Solution: What is Decomposable Tensor?

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In summary: I understand this completely. So is it that we can think of any tensor as a matrix? I mean even if the there are three or more indices?No, you can't think of any tensor as a matrix, because the number of indices in a tensor is always the same.
  • #1
Sudharaka
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Hi everyone, :)

Here's a question I am trying to solve at the moment. I want to know what is meant by decomposable in this context. Really appreciate any input. :)

Problem:

Let \(V\) be a vector space over a field \(F\), \(\{e_1,\,e_2,\,\cdots,\,e_n\}\) a basis in \(V\) and \(\{e^1,\,\cdots,\,e^n\}\) adjoint basis in \(V^*\). Which of the following tensors, given by their coordinates are decomposable:

a) \[t^k_{ij}=2^{i+j+k^2}\]

b) \[t^{ij}=i+j\]
 
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  • #2
I don't know for sure, but I'm guessing that a decomposable tensor would be what I would call a rank-one tensor, in other words one of the form $x\otimes y$ for some $x\in V$, $y\in V^*$. (One of those two given tensors has rank 1, the other one does not.)
 
  • #3
Opalg said:
I don't know for sure, but I'm guessing that a decomposable tensor would be what I would call a rank-one tensor, in other words one of the form $x\otimes y$ for some $x\in V$, $y\in V^*$. (One of those two given tensors has rank 1, the other one does not.)

Thanks for the reply, but isn't \(t^{k}_{ij}\) a rank three tensor and \(t^{ij}\) a rank two tensor. Sorry, but I am new to tensors and I always view the rank as the number of indices. :)
 
  • #4
Sudharaka said:
Thanks for the reply, but isn't \(t^{k}_{ij}\) a rank three tensor and \(t^{ij}\) a rank two tensor. Sorry, but I am new to tensors and I always view the rank as the number of indices. :)
One of the difficulties with tensor products is that they crop up in very different contexts and people use different notations and terminology. I think of a tensor $t^{ij}$ as being represented by a matrix. When I call it a rank-one tensor I am thinking of the rank of the associated matrix, which is obviously different from your usage of the word "rank" here.

The reason that I instinctively feel that your tensor $t_{ij}^k = 2^{i+j+k^2}$ is decomposable is that it "decomposes" as a product $t_{ij}^k = 2^{i}\times2^{j+k^2}$. But since I am unsure about the notation you are using, I won't attempt to say more.
 
  • #5
Opalg said:
One of the difficulties with tensor products is that they crop up in very different contexts and people use different notations and terminology. I think of a tensor $t^{ij}$ as being represented by a matrix. When I call it a rank-one tensor I am thinking of the rank of the associated matrix, which is obviously different from your usage of the word "rank" here.

The reason that I instinctively feel that your tensor $t_{ij}^k = 2^{i+j+k^2}$ is decomposable is that it "decomposes" as a product $t_{ij}^k = 2^{i}\times2^{j+k^2}$. But since I am unsure about the notation you are using, I won't attempt to say more.

I understand this completely. So is it that we can think of any tensor as a matrix? I mean even if the there are three or more indices?
 

1. What is a decomposable tensor?

A decomposable tensor is a type of tensor in mathematics and physics that can be represented as a combination of simpler tensors. It is also known as a separable tensor because it can be separated into simpler components.

2. How is a decomposable tensor different from other types of tensors?

Unlike other types of tensors, a decomposable tensor can be broken down into simpler components, making it easier to work with in mathematical calculations and physical applications. This property also allows for the decomposition of complex systems into simpler parts.

3. What are the applications of decomposable tensors?

Decomposable tensors have various applications in mathematics, physics, and engineering. They are commonly used in the study of geometric algebra, differential geometry, and quantum mechanics. They also have practical applications in signal processing, image recognition, and machine learning.

4. How are decomposable tensors calculated?

The calculation of decomposable tensors involves breaking down the tensor into simpler components and then using mathematical operations such as multiplication, addition, and contraction to manipulate these components. The final result is a combination of the simpler tensors, representing the original decomposable tensor.

5. Can decomposable tensors be used in real-world problems?

Yes, decomposable tensors have many applications in real-world problems. They are commonly used in fields such as physics, engineering, and computer science to model and solve complex systems. They can also be used to analyze and manipulate data in various industries, making them a valuable tool in problem-solving.

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