Non-commutaive Gelfand-Naimark theorem and dimension of Hilbert space It is well known that using non-commutative Gelfand-Naimark theorem for finite dimensional $C^∗$-algebra we can obtain isometric representation on finite dimensional Hilbert space.
My question is : Can we obtain exactly the dimension of this Hilbert space in general case ? If not, can we estimate it by some properties of given $C^∗$-algebra (cardinality etc.) ?
 A: The GNS construction doesn't have a canonical way of choosing "the right amount" of states. For an arbitrary algebra one uses all  states (see for instance Corollary I.9.11 in Davidson's C$^*$-Algebras by Example). But this is often way too much. 
Let's think of the easiest concrete example: let $A=M_2(\mathbb C)$. If we follow the abstract recipe, we need to consider every possible state. The states are given by all positive $a\in A$ of norm one via the duality  $$b\longmapsto \text{Tr}(ab).$$
The positive elements of $A$ are those 
$$
a=\begin{bmatrix}\alpha&\beta\\ \overline\beta&\gamma\end{bmatrix}
$$
such that $\alpha,\gamma\geq0$ and $|\beta|^2\leq\alpha\gamma$. To get a state, we need to add the condition $\text{Tr}\,(a)=1$, that is $\alpha+\gamma=1$. So we have
$$
a=\begin{bmatrix}\alpha&\beta\\ \overline\beta&1-\alpha\end{bmatrix}
$$
for $\alpha\in[0,1]$, $|\beta|^2\leq\alpha-\alpha^2$. Still, an infinite family depending on a real parameter and a complex one. A
The family $a(\alpha,\beta)$ above is uncountable, so the plain universal representation of this 4-dimensional C$^*$-algebra gives a non-separable Hilbert space. One usually avoids this for a separable C$^*$-algebra by using, instead of all states, only those that realize the norm on a countable dense subset of the unit ball, and so we can get a separable Hilbert space. 
Still, of course, for the algebra $A$ above one can simply use the two states $a\longmapsto a_{11}$ and $a\longmapsto a_{22}$ to obtain the canonical representation on $\mathbb C^2$. 
The same idea applies to any finite-dimensional C$^*$-algebra. One can show that if $A$ is finite-dimensional, then $A\simeq\bigoplus_{k=1}^m M_{n_k}(\mathbb C)$. One identifies the blocks via the minimal central projections, and the size of each block by the number of pairwise orthogonal minimal projection that each minimal central projection majorizes. That way, one could identify the "correct" dimension of the Hilbert space, namely $n_1+\cdots+n_m$. 
For a general C$^*$-algebra, things are simply if it is separable (the only ones people usually pay attention to), because then one can represent them as acting on a separable Hilbert space. For non-separable C$^*$-algebras, things are muddier: there are plenty of them acting on a separable Hilbert space (namely, all the von Neumann algebras with separable predual).
As a final comment, you cannot get anything at all from cardinality. For instance, $\mathbb C$ (the one dimensional C$^*$-algebra) and $C_r(\mathbb F_\infty)$ (the universal separable C$^*$-algebra) have the same cardinality.
