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Background:

Let $V$ be a vector space over a field $\Bbbk$. Let $V^*=Hom_\Bbbk(V,\Bbbk)$ be the vector space of linear functionals on $V$. Consider the map $V^*\times V\to End_\Bbbk(V)$ given by $(f,v_0)\to\phi$ where $\phi(v)=f(v)\cdot v_0$. This map is obviously bilinear and thus induces a linear map $V^*\otimes_\Bbbk V\to End_\Bbbk V$.

If $V$ is finite-dimensional, then a choice of basis $\{e_1,\dots,e_n\}$ and a dual basis $\{e_1^*,\dots,e_n^*\}$ establishes that the map is injective since evaluations at $e_i$ for the image of each basis element $e_i^*\otimes e_j$ of $V^*\otimes V$ shows that the corresponding endomorphisms are linearly independent. Identifying $End_\Bbbk(V)$ with the $n^2$-dimensional space of $n\times n$ matrices and the fact that $\dim V^*\otimes V$ is $n^2$ shows that the map is in fact an isomorphism.


Now onto my question. If $V$ is an infinite-dimensional vector space over $\Bbbk$, then:

  1. is the above (obvious?) map $V^*\otimes_\Bbbk V\to End_\Bbbk(V)$ still injective? If not, then under what conditions is it?
  2. given a $\phi\in End(V)$, is there a criterion to determine whether $\phi$ comes from an element of $V^*\otimes V$?
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When you say the space of linear functionals, do you mean continuous linear functionals? There is a distinction in the infinite case. –  Asaf Karagila Feb 27 '11 at 7:51
    
@Asaf: I assume he means just what is asked; there is no mention of topology or continuity. (But if there were, then my answer to question 2. would have to be modified to show that each $c_j$ is continuous.) –  Jonas Meyer Feb 27 '11 at 8:07
    
Jonas: of course. I'm just being petty. :-) –  Asaf Karagila Feb 27 '11 at 10:07

1 Answer 1

up vote 5 down vote accepted
  1. Yes. If $\sum_{j=1}^n f_j\otimes e_j$ is an element of $V^*\otimes_k V$, you can assume WLOG that $\{e_1,\ldots,e_n\}$ is linearly independent. If $\sum_{j=1}^n f_j(v)e_j=0$ for all $v\in V$, then for each $j$, $f_j(v)=0$ for all $v\in V$, so that $f_1=f_2=\cdots=f_n=0$.

  2. They are the linear transformations with finite dimensional range. If the range of $\phi\in \mathrm{End}(V)$ has basis $e_1,\ldots,e_n$, then you can write $\phi(v)=\sum_{j=1}^n c_j(v)e_j$, and each coefficient map $c_j:V\to k$ is linear. Thus $\phi$ comes from $\sum_{j=1}^n c_j\otimes e_j$.

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Thanks: this shows that the explicit finite-dimensionality (i.e. choosing basis for $V$) in my proof was really a red herring. –  Vladimir Sotirov Feb 27 '11 at 15:47

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