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Confusion with tensors

  1. Apr 8, 2015 #1
    First let me give the definition of tensor that my book gives:

    If [itex] V [/itex] is a finite dimensional vector space with [itex] dim(V) = n [/itex] then let [itex] V^{k} [/itex] denote the k-fold product. We define a k-tensor as a map [itex] T: V^{k} \longrightarrow \mathbb{R} [/itex] such that [itex] T [/itex]is multilinear, i.e. linear in each variable if all the other variables are held fixed. I know there are more general definitions but since this is the one I am using in my book, let's stick with this one.

    Okay, now here is my problem. First off, assume from this point on that [itex] \mathbb{R}^n [/itex] has the usual basis. If [itex] L^{2}(\mathbb{R}^{n}) [/itex] is the set of all 2-tensors on [itex] \mathbb{R}^n [/itex] then it has a dimension of [itex] n^2 [/itex]. If [itex] M(n,n) [/itex] is the set of all n by n matrices with real entries then we have [itex] L^2(\mathbb{R}^n) \cong M(n,n) [/itex].

    However, let [itex] L(\mathbb{R}^n, \mathbb{R}^n) [/itex] be the set of all linear transformations [itex] f: \mathbb{R}^n \longrightarrow \mathbb{R}^n [/itex]. It seems obvious to me that [itex] L(\mathbb{R}^n, \mathbb{R}^n) \cong M(n,n) [/itex]. But then this would imply that [itex] L^2(\mathbb{R}^n) \cong L(\mathbb{R^n}, \mathbb{R}^n) [/itex] which is impossible since [itex] dim(L^2(\mathbb{R}^n)) = n^2 [/itex] and [itex] dim(L(\mathbb{R}^n, \mathbb{R}^n)) = n [/itex] (I proved its dimension is [itex] n [/itex] and I am sure the proof is correct).

    Where have I gone wrong?
     
  2. jcsd
  3. Apr 8, 2015 #2

    Fredrik

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    The correspondence between ##L(\mathbb R^n,\mathbb R^n)## and ##M(n,n)## is bijective. So these vector spaces must have the same dimension.
     
  4. Apr 8, 2015 #3
    That's what I thought as well. So this must just mean my "proof" that [itex] dim(L(\mathbb{R}^n, \mathbb{R}^n)) = n [/itex] is incorrect.
     
  5. Apr 8, 2015 #4

    Fredrik

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    Yes, I agree.
     
  6. Apr 8, 2015 #5

    WWGD

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    I think you need to specify the target space of ##L^2 (\mathbb R^n, \mathbb R^n) ## , i.e., bilinear maps into what space? If it is into the base field ## \mathbb R ## of dimension 1 as a v.space over itself, then the dimension is ## n^2## as you said, if the target space is another vector space V, then ## Dim L^2 (\mathbb R^n , \mathbb R^n, V ) ## is the product of the individual dimensions. Sorry if this is obvious and I misread what you meant.
     
  7. Apr 8, 2015 #6

    Fredrik

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    He never mentioned ##L^2(\mathbb R^n,\mathbb R^n)##. The question was about ##L(\mathbb R^n,\mathbb R^n)## (linear operators on ##\mathbb R^n##) and ##L^2(\mathbb R^n)## (bilinear maps from ##\mathbb R^n\times\mathbb R^n## into ##\mathbb R##).
     
  8. Apr 8, 2015 #7

    WWGD

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    OK. Is ##L^2(\mathbb R^n)## supposed to be maps into ##\mathbb R##, or into any vector space ( as in the case of matrix multiplication)?
     
  9. Apr 8, 2015 #8
    [itex] L^2(\mathbb{R}^n) [/itex] maps into the reals. I see the mistake in my "proof" that led to this confusion. It all makes sense now.
     
  10. Apr 8, 2015 #9

    WWGD

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    Just curious, what was it?
     
  11. Apr 8, 2015 #10
    ***Sorry about the crappy LaTex ***

    Well the correct proof would be to fix the usual basis for [itex] \mathbb{R}^n [/itex]. Let [itex] T \in L(\mathbb{R}^n,\mathbb{R}^n) [/itex]. Let [itex] \phi_{j,i} (e_i) = e_j [/itex] but be 0 on all other basis vectors. Note that [itex] 1 \le j \le n, 1 \le i \le n [/itex]. We show that the [itex] \phi_{j,i} [/itex] span [itex] L(\mathbb{R}^n,\mathbb{R}^n) [/itex]. We know that [itex] T(e_j) = \sum_{j = 1}^n d_{j,i} e_j [/itex] for some scalars [itex] d_{j,i} \in \mathbb{R} [/itex].

    So [itex] T(e_1) = d_{1,1} \phi_{1,1} (e_1) + d_{2,1} \phi_{2,1} (e_1) + d_{3,1} \phi_{3,1} (e_1) + \cdots + d_{n,1} \phi_{n,i} (e_n) [/itex]

    We do this for each [itex] e_i [/itex]. Clearly then, [itex] T [/itex] is a linear combination of the [itex] \phi_{j,i} [/itex]. It is also clear that there are [itex] n^2 [/itex] many of the [itex] \phi_{j,i} [/itex] and Linear independence is obvious.

    The mistake I was making before was only allowing one index, so I ended up with only [itex] n [/itex] many of the [itex] \phi [/itex]
     
  12. Apr 8, 2015 #11

    WWGD

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    Thanks; your Latex is fine.
     
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