# Why does $e_i^{**}(e_j^*) = e_j^*(e_i)$ imply $ev(e_i) = e_i^{**}$?

I am following some notes on abstract algebra and I was confused when reading the following:

($$V$$ is a vector space over the field $$k$$ and $$V^*$$ is its dual space.)

Consider the natural map $$V \xrightarrow{ev} V^{**}$$ that maps $$v \mapsto ev_v:V^* \rightarrow k : l \mapsto l(v)$$.
If $$V$$ is finite-dimensional, then working in bases $$\{e_1, \ldots, e_n\}$$, dual basis $$\{e_1^*, \ldots, e_n^*\}$$ and double dual basis $$\{e_1^{**}, \ldots, e_n^{**}\}$$, we see that $$e_i^{**}(e_j^*) = e_j^*(e_i)$$ and so $$ev(e_i) = e_i^{**}$$, hence $$ev$$ is an isomorphism.

I understand why $$ev(e_i) = e_i^{**}$$ implies that $$ev$$ is isomorphic however I would appreciate if someone could explain why $$e_i^{**}(e_j^*) = e_j^*(e_i)$$ implies $$ev(e_i) = e_i^{**}$$.

I tried the following, for any $$l \in V^*$$, $$(ev(e_i))(l) = l(e_i) = \sum_{j=1}^n l(e_j)e_i^{**}(e_j^*) = e_i^{**}(l)$$ so I can see that the equality is true but I didn't need to use the fact that $$e_i^{**}(e_j^*) = e_j^*(e_i)$$.

It might be a silly question but thanks in advance.

• How do you conclude the equality of $\sum_{j=1}^n l(e_j)e_j^{**}(e_i^*) = e_i^{**}(l)$? Commented Aug 28, 2022 at 13:40
• Set $\xi=\mathrm{ev}(e_i)$. This is the map that sends $f\in V^\ast$ to $f(e_i)$. In particular, $e_j^\ast\mapsto e_j^\ast(e_i)$, so that $\xi$ and $e_i^{\ast\ast}$ act in the same way on the basis $e_j^\ast$ of $V^\ast$, and so agree everywhere. Thus $\xi=e_i^{\ast\ast}$. Commented Aug 28, 2022 at 13:57
• @onriv I actually meant $\sum_{j=1}^n l(e_j)e_i^{**}(e_j^*) = e_i^{**}(l)$ since $e_i^{**}(l) = e_i^{**}(\sum_{j=1}^n l(e_j) e_j^*) = \sum_{j=1}^n l(e_j) e_i^{**}(e_j^*)$ Commented Aug 28, 2022 at 14:57

Using Kronecker delta symbol, we have by definition of the double dual basis $$e_i^{**}(e_j^*)=\delta_{ij}$$ for any $$i,j \in \{1, \dots, n\}$$. As we also have $$e_i^{*}(e_j)=\delta_{ij}$$ by definition of the dual basis, we can indeed conclude that $$e_i^{**}(e_j^*) = e_j^*(e_i)$$.
Now, by definition of the $$ev$$ map, $$ev(e_j)$$ is the map of $$V^{**}$$
$$\begin{array}{l|rcl} ev(e_j) : & V^* & \longrightarrow & k \\ & l & \longmapsto & l(e_j) \end{array}$$ In particular
$$ev(e_j)(e_i^*) = e_i^*(e_j)=\delta_{ij}=e_j^{**}(e_i^*).$$ As $$ev(e_j)$$ and $$e_j^{**}$$ takes the same values on the basis $$\{e_1^*, \dots, e_n^*\}$$, those two elements of $$V^{**}$$ are equal. We can conclude to the desired result.