## 2012-12-06

### Authors in a Markov matrix: Which author do people find most inspiring? (11)

Let's consider a case in which we have three people. The number of elements becomes $$3^2 = 9$$, means there are nine relationships. Let me show you this with the example of Figure 10.
 Figure 10: Examples graphs for relationships among three nodes.
A set of three-node relationships is represented by the following 3x3 adjacency matrix:
\begin{eqnarray*} \left[ \begin{array}{ccc} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \\ \end{array} \right] \end{eqnarray*}
Figure 10 (a) has an edge from node 1 to node 2. According to the definition of an adjacency matrix, the element $$a_{12}$$ is 1. The adjacency matrix is the following: \begin{eqnarray*} M_{(a)} = \left[ \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{array} \right] \end{eqnarray*} The adjacency matrices of Figures 10 (b), (c) are the following: \begin{eqnarray*} M_{(b)} = \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{array} \right] \end{eqnarray*} \begin{eqnarray*} M_{(c)} = \left[ \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 1 & 0 \\ \end{array} \right] \end{eqnarray*} Figures 10 (d), (e), (f) are undirected graphs. There are no arrows since these relationships are mutual. The adjacency matrices are the following: \begin{eqnarray*} M_{(d)} = \left[ \begin{array}{ccc} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \\ \end{array} \right] \end{eqnarray*} \begin{eqnarray*} M_{(e)} = \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{array} \right] \end{eqnarray*} \begin{eqnarray*} M_{(f)} = \left[ \begin{array}{ccc} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \\ \end{array} \right] \end{eqnarray*} In an undirected graph (a graph that only has undirected edges), all the relationships are mutual, therefore, the matrix has symmetry. As we saw with the symmetric matrix of Alice and Cheshire Cat, if you treat the diagonal (the elements from top left to bottom right) as a mirror, the matching elements are the same on either side of this line. Let me explain a bit more about a symmetric matrix. The main diagonal of matrix is all diagonal elements from top left to bottom right. The main diagonal of 3x3 matrix is formed by the d'' elements in the following: \begin{eqnarray*} \left[ \begin{array}{ccc} d & & \\ & d & \\ & & d \\ \end{array} \right] \end{eqnarray*} A symmetric matrix has symmetry with respect to the main diagonal. No matter what kind of elements a matrix contains, it will be symmetrical if the corresponding elements across the diagonal are equal. For example, the following matrix has three pairs (a'-pair, b'-pair, and c'-pair) in the matrix. When each pair is the same number, the matrix is called symmetric. \begin{eqnarray*} \left[ \begin{array}{ccc} & \mbox{a} & \mbox{b} \\ \mbox{a} & & \mbox{c} \\ \mbox{b} & \mbox{c} & \\ \end{array} \right] \end{eqnarray*} For example, you can see the symmetry in $$M_{(f)}$$ as the following: \begin{eqnarray*} M_{(f)} = \left[ \begin{array}{ccc} & 1 & 0 \\ 1 & & 1 \\ 0 & 1 & \\ \end{array} \right] \end{eqnarray*} Another symmetric matrix example is the following: \begin{eqnarray*} \left[ \begin{array}{ccc} 1 & 6 & 7 \\ 6 & 2 & 8 \\ 7 & 8 & 3 \\ \end{array} \right] \end{eqnarray*}

#### 1 comment:

Shitohichi said...

Note: A reader pointed out the
self reference cannot be directional. He is correct. Because a directional relationship is only either A likes B'' or B likes A.'' If both are true, this is non directional. But the self link is A = B. It cannot be both Alice like Alice'' and `Alice doesn't like Alice'' are true at once.

Thanks for the feedback.

2013-1-23(Wed)