First let me give the definition of tensor that my book gives:(adsbygoogle = window.adsbygoogle || []).push({});

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?

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# Confusion with tensors

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