Tensor product and basis

**Sogan** · January 21st 2011, 02:57 PM

Hello,

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

Now i want to show that $(e_i \otimes g_j) ,1\leq i,j\leq m,n$ is a basis for the tensor product $V \otimes W$ .
(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 $V\otimes W$ . This wasn't so difficult. But i couldn't show that they are linearly independent.

I also try to show that $V\otimes W$ 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

**tonio** · January 21st 2011, 03:39 PM

Originally Posted by Sogan

Hello,

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

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

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

I also try to show that $V\otimes W$ 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 $\mathbb{F}$ are

isomorphic, so this answers your second question.

About the first one check the following is true:

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

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

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

Tonio

**Sogan** · January 21st 2011, 03:42 PM

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

Thanks

**tonio** · January 21st 2011, 03:47 PM

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 $V,W$ ...

Tonio

**Sogan** · January 21st 2011, 03:51 PM

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

**tonio** · January 21st 2011, 03:59 PM

Originally Posted by Sogan

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

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

Tonio

**Sogan** · January 21st 2011, 04:18 PM

Originally Posted by tonio

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

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

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

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

Tonio

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

**Drexel28** · January 21st 2011, 05:11 PM

Originally Posted by Sogan

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

What are you talking about? In general as has been pointed out $\displaystyle \dim\left(V\otimes W\right)=\dim\left(V\right)\dim\left(W\right)$ and thus in particular

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

And this is certainly not equal to $\dim(V)\dim(W)\dim(U)^2$ 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 $V$ has basis $\{v_1,\cdots,v_n\}$ and $W$ has basis $\{w_1,\cdots,w_m\}$ then try considering the unique linear homomorphism $T:V\otimes W\to\text{Hom}\left(V,W\right)$ such that $v_i\otimes w_j\mapsto T_{i,j}$ where $T_{i,j}:V\to W$ is the unique element of $\text{Hom}\left(V,W\right)$ with $v_i\mapsto w_j$ and $v_k\mapsto \bold{0},\text{ }k\in[n]-\{i\}$ . 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 $V\otimes W$ as $\text{Bil}\left(V,W\right)^\ast$ where $\text{Bil}(V,W)$ is the space of bilinear forms on $V\oplus W$ .

**tonio** · January 21st 2011, 11:54 PM

Originally Posted by Sogan

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

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

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

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

gave you in my first message.

Tonio

**TheArtofSymmetry** · January 22nd 2011, 04:33 AM

Originally Posted by Sogan

Hello,

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

Now i want to show that $(e_i \otimes g_j) ,1\leq i,j\leq m,n$ is a basis for the tensor product $V \otimes W$ .
(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 $V\otimes W$ . This wasn't so difficult. But i couldn't show that they are linearly independent.

I also try to show that $V\otimes W$ 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.

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

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

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

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

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

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

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

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

**Sogan** · January 22nd 2011, 06:19 AM

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 $\mathbb{R}^2 \otimes \mathbb{R}^2$ .
I think tensors like $v=e_1\otimes \e_2 + e_2\otimes e_1$ are not decomposable, because they are linearly independent. Is this argument true?

Regards.

**TheArtofSymmetry** · January 24th 2011, 05:13 AM

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 $\mathbb{R}^2 \otimes \mathbb{R}^2$ .
I think tensors like $v=e_1\otimes \e_2 + e_2\otimes e_1$ are not decomposable, because they are linearly independent. Is this argument true?

Regards.

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

**Sogan** · January 24th 2011, 06:32 AM

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

Regards

**TheArtofSymmetry** · January 24th 2011, 02:47 PM

Originally Posted by Sogan

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

Regards

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

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