I am studying Humphreys "Introduction to Lie algebras and Representation theory" and I am stuck with a remark in fact concerning linear algebra more (see a piece of text below).
I need to show that for $V$ finite dimensional vector space it is true that $V^* \otimes V$ is isomorphic to End$(V)$.
In Humphreys they suggest to show that a map sending a generator $f \otimes v$ to an endomorphism $w \mapsto f(w)v$ is an epimorphism, i.e. that $ \forall a\in End(V) \ \exists \ b \in V^* \otimes V$ such that it maps to $a$, and then use the fact, that spaces have equal dimensions.
I am having troubles with tensor products, and maybe this is the problem of my attempt to show this, but here is what I tried: let $\operatorname{dim}V = n$, and $\{e_j\}_{j=1}^n$ be a basis of $V$. For every $a\in End(V)$ I need to find an element of $V \otimes V^*$, that maps to it. Every element of $V^* \otimes V$ can be written as $\sum_{i} \alpha_{i_1, i_2} f_{i_1} \times v_{i_2}$ (right?), so we need:
$$\sum_{i} \alpha_i f_i(w)v_i = a(w)$$
We write everything in coordinates: $v_i=\sum v_i^{(k)} e_k$, $w = \sum w_j e_j$, $f_i=\sum f_i^{(j)} e^j$ (standard dual basis associated to $\{e_j\}_{j=1}^n$); then we obtain a large sum:
$$\sum_{i_1,i_2,j,k} \alpha_{i_1,i_2} f_{i_1}^{(j)} w_j v_{i_2}^{(k)} e_k = \sum_{j,k} a_{j,k} w_j e_k$$
Hence
$$\sum_{i_1,i_2,j} \alpha_{i_1,i_2} f_{i_1}^{(j)} w_j v_{i_2}^{(k)}= \sum_{j} a_{j,k} w_j, \ \forall k$$
which has to be true for every $w$. That's why we need
$$\sum_{i_1,i_2} \alpha_{i_1,i_2} f_{i_1}^{(j)} v_{i_2}^{(k)}= \sum_{j,k} a_{j,k}, \ \forall k,j$$
which seems to be true, because there so many things I can control (I mean coefficients $a_{j,k}$ are the only one given),but I don't know how to justify it properly.
Also, if there is a simpler way to do this, I would like to know.
Best Answer
I guess you know that the $e^i \otimes e_j$ with $i, j = 1,\ldots n$, form a basis of $V^* \otimes V$. You send $e^i \otimes e_j$ to $$e_{ij} : V \to V, e_{ij}(w) = e^i(w)e_j = w_ie_j ,$$ where $w = \sum_{k=1}^n w_k e_k$.
A linear map $f : V \to V$ is determined by its values on the basis $e_1,\ldots, e_n$. Writing $f(e_i) = \sum_{j=1}^n f_{ij}e_j$ we get $$f(w) = \sum_{i=1}^n w_if(e_i) = \sum_{i=1}^n w_i(\sum_{j=1}^n f_{ij}e_j) = \sum_{i,j=1}^n f_{ij} w_ie_j = \sum_{i,j=1}^n f_{ij} e_{ij}(w) .$$ This shows that $f$ is a linear combination of the $e_{ij}$ and shows that your assigment is an epimorphism.
Alternatively you can also directly show that the $e_{ij}$ are linearly independent and therefore form a basis of $\text{End}(V)$. Let $\sum_{i,j=1}^n f_{ij} e_{ij} = 0$. We have $e_{ij}(e_k) = \delta_{ik}e_j$ with the usual $\delta_{ik}$-notation. Then for each $k$ $$(\sum_{i,j=1}^n f_{ij} e_{ij})(e_k) = \sum_{i,j=1}^n f_{ij} \delta_{ik}e_j = \sum_{j=1}^n f_{kj} e_j = 0 ,$$ thus $f_{kj} = 0$ for $j= 1, \ldots, n$.