Suppose we have $4$ sports teams, $\{T_1, T_2, T_3, T_4\}$, and we have to following records:
$T_1$ beat $T_2$ by a score of 4
$T_1$ beat $T_4$ by a score of 2
$T_2$ beat $T_3$ by a score of 1
$T_2$ loses to $T_4$ by a score of 7
$T_3$ and $T_4$ tied
We want to know what is the ranking of these teams (from best team to worst team).
Consider the Kenneth Massey's team ranking model, a very simply model that assigns $r_i$ to team $T_i$ to consider the difference between team scores. So we have the linear system:
$$r_1 - r_2 =4, r_1 -r_4 = 2, r_2 - r_3 = 1, r_2 - r_4 = -7, r_3 - r_4 = 0$$
that can be represented by the matrix:
$$\begin{bmatrix} 1 & -1 & 0 & 0 \\ 1 & 0 & 0 & -1 \\ 0 & 1 & -1 & 0 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & -1 \end{bmatrix} \cdot \begin{bmatrix}r_1 \\ r_2 \\ r_3 \\ r_4 \end{bmatrix} = \begin{bmatrix}4 \\ 2 \\ 1 \\ -7 \\ 0 \end{bmatrix}$$
We solve for the least-square solution to $r_i$ and get:
$$\begin{bmatrix}r_1 \\ r_2 \\ r_3 \\ r_4 \end{bmatrix} = \begin{bmatrix} 1.125 \\ -3.75 \\ -2.375 \\ 0 \end{bmatrix} + r_4 \begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \end{bmatrix}$$
So we have a team ranking of $T_1 > T_4 > T_3 > T_2$
I can't help but to think of the problem in terms of an adjacency matrix for a graph. Consider this approach:
Initialize a graph with $4$ vertices, each representing its respective team $T_i$, draw an directed edge with edge weight $w_{ji}$ from $T_i$ to $T_j$ if $T_i$ beat $T_j$ by $w_{ji}$. If a team lost, reverse the direction of the edge so that all edges have positive weight (ignore ties).
I notices if we consider the sum of each vertex's outgoing edges minus its incoming edges, and rank all team accordingly, we get the same ranking. That is $T_1$ have $2+4 = 6$, $T_2$ have $1 - 4 - 7 = -10$, and so on. Sorting these gives us $T_1 > T_4 > T_3 > T_2$, same solution as above.
So I'm wondering if this these two "models" will always produce the same result? My hypothesis is that it does, but I'm not certain.
Not necessarily. Consider the (rather boring) set of outcomes:
In the "graph theory" model, $T_1$ has a score of $0$ and $T_4$ has a score of $1$, so $T_1$ loses to $T_4$.
However, in the team ranking model, the least-squares solution we get is $(\frac18, \frac14, -\frac{11}{8}, 0) + r_4 (1,1,1,1)$, giving $T_1$ a slight edge over $T_4$.
If we add a sixth game between $T_1$ and $T_3$, then this is not possible. In that case, we can compute the answer in terms of a vector of six scores $(s_1,s_2,s_3,s_4,s_5,s_6)$, for both algorithms. The team ranking model then gives us $$ (r_1,r_2,r_3,r_4) = \left(\frac{s_1+s_2+s_6}{4},\frac{-s_1+s_3+s_4}{4}, \frac{-s_3+s_5-s_6}{4}, \frac{-s_2-s_4-s_5}{4}\right) $$ up to adding a multiple of $(1,1,1,1)$, which is just what the graph theory model tells us.
When $T_1$ and $T_3$ don't play a game, the graph theory model acts as if they tie; the team ranking model "makes a guess" about the hypothetical outcome of that game which is a linear combination of the other scores. This gives us a potential discrepancy.
In the example above, it's reasonable to suppose that if $T_1$ and $T_3$ played, then $T_1$ would win. After all, the first two games suggest that $T_1$ is about as good as $T_2$ and $T_4$; the average of the scores of $T_2$ and $T_4$ versus $T_3$ is $\frac32$. This is in fact the "guess" that the team ranking model ends up making about what the score of $T_1$ versus $T_3$ would be if they played.
To elaborate on the "guessing" that I refer to: of course, the team ranking algorithm doesn't actually involve making such a guess. Rather, this is an emergent feature of the model. If we only have five scores $(s_1,s_2,s_3,s_4,s_5)$, then up to adding a multiple of $(1,1,1,1)$ the team ranking model assigns the scores $$ (r_1,r_2,r_3,r_4) = \left(\frac{3s_1 + 3s_2 + s_3 - s_5}{8}, \frac{-s_1+s_3+s_4}{4}, \frac{-s_1-s_2-3s_3+3s_5}{8}, \frac{-s_2-s_4-s_5}{4}\right) $$ which exactly corresponds to setting $s_6 = \frac{s_1+s_2+s_3-s_5}{2}$ in the previous solution. So in some sense, $\frac{s_1+s_2+s_3-s_5}{2}$ is the model's guess at the what would happen if $T_1$ and $T_3$ had played. (It is the average of $s_1+s_3$ and $s_2-s_5$: the guesses at the outcome based on how $T_1$ and $T_3$ both played against $T_2$, and based on how $T_1$ and $T_3$ both played against $T_4$.)