# (2,0) tensor is not a tensor product of two vectors?

• I
Hi. I'm trying to understand tensors and I've come across this problem:

"Show that, in general, a (2, 0) tensor can't be written as a tensor product of two vectors".

Well, prior to that sentence, I would have thought it could... Why not?

## Answers and Replies

wrobel
let a (2,0) tensor be a product of two vectors. Take a coordinate system such that one of the vectors has the form (1,0,...0)

voila
Orodruin
Staff Emeritus
Homework Helper
Gold Member
A (2,0) tensor is a linear combination of such tensor products. You must show that not all sych linear combinations are tensor products of two vectors.

voila
May I state clearly that this is not a problem I must solve for class, this was just an example written somewhere which suggested we did it (thus why I didn't provide an attempt at solving it, just asking why it is that way). I still can't see why.

Oh, I think I'm getting it. Thinking about the matrix representation, that's just like stating that there are such matrices that can't be written as the tensor product of two vectors?

fresh_42
Mentor
Oh, I think I'm getting it. Thinking about the matrix representation, that's just like stating that there are such matrices that can't be written as the tensor product of two vectors?
Yes. You can write all matrices as a sum of (2,0) tensors, but a single tensor ##a \otimes b## will always result in a matrix of rank 1.

voila
The tensor notation such as ##(2,0)## only applies when the tensor is made up of a number of copies of a particular vector space and its dual vector space.
##\mathbf{a} \otimes \mathbf{b}## with ##\mathbf{a},\mathbf{b}\in V## is a ##(2,0)## tensor which has total rank 2. The main point here was all ##(2,0)## tensors cannot be expressed ##\mathbf{a} \otimes \mathbf{b}## .

voila
fresh_42
Mentor
##\mathbf{a} \otimes \mathbf{b}## with ##\mathbf{a},\mathbf{b}\in V## is a ##(2,0)## tensor which has total rank 2.
Just a remark.
Rank in this context is a bit of an ill-fated notation, since it has nothing to do with the rank of linear transformations which are also part of the context. Degree is (IMO) a far better word for it.

voila and MisterX
Here is the sketch of a proof. Let ##V## be a finite dimensional real vector space with dimension greater than 2 (the statement is simply not true for ##\dim V=1##, since any real number ##a## can be written as ##a=1\cdot a##). Let ##(e_j)## be a orthonormal basis and ##T=e_1 \otimes e_2 - e_2 \otimes e_1##. Suppose ##T=a \otimes b##, for some ##a, b \in V##. By equating components, get a contradiction. So there exists ##(2,0)## tensors which cannot be written as the direct product of vectors.

Note: The counterexample above was inspired by the singlet state in Quantum Mechanics.

voila
Thank you all for your answers. I reckon it's a rather simple question, but I was just beginning to study tensors and couldn't get my mind around it.