Dual space of a finite dimensional normed space My lecturer gave us this result today in class, but he didn't give a proof, he said we can prove it ourselves, only I'm really struggling to see how to do it.
Let $E$ be a normed space with dual $E'$. Then $E$ is finite dimensional if and only if $E'$ is finite dimensional, and in fact $\dim{E} =\dim{E'}$.
Would appreciate any input!
 A: Suppose that $E$ is finite dimensional. Let $x_1, \ldots, x_n$ be a basis for $E$. Consider the coordinate functionals $f_1, \ldots, f_n$ given by
$$\langle f_i, \sum_{k=1}^n \lambda_k x_k \rangle = \lambda_i\quad (i\leqslant n)$$
Prove that $f_1, \ldots, f_n$ are linearly independent. Having this done, note that they span $E^\prime$. Indeed, if $f$ is any functional on $E$, then 
$$\langle f, \sum_{k=1}^n \lambda_k x_k\rangle = \sum_{i=1}^n \lambda_k\langle f,x_k\rangle $$
so $f = \sum_{i=1}^n \langle f, x_k \rangle f_i$ because
$$\langle \sum_{i=1}^n \langle f, x_k \rangle f_i, \sum_{k=1}^n \lambda_k x_k\rangle = \sum_{k=1}^n\sum_{i=1}^n \lambda_k \langle f_i, x_k\rangle \langle f, x_k \rangle $$
Note that $\langle f_i, x_k\rangle = 0$ unless $i=k$ in which case this is equal to 1. Consequently, the right hand side is equal to $\sum_{i=1}^n \lambda_k\langle f,x_k\rangle$.
We thus proved that $f_1, \ldots, f_n$ form a basis for $E^\prime$. So $E$ and $E^\prime$ have the same dimension.
A: See if you can figure it out from the following hint:
If $E$ is finite dim, by defn. it has a finite basis. Say $x_1, x_2, \dots x_n$ is a basis of $E$. 
For any $x \in E$, x can be written uniquely as $x = a_1x_1 + a_2x_2 + \dots a_n x_n$. Try to prove the following:
1) For $j \in \{1, \dots n \}$, the maps $x_j^*: X \to \mathbb{R}$ defined by $$
x_j^*(a_1x_1 + a_2x_2 + \dots a_n x_n) = a_j
$$
are in the dual space of $E$.
2) Can you show the $x_j^*$ form a basis of the $E'$?
