ik}P_{jl}T_{kl}$$ if matrix $P$ takes the original frame to the primed (') frame.

In this sense general nonsquare matrices aren't tensors, nor are the matrices that represent just some sets of linear equations instead of objects that are bound to geometry of the physical space.

Tensors are very commonly symmetric (Hermitian).

Moreover, we are talking here about 2nd rank tensors. A tensor can be of any dimension (e.g. tensor of 1st rank is a vector). For instance, force is a vector, but a set of three variables is not, in the context of physics, called a vector. Even magnetic field is not strictly speaking a vector, when you go to relativistic space-time representation because it doesn't transform as one (plus, its parity is wrong under coordinate transformations that invert the space, so even classically, it's called an pseudo vector).

This is a physicists' interpretation of the question what is a tensor and what is not.

If there are any questions, I can elaborate more.

Improve this answer
$\endgroup$
  • 2
    $\begingroup$ In Susskind's "Einstein's General Theory of Relativity, Lecture 4", he gives a very simple concrete example of a matrix that's not a vector: If $v_1$, $v_2$, and $v_3$ are respectively the temperature, pressure, and relative humidity at a point in space, that's just a collection of three scalars and not a vector. His example helped me make sense of your answer, so I thought I'd note it here. $\endgroup$ – Bezewy Apr 28 '14 at 18:39
3
$\begingroup$

I think you have the relationship between matrices and tensors wrong. The tensor $e_i\otimes e_j$ corresponds directly to a matrix (rather than to an entry of a matrix), specifically the matrix $e_ie_j^T$. I am also assuming that by tensor you mean elementary tensor product, i.e. a tensor product of the form $v\otimes w$ for some $v,w\in\mathbb R^n$.

In general, $v\otimes w$ corresponds to the matrix $vw^T$. It is easy to see that this matrix has rank $1$. In fact, the converse is also true: if $M$ has rank $1$ then $M=vw^T$ for some vectors $v,w\in\mathbb R^n$. To see this, let $v$ be a $\mathrm{im}(M)$ and $w$ be a basis for $\ker(M)^\perp$. Then $vw^T=M$ up to a constant factor.

Improve this answer
$\endgroup$
  • $\begingroup$ Yes, I saw "$\mathbf S=S_{ij}e_i\otimes e_j$" on the web, misunderstood what it was saying, and then munged it into something nonsensical in my question. But I think your answer has me back on track. $\mathbf S$ is the tensor and $S$ is a matrix with entries $S_{ij}\in\mathbb{R}$? So any n$\times$n matrix $S$ can be considered a representation of a second order tensor $S_{ij}e_i\otimes e_j$? $\endgroup$ – Bezewy Apr 17 '14 at 7:25
  • $\begingroup$ @Bezewy You need a sum in there. $S$ can be represented as $\sum_{ij} S_{ij}e_i\otimes e_j$. $\endgroup$ – Alex Becker Apr 17 '14 at 8:04
  • $\begingroup$ Since $i$ and $j$ appear twice, I thought they'd be dummy indices, but I guess those rules only apply to terms that are multiplied scalars and not to terms that take tensor products of vectors (or anything else). If $e_i\otimes e_j = e_{ij}$ then the equation would be $\mathbf S=S_{ij}e_{ij}$, with implicit summation? $\endgroup$ – Bezewy Apr 17 '14 at 9:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.