An algebra with a strictly positive state + condition ? to become a C$^*$-algebra. Let $(A,*)$ be an involutive $\mathbb{C}$-algebra i.e. a $\mathbb{C}$-algebra with a semi-linear anti-automorphism i.e.
identically

*

* $(x+y)^*=x^*+y^*$

* $(\alpha.x)^*=\overline{\alpha}.x^*$ 

* $(xy)^*=y^*x^*$

A strictly positive (or faithful) state $\varphi\in A^*$ is such that for all $x\in A\setminus \{0\}$, $\varphi(x^*x)> 0$. 
One can prove easily that $g(x,y):=\varphi(x^*y)$ is a non-degenerate hermitian form. It can be noted that, due to the fact that $g(a.x,y)=g(x,a^*y)$ and with finite dimensionality, one gets that $A$ is semi-simple. So, one can consider $A$ as simple. 
My question is the following.
Q) We suppose $A$ to be finite dimensional. Under which additional condition(s) is the Hilbert space $(A,g)$ a C$^*$-algebra ?
Late edit I insist on the fact that I am in search of conditions for
$(A,g)$ to be a C$^*$-algebra, not only $A$. This means that it should be a C$^*$-algebra for the precise norm $||x||=\sqrt{g(x,x)}$.
Late remark The fact that $A$ be a star-algebra of finite dimension, sum of matrix algebras is by no means sufficient to imply that the projectors on the blocks are *-invariant nor $A\simeq \mathbb{C}$ as shows the following counterexample. Take $B=\mathbb{C}^{n\times n}$ (algebra of complex square matrices of dimension $n>0$) and $A=B\oplus B$ with the anti-automorphism
$(X,Y)^\star=(Y^*,X^*)$. Then $(A,\star)$ is easily checked to be a star algebra. It is of finite dimension, sum of matrix algebras but $dim_\mathbb{C}=2n^2\not= 1$. Indeed, the existence of a strictly positive state is crucial as there is none over $A$.
 A: No  Hilbert space is a $C^*$-algebra in any natural way since the structure of Hilbert spaces (a vector  space with an inner
product) is quite different from that of  a C*-algebra (a Banach space equipped with an involutive algebra structure).
Nevertheless from the data provided by the OP one may give  $A$ the structure of a  $C^*$-algebra in a very natural
way.  The first step is to consider the hermitian form $g$ as an inner product on $A$, so that
$$
  H:= (A,g)
  $$
becomes a Hilbert space
(completeness follows from  finite dimensionality).
One may  then define  a map
$$
  \pi :A\to B(H),
  $$
by
$$
  \pi (a)\xi  = a\xi , \quad\forall a\in  A,\quad \forall \xi \in H,
  $$
(here $a\xi $ is nothing other than  the product of $a$ and $\xi $).
One may easily show that $\pi $ is a *-homomorphism.  It is moreover injective since, for $a\neq 0$, one has that
$\pi (a)a^*=aa^*\neq 0$, as a consequence of $\varphi (aa^*)>0$.
Finally,  identifying $A$ with its image within $B(H)$ via $\pi $, we  have  that $A$ becomes a $C^*$-algebra.

Later edit:
The late edit indeed makes the question trivial and, as noted my Martin, the only example is the complex numbers.
A perhaps more elementary proof of this fact is that every finite dimensional $C^*$-algebra is the direct sum of matrix algebras, so unless $\text{dim}(A)=1$, there exists a pair of mutually orthogonal projections $p$ and $q$.  It follows that  $\|p\pm q\|=1$, hence the parallelogram law fails.
A: The following statements are equivalent:

*

*$A$ is a C$^*$-algebra with the norm induced by $\varphi$


*$\dim A=1$, i.e., $A=\mathbb C$
Proof. if $A=\mathbb C$, then the only state is the identity, which induces the norm.
Conversely, if $A$ is a C$^*$-algebra, its norm satisfies the C$^*$-identity
$$\tag1\|x\|^2=\|x^*x\|.$$ In terms of $\varphi$, this is
$$\tag2\varphi(x^*x)=\varphi((x^*x)^2)^{1/2},\qquad x\in A.$$
So, for each positive $a\in A$,
$$\tag3\varphi(a)^2=\varphi(a^2).$$
As $\varphi$ is a ucp map, the equality $(3)$ says that $a$ is in the multiplicative domain of $\varphi$. Thus the multiplicative domain of $\varphi$ is all of $A$, since a C$^*$-algebra is spanned by its positive elements. So $\varphi$ is a faithful representation of $A$ into (thus onto) $\mathbb C$.
