# Direct product between vectorspaces

by Kontilera
Tags: product, vectorspaces
 P: 101 Hello! Im currently reading the analysis of tensors and have now encountered the tensorproduct, $$\otimes$$. I am wondering about the statement that every vector in: $$V \otimes W$$ (with the basis (v_i) and (w_i)) can be written as a linear combination of the basis: $$v_i \otimes w_i ,$$ but not in general as an element of the form:$$v \otimes w,$$ where w and v are elements i W and V. Which elements can I not reach by the second way? If we set V and W to R^3 it looks like we are comparing a 6-dimensional space to a 9-dimensional space (true?), in that case does it have something to do with the symmetric or antisymmetric components of $$V \otimes W$$ that can not be reached by $$v \otimes w$$? I am thankful for all help possible. :) Best Regards Kontilera
 P: 4,570 Hey Kontilera and welcome to the forums. Forgive me if this is out of place, but are these spaces completely independent or do they have any kind of linear dependence between the spaces? I don't actually know if you can do tensor products on vector spaces that are linearly dependent in some way, but if you can then this is a red flag to make sure that this is handled correctly. (My guess is that these aren't considered that much even if you can have this happen).
 P: 101 Hello and thanks! They are not completely independent. For example we can consider the cotangentspaces for a smooth manifold M at a point p, where: $$T^*_pM \otimes T^*_pM = T^2_pM.$$
Emeritus
PF Gold
P: 16,101

## Direct product between vectorspaces

Let V be a vector space of column vectors, and W be a vector space of row vectors. It's pretty easy to see that $V \otimes W$ is (isomorphic to) a vector space of matrices. The tensor product is just the outer product: $v \otimes w = vw$.

Invoking linear algebra, there is a simple characterization of which matrices are pure tensors -- i.e. of the form $v \otimes w$ -- they are the matrices with rank 0 or 1.
 P: 101 Yeah, I agree that the product of V and W is isomorphic to the space of matrices but I think you miss my question here. Lets say V = W = R^3. Then we get, as expected, nine basis elements: $$e_i \otimes e_j ,$$ where i and j runs up to 3. The elements of this space can however not in general be written as a outer product of two vectors: $$v \in V \;\quad \text{and}\;\quad w \in W .$$ Let us take the standard basis for R^3 in both V and W. We should then be able to find some matrices in $$V \otimes W$$ that confirm this (is not an outer product). As I said - when choosing the vectors v and w for: $$v \in V \quad \text{and} \quad w \in W,$$ we are dealing with 6 degrees of freedom while $$V \otimes W$$ has nine. Or has I missunderstood something basic?
 Emeritus Sci Advisor PF Gold P: 16,101 I think you're counting the degrees of freedom in making a choice from V OR W, rather than from V AND W. You can see the 9 degrees of freedom by seeing the 9 ways to take the tensor product of standard basis vectors. In fact, if S and T are disjoint sets, then if V is the free vector space generated by S and W is the free vector space generated by T, then $V \otimes W$ is the free vector space generated by $S \times T$, whereas the one generated by $S \cup T$ is $V \oplus W$. The free vector space generated by a set S is isomorphic to Rs, where s is the number of elements in S -- just imagine that the indexes are labeled by the elements of S, rather than by the numbers {1, 2, ..., s}. P.S. "direct product" is usually used for $V \oplus W$, not $V \otimes W$.
 P: 101 Hmm, thanks for the responses. I think this needs some thinking. Are you saying that every tensor in $$V \otimes W$$ can be written as a tensorproduct between one vector v in V and another vector w in W? :/ The author said the opposite so I started wondering what the pattern between those tensors of second rank was, for the standard basis.
Emeritus
PF Gold
P: 16,101
 Quote by Kontilera Hmm, thanks for the responses. I think this needs some thinking. Are you saying that every tensor in $$V \otimes W$$ can be written as a tensorproduct between one vector v in V and another vector w in W? :/
No, I'm saying that every tensor can be written as a linear combination of such tensors.

Most of them cannot be written as pure tensors; e.g. any matrix of rank 2 or more in the example I mentioned earlier.
 Emeritus Sci Advisor PF Gold P: 16,101 Are you asking about how the set of pure tensors looks? It's not a vector space, but it is a topological space and you can ask about it's geometry. I think the space is six-dimensional in your example, but that has nothing to do with $V \otimes W$. (I am still a little confused about what you're asking. I thought you originally were asking the question that is answered by my mention of rank 2+ tensors, but then you said you weren't. But your latest post sounds like you're asking that question again)
P: 101
 Quote by Hurkyl Are you asking about how the set of pure tensors looks? It's not a vector space, but it is a topological space and you can ask about it's geometry. I think the space is six-dimensional in your example, but that has nothing to do with $V \otimes W$.
Yeah! I agree, its the once that can not be written as a outer product that Im intrested in. As said in the first post,

"I am wondering about the statement that every vector in: $$V \otimes W$$ (with the basis (v_i) and (w_i)) can be written as a linear combination of the basis: $$v_i \otimes w_i ,$$ but not in general as an element of the form:$$v \otimes w,$$ where w and v are elements i W and V.

Which elements can I not reach by the second way?" :)

Is there any way of determining if a 2-rank tensor can be written as a outerproduct?
 Emeritus Sci Advisor PF Gold P: 16,101 A matrix is a pure tensor if and only if it has rank 0 or 1.
P: 101
 Quote by Hurkyl A matrix is a pure tensor if and only if it has rank 0 or 1.
Ah okey, thanks.
I will see if I can find some good lecture notes on this stuff. The concept of 'pure tensor' seems to be the key. Ill come back with questions if there is something more. :)

All the best!

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