Linear Algebra and Its Applications, Exercise 2.5.17

Exercise 2.5.17. In the scheme described in section 2.5 for ranking football teams, is strength of opposition accounted for, or must it be considered separately?

Answer: The ranking system described calculates potential values for each team based on score differences for the games each team plays. Consider the situation where the graph of teams is not connected; in other words, at least one group of teams has games only among members of the group, with teams in the group never playing teams in other groups. (For example, consider a football conference in which the teams never play non-conference games.)

In that case the calculated potentials can be used to rank teams within the group, but cannot be used to rank a team within the group relative to a team outside the group, or vice versa. If you wanted to create a ranking of all teams then you would have to explicitly incorporate some measure of the strength of the opposition, e.g., based on subjective assessments of the talent of the teams in each conference.

However if the graph is connected then the estimated potential for a team should already reflect the relative strength of the team’s opponents. Consider the following simple example with four teams (1 through 4):

Team 1 defeats team 2, 14-8.
Team 2 defeats team 3, 27-18.
Team 4 defeats team 3, 14-0.

If we assume that in each case the winning team is the home team, this corresponds to the incidence matrix

$A = \begin{bmatrix} 1&-1&0&0 \\ 0&1&-1&0 \\ 0&0&-1&1 \end{bmatrix}$

with the system $Ax = b$ expressed in matrix form as

$Ax = \begin{bmatrix} 1&-1&0&0 \\ 0&1&-1&0 \\ 0&0&-1&1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} 6 \\ 9 \\ 14 \end{bmatrix}$

where 6, 9, and 14 are the respective differences in the scores. This in turn corresponds to the system of equations

$\begin{array}{rcrcrcrcr} x_1&-&x_2&&&&&=&6 \\ &&x_2&-&x_3&&&=&9 \\ &&&&-x_3&+&x_4&=&14 \end{array}$

If we arbitrarily ground the network at team 4 by setting $x_4 = 0$ then from the third equation we have $x_3 = -14$ . Substituting the value of $x_3$ into the second equation we have $x_2 = -5$ . Finally, substituting the value of $x_2$ into the first equation we have $x_1 = 1$ .

Note that team 1 (with potential $x_1 = 1$ ) is ranked higher than team 4 (with potential $x_4 = 0$ ) even though team 4 won its game against team 3 in a more convincing fashion. That’s because based on the game results team 1’s opponent team 2 is considered a stronger opponent than team 4’s opponent team 3, and team 1 won its game with just enough points to have its calculated potential be higher than that of team 4.

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.

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2 Responses to Linear Algebra and Its Applications, Exercise 2.5.17

fatma elbialy says:

March 14, 2013 at 12:33 am

Thank you , please solve anther proplems very quicly

- hecker says:
  
  March 14, 2013 at 2:03 am
  
  I am glad that you like these posts. Unfortunately I do not have much time to work on the problems and publish the answers. You should expect to see new posts about once a week, maybe less often if I don’t have time.