Solution: What is Decomposable Tensor?

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Discussion Overview

The discussion revolves around the concept of decomposable tensors within the context of vector spaces and tensor algebra. Participants are exploring the definitions and characteristics of decomposable tensors, particularly in relation to given examples of tensors with specific coordinate representations.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant suggests that a decomposable tensor is akin to a rank-one tensor, defined as a tensor of the form \(x \otimes y\) for some \(x \in V\) and \(y \in V^*\).
  • Another participant notes that \(t^{k}_{ij}\) is a rank three tensor and \(t^{ij}\) is a rank two tensor, expressing uncertainty about the terminology used in the discussion.
  • There is a mention of different interpretations of "rank" depending on the context, with one participant relating it to the rank of the associated matrix representation of the tensor.
  • A participant feels that the tensor \(t_{ij}^k = 2^{i+j+k^2}\) appears decomposable because it can be expressed as a product \(t_{ij}^k = 2^{i} \times 2^{j+k^2}\), but acknowledges uncertainty regarding the notation used by others.
  • Another participant questions whether all tensors, regardless of the number of indices, can be conceptualized as matrices.

Areas of Agreement / Disagreement

Participants express differing views on the definition and implications of decomposable tensors, particularly regarding the relationship between tensor rank and matrix representation. No consensus is reached on the definitions or the examples provided.

Contextual Notes

Participants highlight the variability in terminology and notation used in tensor algebra, which may lead to confusion. The discussion reflects differing interpretations of tensor rank and decomposability without resolving these differences.

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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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.)
 
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. :)
 
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.
 
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?
 

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