# Thread: Tensor product and basis

1. ## Tensor product and basis

Hello,

Let V,W be some finite dim. vector spaces and denote $\displaystyle (e_i)_{i=1,..,n}, (g_j)_{j=1,..,m}$ the bases of V,W respectively.

Now i want to show that $\displaystyle (e_i \otimes g_j) ,1\leq i,j\leq m,n$ is a basis for the tensor product $\displaystyle V \otimes W$.
(We have defined the tensor product of vector spaces as a quotient vector space, i hope there is only one such def. such that you are knowing what i'm talking about)

I could show, that this set is a generating system fo $\displaystyle V\otimes W$. This wasn't so difficult. But i couldn't show that they are linearly independent.

I also try to show that $\displaystyle V\otimes W$ is isomorphic to another vector space with dimesnion n*m, like Hom(V,W). But i have difficulties with this.
Can you give me a hint?

Regards

2. Originally Posted by Sogan
Hello,

Let V,W be some finite dim. vector spaces and denote $\displaystyle (e_i)_{i=1,..,n}, (g_j)_{j=1,..,m}$ the bases of V,W respectively.

Now i want to show that $\displaystyle (e_i \otimes g_j) ,1\leq i,j\leq m,n$ is a basis for the tensor product $\displaystyle V \otimes W$.

I could show, that this set is a generating system fo $\displaystyle V\otimes W$. This wasn't so difficult. But i couldn't show that they are linearly independent.

I also try to show that $\displaystyle V\otimes W$ is isomorphic to another vector space with dimesnion n*m, like Hom(V,W). But i have difficulties with this.
Can you give me a hint?

Regards

Any two vectors spaces of the same dimension (finite or infinite) over the same field $\displaystyle \mathbb{F}$ are

About the first one check the following is true:

$\displaystyle V\cong V\otimes \mathbb{F}\,,\,\,(V\oplus W)\otimes U\cong (V\otimes U)\oplus (W\otimes U)$ ,

from which you obtain that $\displaystyle V\otimes W$ is, as vector space, the direct sum of $\displaystyle n+m$

copies of $\displaystyle \mathbb{F}$ and, thus, its dimension is $\displaystyle n+m$ .

Tonio

3. What do you mean by U?
Is it just a subset of the field?

Thanks

4. Originally Posted by Sogan
What do you mean by U?
Is it just a subset of the field?

Thanks

Any other vector space over the same field as $\displaystyle V,W$...

Tonio

5. Excuse me but i'm a bit confused. The Dimension of $\displaystyle V\otimes W$ isn't n+m, as you have written above. it is m*n. isn't it?

6. Originally Posted by Sogan
Excuse me but i'm a bit confused. The Dimension of $\displaystyle V\otimes W$ isn't n+m, as you have written above. it is m*n. isn't it?

Of course, my bad: it must say $\displaystyle m\cdot n$ in both places where $\displaystyle m+n$ was written.

Tonio

7. Originally Posted by tonio
Any two vectors spaces of the same dimension (finite or infinite) over the same field $\displaystyle \mathbb{F}$ are

$\displaystyle V\cong V\otimes \mathbb{F}\,,\,\,(V\oplus W)\otimes U\cong (V\otimes U)\oplus (W\otimes U)$ ,

from which you obtain that $\displaystyle V\otimes W$ is, as vector space, the direct sum of $\displaystyle n+m$

copies of $\displaystyle \mathbb{F}$ and, thus, its dimension is $\displaystyle n+m$ .

Tonio
hmmm. I really don't see it. Let us assume that the vs above are isomorphic.
The dimensions are: dim($\displaystyle (V\times W) \otimes U)=dim(V \otimes U)*dim(W \otimes U)$ how do you seperate the LHS?

8. Originally Posted by Sogan
hmmm. I really don't see it. Let us assume that the vs above are isomorphic.
The dimensions are: dim($\displaystyle (V\times W) \otimes U)=dim(V \otimes U)*dim(W \otimes U)$ how do you seperate the LHS?
What are you talking about? In general as has been pointed out $\displaystyle \displaystyle \dim\left(V\otimes W\right)=\dim\left(V\right)\dim\left(W\right)$ and thus in particular

\displaystyle \begin{aligned}\dim\left(\left(V\oplus W\right)\otimes U\right) &=\dim\left(V\oplus W\right)\dim\left(U\right)\\ &=\left(\dim\left(V\right)+\dim\left(W\right)\righ t)\dim\left(U\right)\\ &=\dim(V)\dim(U)+\dim(W)\dim(U)\end{aligned}

And this is certainly not equal to $\displaystyle \dim(V)\dim(W)\dim(U)^2$ as you seemed to indicate. Now, back to your original question if you don't want to appeal to a dimensional argument and explicitly want to write down an isomorphism try taking the obvious one. Namely, if $\displaystyle V$ has basis $\displaystyle \{v_1,\cdots,v_n\}$ and $\displaystyle W$ has basis $\displaystyle \{w_1,\cdots,w_m\}$ then try considering the unique linear homomorphism $\displaystyle T:V\otimes W\to\text{Hom}\left(V,W\right)$ such that $\displaystyle v_i\otimes w_j\mapsto T_{i,j}$ where $\displaystyle T_{i,j}:V\to W$ is the unique element of $\displaystyle \text{Hom}\left(V,W\right)$ with $\displaystyle v_i\mapsto w_j$ and $\displaystyle v_k\mapsto \bold{0},\text{ }k\in[n]-\{i\}$. This is evidently an isomorphism since it's linear and a bijection between bases.

P.S. There are multiple ways of thinking about the tensor product. I was taught initially, before I was given the definition you mentioned, to think of $\displaystyle V\otimes W$ as $\displaystyle \text{Bil}\left(V,W\right)^\ast$ where $\displaystyle \text{Bil}(V,W)$ is the space of bilinear forms on $\displaystyle V\oplus W$.

9. Originally Posted by Sogan
hmmm. I really don't see it. Let us assume that the vs above are isomorphic.
The dimensions are: dim($\displaystyle (V\times W) \otimes U)=dim(V \otimes U)*dim(W \otimes U)$ how do you seperate the LHS?

It's a little confusing so far: first I made that stupid mistake and instead product I wrote sum, and now you use $\displaystyle V\times W$

when you mean, apparently, $\displaystyle V\oplus W$...

Anyway, as $\displaystyle V\cong \bigoplus\limits^n_{k=1} \mathbb{F}\,,\,\,W\cong \bigoplus\limits^m_{i=1}\mathbb{F}$ , we get that

$\displaystyle V\otimes W\cong \bigoplus\limits^n_{k=1} \mathbb{F}\otimes \bigoplus\limits^m_{k=1} \mathbb{F}$ , and now use the hints I

gave you in my first message.

Tonio

10. Originally Posted by Sogan
Hello,

Let V,W be some finite dim. vector spaces and denote $\displaystyle (e_i)_{i=1,..,n}, (g_j)_{j=1,..,m}$ the bases of V,W respectively.

Now i want to show that $\displaystyle (e_i \otimes g_j) ,1\leq i,j\leq m,n$ is a basis for the tensor product $\displaystyle V \otimes W$.
(We have defined the tensor product of vector spaces as a quotient vector space, i hope there is only one such def. such that you are knowing what i'm talking about)

I could show, that this set is a generating system fo $\displaystyle V\otimes W$. This wasn't so difficult. But i couldn't show that they are linearly independent.

I also try to show that $\displaystyle V\otimes W$ is isomorphic to another vector space with dimesnion n*m, like Hom(V,W). But i have difficulties with this.
Can you give me a hint?

Regards
Suppose V, W are finite dimensional vector spaces over F.

$\displaystyle Hom_F(V,W)$ is "naturally isomorphic" to $\displaystyle V^* \otimes W$.

Recall that an element of $\displaystyle Hom_F(V,W)$ is a linear transformation from V to W. An input vector $\displaystyle v \in V$ is transformed into a vector $\displaystyle w \in W$ by the linear transformation.

Let $\displaystyle V^*=Hom_F(V,F)$ and $\displaystyle \phi \in V^*$.

Let $\displaystyle \{v_1, v_2, \ldots, v_n\}$ be a basis for $\displaystyle V$ and $\displaystyle \{v_1^*, v_2^*, \ldots, v_n^*\}$ be a basis for $\displaystyle V^*$.

Consider the map $\displaystyle f:V^* \otimes W \rightarrow Hom_F(V, W)$ given by $\displaystyle f(\phi \otimes w)(v)=\phi(v)(w)$.

To show surjectivity, we let $\displaystyle h \in Hom_F(V,W)$. For $\displaystyle v \in \sum_{i=1}^nc_iv_i \in V$, we have

$\displaystyle f(\sum_{i=1}^nv_i^* \otimes h(v_i))(v)=\sum_{i=1}^n{v_i^*(v)h(v_i)=\sum_{i=1}^ nc_ih(v_i)=h(v)$.

We have shown that $\displaystyle f(\sum_{i=1}^nv_i^* \otimes h(v_i))=h$. Thus, f is surjective. To show injectivity and check further details, refer the book "Groups and Representation" by Alperin p113.

11. Thank you for your help!! I have solved this problem. Now i have a short question about decomposable tensors:
I was just thinking about decomposable tensors in $\displaystyle \mathbb{R}^2 \otimes \mathbb{R}^2$.
I think tensors like $\displaystyle v=e_1\otimes \e_2 + e_2\otimes e_1$ are not decomposable, because they are linearly independent. Is this argument true?

Regards.

12. Originally Posted by Sogan
Thank you for your help!! I have solved this problem. Now i have a short question about decomposable tensors:
I was just thinking about decomposable tensors in $\displaystyle \mathbb{R}^2 \otimes \mathbb{R}^2$.
I think tensors like $\displaystyle v=e_1\otimes \e_2 + e_2\otimes e_1$ are not decomposable, because they are linearly independent. Is this argument true?

Regards.
Do you mean $\displaystyle v=e_1\otimes e_2 + e_2\otimes e_1$?
$\displaystyle v=e_1\otimes e_2 + e_2\otimes e_1=2(e_1 \otimes e_2)=2e_1 \otimes e_2$. So, I think v is a decomposable tensor.

13. Why do $\displaystyle e_1\otimes e_2$commute?. The tensor product isn't commutative in general. It is associative, but you can't write w $\displaystyle \otimes v$=$\displaystyle v\otimes w$ in general.

Regards

14. Originally Posted by Sogan
Why do $\displaystyle e_1\otimes e_2$commute?. The tensor product isn't commutative in general. It is associative, but you can't write w $\displaystyle \otimes v$=$\displaystyle v\otimes w$ in general.

Regards
It's my mistake. The tensor product of two vector spaces over $\displaystyle \mathbb{Re}$ are commutative $\displaystyle V \otimes W \cong W \otimes V$ up to isomorphism, but it does not mean that a tensor product of two vectors is commutative in general, i.e., $\displaystyle v \otimes w \neq w \otimes v$ in general (unless $\displaystyle v=w$).