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Math Help - Tensor product and basis

  1. #1
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    Tensor product and basis

    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
    Last edited by Sogan; January 21st 2011 at 03:33 PM.
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  2. #2
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    Quote Originally Posted by Sogan View Post
    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
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    What do you mean by U?
    Is it just a subset of the field?

    Thanks
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    Quote Originally Posted by Sogan View Post
    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
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    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?
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    Quote Originally Posted by Sogan View Post
    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
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    Quote Originally Posted by tonio View Post
    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?
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  8. #8
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    Quote Originally Posted by Sogan View Post
    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.
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    Quote Originally Posted by Sogan View Post
    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
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  10. #10
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    Quote Originally Posted by Sogan View Post
    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.
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  11. #11
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    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.
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    Quote Originally Posted by Sogan View Post
    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.
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    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
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    Quote Originally Posted by Sogan View Post
    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).
    Last edited by TheArtofSymmetry; January 24th 2011 at 04:22 PM.
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