How determine that a matrix is a tensor?

In summary, the conversation is about proving that a given matrix is a tensor. The context is rotations within the xy plane and the corresponding coordinate transformation. The issue of whether the matrix is a tensor or not is discussed, with the conclusion that it depends on the context and the type of tensor being considered. The conversation also touches on the use of partial derivatives and covariant derivatives, as well as the importance of specifying the context when dealing with tensors. Finally, there is a request for the person to show their work and any assumptions made in order to provide guidance.
  • #1
GingFritz
3
0
Homework Statement
Hi! I need help with my homework.
I need to prove that the matrix below is a tensor.
Relevant Equations
## T'_{ij} = T_{rs}\frac{\partial x^r}{\partial x'^{i}}\frac{\partial x^s}{\partial x'^{j}} ##
I have the matrix

$$
A = \left(\begin{array}{cc}

y^2 & -xy\\

-xy & x^2

\end{array} \right)

$$

I know that to prove that the matrix is a tensor, it transform their elements in another base. But I still without how begin this problem.

Help please! Thanks.
 
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  • #2
You must specify the particular group under whose action this matrix transforms tensorially (not least to evaluate the Jacobian entries).

I guess it’s rotations within the plane. What’s the corresponding coordinate transformation?
 
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  • #3
I guess the notation is that ##x^1=x## and ##x^2=y##. Then you can build ##^{\dagger}x_k=\epsilon_{kl} x^l## with the Levi-Civita symbol ##\epsilon_{kl}##. Then investigate ##^{\dagger} x_k ^{\dagger}x_l## concerning the question, for which transformations it behaves like the covariant components of a tensor.
 
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  • #4
I find this problem strange. A matrix may be a tensor or may not be a tensor. It depends on a genesis of the matrix. For example if ##f(x)## is a function then a matrix $$\frac{\partial^2 f}{\partial x^i\partial x^j}$$ is not a tensor. But a matrix
$$\frac{\partial f}{\partial x^i}\frac{\partial f}{\partial x^j}$$ is a tensor.
If you are given with a matrix in only one coordinate frame and that is all then you can not know whether it is a tensor or not.Moreover there are 3 different types of tensors that a matrix can present
$$(0,2),\quad (2,0),\quad (1,1)$$
 
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  • #5
If ##f## is a scalar field, then ##\partial_i \partial_j f## are covariant components of a 2nd-rank tensor field under arbitrary basis transformations:
$$\partial_i' \partial_j' f=\frac{\partial x^k}{\partial x^{\prime i}} \frac{\partial x^l}{\partial x^{\prime j}} \partial_k \partial_l f,$$
and this is as covariant 2nd-rank tensor components transform.

It's of course not providing tensor components under general transformations (diffeomorphisms) of a differentiable manifold. For this you need a connection defining a covariant derivative. That's why it's important to tell the context you look at, as already stated in #2.
 
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  • #6
vanhees71 said:
If ##f## is a scalar field, then ##\partial_i \partial_j f## are covariant components of a 2nd-rank tensor field under arbitrary basis transformations:
$$\partial_i' \partial_j' f=\frac{\partial x^k}{\partial x^{\prime i}} \frac{\partial x^l}{\partial x^{\prime j}} \partial_k \partial_l f.$$
nope. Do not confuse a partial derivative ##\frac{\partial^2 f}{\partial x^i\partial x^j}## with a covariant derivative ##\nabla_i\nabla_j f##. In the last case you need to have a connection
 
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  • #7
As I wrote, it depends on the context. If you are on an affine space and you consider only basis transformations, the partial derivatives give tensor components. If you refer to a differentiable manifold and look at general diffeomorphisms you need a connection defining a covariant derivative.
 
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  • #8
vanhees71 said:
If you are on an affine space and you consider only basis transformations, the partial derivatives give tensor components. If you refer to a differentiable manifold
an affine space is a differentiable manifold as well:) I have a habit to consider general case until anything else has not been specified explicitly
 
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  • #9
wrobel said:
nope. Do not confuse a partial derivative ##\frac{\partial^2 f}{\partial x^i\partial x^j}## with a covariant derivative ##\nabla_i\nabla_j f##. In the last case you need to have a connection
Sorry to nitpick, but aren't you assuming the use of a trivial connection( identity) when using " standard" partial derivatives?
 
  • #10
Hi @GingFritz.

The rules here require you to show some attempt before we offer guidance, so I’m bending the rules a bit.

As already noted by @ergospherical in Post #2, assume that the intended problem is to show that matrix A behaves as a tensor under rotation in the xy (Cartesian) plane about the origin.

If this is written work to be handed-in, state the assumption.

You didn’t answer the question in Post #2!

Find the following (just elementary trig’ and calculus):
##x’## = some expression with ##x, y## and ## θ##
##\frac{\partial x’}{\partial x}## and ##\frac{\partial x’}{\partial y}##

Repeat for ##y’##.

You might prefer to use indexed (contravariant preferred) notation (##x^1, x^2, x’^1## and ##x’^2##) rather than ##x, y, x’## and ##y’##. But I've used the notation in the question for the moment.

Show us your work and maybe we can guide you from there.
 
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  • #11
Steve4Physics said:
Hi @GingFritz.

The rules here require you to show some attempt before we offer guidance, so I’m bending the rules a bit.

As already noted by @ergospherical in Post #2, assume that the intended problem is to show that matrix A behaves as a tensor under rotation in the xy (Cartesian) plane about the origin.

If this is written work to be handed-in, state the assumption.

You didn’t answer the question in Post #2!

Find the following (just elementary trig’ and calculus):
##x’## = some expression with ##x, y## and ## θ##
##\frac{\partial x’}{\partial x}## and ##\frac{\partial x’}{\partial y}##

Repeat for ##y’##.

You might prefer to use indexed (contravariant preferred) notation (##x^1, x^2, x’^1## and ##x’^2##) rather than ##x, y, x’## and ##y’##. But I've used the notation in the question for the moment.

Show us your work and maybe we can guide you from there.
I think the same. I will perform a solution with a rotation through the z axis and share it here. Thanks!
 
  • #12
WWGD said:
Sorry to nitpick, but aren't you assuming the use of a trivial connection( identity) when using " standard" partial derivatives?
Actually not.
There is no a connection such that the equality
$$\nabla_i\nabla_j f=\frac{\partial^2 f(x)}{\partial x^i\partial x^j}$$
holds in each coordinate frame
 
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  • #13
wrobel said:
Actually not.
There is no a connection such that the equality
$$\nabla_i\nabla_j f=\frac{\partial^2 f(x)}{\partial x^i\partial x^j}$$
holds in each coordinate frame
Aren't we working in Euclidean space with Cartesian coordinates? In that case, Christopher symbols are trivial. But I guess I'm missing something.
 
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  • #14
WWGD said:
Aren't we working in Euclidean space with Cartesian coordinates? In that case, Christopher symbols are trivial.
We can calculate ##\partial^2f/(\partial x^i\partial x^j)## in any coordinate frame. If your idea is to declare any coordinate frame to be a Cartesian frame with trivial Christoffel symbols then you define not a connection but an infinite set of connections: each coordinate frame has its own connection.
That is why ##\partial^2f/(\partial x^i\partial x^j)## is not a tensor
 
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  • #15
GingFritz said:
...
But I still without how begin this problem.
...
Best if you write the problem exactly as given with all the details, without omitting what you thought was not important.
 
  • #16
wrobel said:
We can calculate ##\partial^2f/(\partial x^i\partial x^j)## in any coordinate frame. If your idea is to declare any coordinate frame to be a Cartesian frame with trivial Christoffel symbols then you define not a connection but an infinite set of connections: each coordinate frame has its own connection.
That is why ##\partial^2f/(\partial x^i\partial x^j)## is not a tensor
No that's for sure not the intention. The point is that the problem was not stated correctly, and we shouldn't have made any attempt to answer, because there is no correct answer to an ill-posed problem.
 
  • #17
martinbn said:
Best if you write the problem exactly as given with all the details, without omitting what you thought was not important.
I wrote the problem exactly as in the worksheet given.
 
  • #18
GingFritz said:
I wrote the problem exactly as in the worksheet given.
It remains that the problem, as stated without context, is ill-defined. It may have a single interpretation within the limitations and confines of your course, but as should be painfully clear from the discussion here, the general case is more involved. In order to appropriately answer your question, we need to know what definition your course uses for a tensor (hoping against hope that it is not the all too common ”a tensor transforms as a tensor”) as well as what type of spaces you are considering (ie, Euclidean space with Cartesian coordinates, general coordinates, or differentiable manifolds).
 
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  • #19
To add to that, my interpretation assuming a restriction to Cartesian coordinates on the Euclidean plane would be ”to show that the object with components given by ##(f_{ij}(x,y)) = \ldots## in any Cartesian coordinate system ##(x,y)## transforms as the components of a rank 2 tensor under rotations”. I suspect this is the actual question but we cannot be sure without confirmation from OP. Post #3 by @vanhees71 covers this case adequately.
 
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  • #20
GingFritz said:
I wrote the problem exactly as in the worksheet given.
What textbook are you using?
 

FAQ: How determine that a matrix is a tensor?

What is a tensor?

A tensor is a mathematical object that describes the relationships between vectors and other tensors. It is represented as a multi-dimensional array of numbers, and its properties can be used to model physical phenomena in fields such as physics, engineering, and computer science.

How is a matrix different from a tensor?

A matrix is a special type of tensor that has two dimensions (rows and columns), while a tensor can have any number of dimensions. A matrix can be thought of as a specific case of a tensor, where the number of dimensions is limited to two.

How can I determine if a matrix is a tensor?

A matrix is a tensor if it follows the rules of tensor algebra and can be used to model relationships between vectors. These rules include properties such as transformation under coordinate changes, addition and multiplication, and contraction. If a matrix satisfies these properties, it can be considered a tensor.

Can a tensor be represented as a matrix?

Yes, a tensor can be represented as a matrix if it has two dimensions. However, not all tensors can be represented as matrices, as they can have any number of dimensions. In these cases, other mathematical representations such as multi-dimensional arrays or higher-order tensors may be used.

What are some real-world applications of tensors?

Tensors have numerous applications in fields such as physics, engineering, computer science, and data analysis. Some examples include modeling stress and strain in materials, analyzing image and signal data, and performing operations in deep learning algorithms. Tensors are also used in general relativity to describe the curvature of spacetime.

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