I have the expression for the rank ($x_{i}$) of a page $i$ in an internet with $n$ sites, each site contains $n_{i}$ links to other sites and is linked to by the pages $L_{i}\subset\{1,\dots,n\}$. The expression is:
$$x_{i}=\sum_{j\in L_{i}}\frac{x_{j}}{n_{j}}\tag{1}$$
We can express this in matrix form as:
$$x_{i}=\sum_{j\in L_{i}}\frac{x_{j}}{n_{j}}\implies x_{i}-\sum_{j\in L_{i}}\frac{x_{j}}{n_{j}}=0$$
And therefore we can write the system of linear equations as: $\mathbf{A}\vec{x}=\vec{0}$, with:
$$A_{ij}=\begin{cases}1 & i=j \\ -\frac{1}{n_{j}} & j\in L_{i} \\ 0 & j \notin L_{i} \cup \{i\}\end{cases}$$
I have two questions:
- Is there a better way to write the components of the matrix which doesn't involve a conditional expression?
- Can we show that there always $\exists \vec{x}\mid \vec{x}\neq \vec{0}$?
EDIT: I have realised that by the Rank-nullity Theorem we can satisfy question 2 by determining that:
$$\operatorname{rank}{\mathbf{A}}\lt n$$
Thus proving that:
$$\dim_{\mathbb{R}}(\ker(\mathbf{A}))\geq1$$
And therefore that there exist $1$ or more non-trivial solutions to $\mathbf{A}\vec{x}=\vec{0}$.
