Isometry between finite-dimensional space and its topological dual 
Let $(X, \|\cdot \|_X)$ be a normed, finite dimensional vector space. Show that its topological dual $X'$ is isometrically isomorphic to $X$.

The general form of such a linear map between those spaces should be: $$l : X \to X', x_o \mapsto l_{x_o} :=x_0^T A$$ for an $A \in \mathbb{R}^{n \times n}$ with $\dim X = n$, and $l_{x_0}(x) = x_0^T A x$ (representing $x_0$ and $x$ in $\mathbb{R}^n$, which should be possible by isomorphy). For this to be an isometry, I tried to find $A$ such that
$$\|x_0\|_X \stackrel{!}{=} \|l_{x_0}\| = \sup_{x \in X \setminus \{0\}} \frac{|x_0^T A x|}{\|x\|_X},$$
the latter terms being the operator norm. If all my assumptions up to this point are correct, this should eventually lead me to the solution, but I just cannot come up with any information about the matrix $A$ that I need to find. It's probably really easy, but I just started on the subject.
Thank you in advance!
 A: Okay, after some more research and thought, this statement seems to be wrong. Some proofs I've seen now argue equivalency of norms, which just doesn't work when you explicitly try to apply it; equivalency of norms is only useful when you talk about inequalities of norms. In general, $\|x\|_a = \|y\|_a  \nRightarrow \|x\|_b = \|y\|_b$ even when the norms are equivalent (consider for example $(1,1),(1,0) \in \mathbb{R}^2$ with the sup norm and the $1$-Norm).
$(\mathbb{R}^2 , \|\cdot\|_p)$ with its dual space $(\mathbb{R}^2, \|\cdot \|_q)$ (with $\frac{1}{p} + \frac{1}{q} = 1$ ) provieds a counterexample for $p,q \notin \{ 1, 2, \infty \}$, as these spaces can never be isometric. The proof is a little difficult, but can be found here.
An intuitive approach: A linear map between two unit circles needs to conserve the "corners", that means for $p,q \notin \{ 1, 2, \infty \}$ for example $(1,0)$ would need to be mapped to $(\pm 1, 0)$ or $(0, \pm 1)$, as this is where the "corners" on the respective unit circles are placed. For $p \in \{ 1, \infty\} \implies q \in \{\infty, 1\}$ this is not a problem: Both unit circles are squares, which can be transformed into each other by rotation and scaling, and the case $p = 2 \implies q = 2$ is trivial. 
But in all the other cases, this "preserving corners" condition gives us a contradiction when we map the standard basis $\{(1,0), (0,1)\}$ to any of the possibilites, because then the other points on the circle will not be mapped correctly.
A: X is finite dimensional. Suppose $A=\{x_1,...,x_n\}$ is its basis. Define $l_i:X\to \Bbb C$ such that $l_i(x_j)=\delta_{ij}$ for $x_j\in A$. Clearly $l_i$ is linear and bounded(X is finite dimensional). Show that $\{l_i\}$ is a basis for $X^*$. 
Define $\phi:X\to X^*$ such that $\phi(x)=l_x$. Every $x\in X$ has the representation $x=\sum_{I=1}^n \alpha_i x_i$, so $l_x= \sum_{I=1}^n \alpha_il_i$. Clearly $\phi$ is an isomorphism. To show it's an isometry, we know all norms are equivalent in finite dimensional space. Thus using $||.||_\infty$, we have $\|x\|=\|l_x\|$.
