2
$\begingroup$

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.

$\endgroup$
3
  • 1
    $\begingroup$ How do you conclude the equality of $\sum_{j=1}^n l(e_j)e_j^{**}(e_i^*) = e_i^{**}(l)$? $\endgroup$
    – onRiv
    Commented Aug 28, 2022 at 13:40
  • 1
    $\begingroup$ 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 $\xi=e_i^{\ast\ast}$. $\endgroup$ Commented Aug 28, 2022 at 13:57
  • $\begingroup$ @onriv I actually meant $\sum_{j=1}^n l(e_j)e_i^{**}(e_j^*) = e_i^{**}(l)$ since $e_i^{**}(l) = e_i^{**}(\sum_{j=1}^n l(e_j) e_j^*) = \sum_{j=1}^n l(e_j) e_i^{**}(e_j^*)$ $\endgroup$
    – Torrente
    Commented Aug 28, 2022 at 14:57

1 Answer 1

1
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .