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!

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 $\|x\|=\|l_x\|$.
