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Decomposable Tensors

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
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) \[t^{ij}=i+j\]
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,718
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.)
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
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. :)
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,718
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.
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
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?