Let the adjacency matrix of our network be $A∈\{0,1\}^{n×n}$ with an empty diagonal ($A_{ii} = 0 ∀i$).
Direct approach
Let’s start with the approach that a node’s centrality ($C_i$) shall be proportional to the sum of the centralities of its neighbours with a proportionality constant $\frac{1}{λ}$ (chosen thusly with some foresight):
$$ C_i = \frac{1}{λ} \sum_{j=1}^{n} A_{ij} C_j.$$
This is nothing but a line-wise formulation of the matrix–vector multiplication:
$$ λ \vec{C} = A·\vec{C},$$
which is exactly the definition of an eigenvector. Now, to see why the largest eigenvector is chosen, we can turn to the Perron–Frobenius theorem, which tells us that for this eigenvalue (and in case of a connected network only for this eigenvalue), we can find an eigenvector all of whose components, i.e., the eigenvector centralities are positive.
Iterative approach
Alternatively, we can interpret the above ansatz iteratively:
- Assign random positive values to the components of $\vec{C}$.
Update these values according to:
$$\vec{C} ← \frac{A·\vec{C}}{\left| A·\vec{C} \right|}.$$
This means that each component is updated according to:
$$C_i ← \frac{1}{\left| A·\vec{C} \right|} \sum_{j=1}^{n} A_{ij} C_j,$$
i.e., you say the new centrality of a node is the sum of the centralities of its neighbours – times some normalisation to avoid values getting very big.
- Repeat Step 2 until the centralities converge. The idea is that if this converges to a unique result, this result does not only fulfil $C_i \propto \sum_{j=1}^{n} A_{ij} C_j$ but is also robust in that respect.
For almost all initial choices of $\vec{C}$, this will converge to the positive, length-1 eigenvector to the largest eigenvalue (which exists and is unique for a connected network, see above). The reason for this is that components along the eigenvector to the largest eigenvalue will be most enlarged by the multiplication and thus dominate the others over the iterations. (Having no components along the eigenvector to the largest eigenvalue does not happen in reality and is the reason why it’s almost all above.)
Note that such an iteration can also be used to numerically determine the largest eigenvalue.