Question

Given an undirected graph H, let C be the adjacency matrix of H. What is the meaning of entries in \(C^n\)?

Answer 1

If I'm remembering correctly...and I may not be, the \(i,j\)-th entry in \(C^n\) is the number of paths of length \(n\) between the \(i\)-the vertex and the \(j\)-th vertex.

Answer 2

How do you know? :O

Answer 3

I think I saw this in a combinatorics class I had last year. I could probably dig out the proof if you want.

Answer 4

Ah! I would love to know the proof, at least to understand why. :)

Answer 5

By the way, if the proof involves MI, please, please do not post it...

Answer 6

What do you mean by MI?

Answer 7

Ehm, sorry, induction.

Answer 8

Ah. Well my proof does do it by induction...

Answer 9

King George is correct. Wahid yes ;)

Answer 10

Haha, because the hint of the proof is to prove by induction, and I don't want to copy your work, lol!

Answer 11

Would you mind explaining the intuition behind?

Answer 12

Well, it's actually a pretty short proof. The main part that uses induction, is where you write\[C^n=C^{n-1}C,\]and for a given entry \(C_{ij}\), you can write\[(C^{n-1}C)_{ij}=\sum_{k}(C^{n-1})_{ik}C_{kj}\]

Answer 13

After that, it's a counting argument.

Answer 14

This is really rough but you can think of it like this: the i,jth entry of C is 1 only if i and j are connected. Now if we multiply C by itself, to get the i,kth entry (of C^2), we would be multiply the ith row of C by the kth column of C

now the ith row would be 1's in places where i is adjacent to some other node and 0's elsewhere

and the kth column would be the same, 1's only where k was adjacent to some node, and 0's elsewhere

so to get the i,kth entry of C^2, it would only be non-zero if i and k were adjacent to some other node, which would imply that there was a path of length 2 from i to k, and everytime they shared a common node, it would add 1 to the entry, so the entry counts the number of patsh of length 2

and then u go up from there, each power basically extends the path by 1 edge if it is possible, (if it not possible the product would just be 0)

Answer 15

So I guess the underlying idea of the proof, is to count the number of paths of length \(n-1\) from a vertex \(i\) to a vertex \(k\), and then the number of paths of length \(1\) from the vertex \(k\) to the vertex \(j\). And then let \(k\) vary over the whole graph.

Answer 16

Not sure if it looks good.

Base case:
When n=1,
\(A^1 = A\), which shows the number of path to vertex j, \(v_j\), to vertex i, \(v_i\), with the path length being 1.
So, it is true for n=1.

Induction step:
Assume it is true for n=m-1. Consider the i,j-th entry of \(A^m\),
\[(A^m)_{ij}\]\[ = (AA^{m-1})_{ij} \]\[= A_{i1}(A^{m-1})_{1j} + A_{i2}(A^{m-1})_{2j} +...+A_{in}(A^{m-1})_{nj}\]\[=\sum_{k}A_{ik}(A^{m-1})_{kj}\]
Consider \( A_{i1}(A^{m-1})_{1j}\),
\( A_{i1}(A^{m-1})_{1j}\) is the number of paths of length 1 from \(v_i\) to \(v_1\) times the number of paths of length m-1 from \(v_1\) to \(v_j\), which is equal to the number of paths of length m from \(v_i\) to \(v_j\) with \(v_1\) being the second vertex visited.

This argument follows for each term \( A_{ik}(A^{m-1})_{kj}\), where \( A_{ik}(A^{m-1})_{kj}\) is the number of paths of length m from \(v_i\) to \(v_j\) with \(v_k\) being the second vertex visited.

Hence, \(\sum_{k}A_{ik}(A^{m-1})_{kj}\) is the total number of all possible paths with length m from \(v_i\) to \(v_j\).

Answer 17

let's take this graph for example
http://en.wikipedia.org/wiki/Adjacency_matrix#Examples

also let's take that we are at vertex 3 of that graph.

Answer 18

from 3, you can go to vertex 2 or vertex 4.

now let us represent the vertex 3 by
\[ V = \begin{bmatrix}
0\\
0\\
1\\
0\\
0\\
0
\end{bmatrix} \]
Let the adjacency matrix be A.
\[ A V = \begin{bmatrix}
0\\
1\\
0\\
1\\
0\\
0
\end{bmatrix} \]

Answer 19

it means that ... after one step, either you end up in vertex 2 or vertex 4. therefore it makes sense to construct adjacency matrix this way, and represent state by vectors.

Answer 20

now let's look at
\[ A^2 V = A \begin{bmatrix}
0\\
1\\
0\\
1\\
0\\
0
\end{bmatrix} = A \left( \begin{bmatrix}
0\\
1\\
0\\
0\\
0\\
0
\end{bmatrix} +\begin{bmatrix}
0\\
1\\
0\\
0\\
0\\
0
\end{bmatrix} \right)\]