First, we can rephrase your question in the following way:
Given a linear $u^*: \mathcal L \to \mathbb R$ and an expected utility preference $\succsim$, what is the minimal subset of $\succsim$ needed to ensure that it is represented by $u^*$.$^1$
You can do it in $|X| - 1$ observations, so long as every dimension of $\mathcal L$ is "represented" in the observations. Let's explore how this can be done, and hopefully in the process, make my claim coherent.
For the moment assume that $X$ is finite. Let $b,w$ be the best and worst elements of $X$ according to $u^*$; wlog assume that $u^*(b) = 1$ and $u^*(w) = 0$. Of course it must be that $b \succsim w$, or we have a counter-example to the hypothesis that $\succsim$ is represented by $u^*$. Now for each other element $x \in X$ we need only to know if
$$u^*(x)b + (1-u^*(x))w \sim x$$
If this is true then by the properties of EU, any representation (also normalized on $b,w$) must set $u(x)$ to be $u^*(x)u(b) + (1-u^*(x))u(w) = u^*(x)1 + (1-u^*(x))0 = u^*(x)$.
Now, if $X$ is infinite, there may not be a best and worst element, you will have to deal with cases where the element $x$ is above or below the bounds of the normalization. This issue arrises also in the proof of the mixture space theorem and is dealt with in the same way---take mixtures with the extreme elements, and figure out the weights accordingly.
It is easy to show that any less data than the above leaves a degree of freedom which can be exploited. In other words, you must check each dimension but it suffices to do this with a single observation.
For what its worth this is a specific case of a much more general result: a linear function is completely determined by its behavior on basis vectors! $X$ is your basis for $\mathcal L$.
 I could be getting your question wrong. Perhaps you mean that $u^*$ is over $X$ not $\mathcal L$ and you do not have access to cardinal properties of $u^*$. If this is the case, the answer still stands, but the question seems pretty meaningless.