
Number of paths
http://img826.imageshack.us/img826/3527/graphk.jpg
a) Write down the adjacency matrix, F.
This part I can do. (Using the convention that loops count as 1 edge)
$\displaystyle F = \begin{bmatrix} 1 & 0 & 1 & 1 \\ 0 & 1 & 1 & 1 \\ 1 & 1 & 1 & 0 \\ 1 & 1 & 0 & 1 \end{bmatrix}$
The ordering of the vertices are $\displaystyle v_1, v_2, v_3, v_4$ for both rows and columns.
b) Explain why the number of paths of length $\displaystyle 2^n$ which begin and end at $\displaystyle v_1$ is always one more than the number of walks of length $\displaystyle 2^n$ which begin at $\displaystyle v_1$ and end at $\displaystyle v_2$.
I'm stuck on this part.
This is what I have done so far:
So I have computed $\displaystyle F^2 = \begin{bmatrix} 1 & 0 & 1 & 1 \\ 0 & 1 & 1 & 1 \\ 1 & 1 & 1 & 0 \\ 1 & 1 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 1 & 1 \\ 0 & 1 & 1 & 1 \\ 1 & 1 & 1 & 0 \\ 1 & 1 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 3 & 2 & 2 & 2 \\ 2 & 3 & 2 & 2 \\ 2 & 2 & 3 & 2 \\ 2 & 2 & 2 & 3 \end{bmatrix}
$
And I have also computed $\displaystyle F^4 = \begin{bmatrix} 3 & 2 & 2 & 2 \\ 2 & 3 & 2 & 2 \\ 2 & 2 & 3 & 2 \\ 2 & 2 & 2 & 3 \end{bmatrix} \begin{bmatrix} 3 & 2 & 2 & 2 \\ 2 & 3 & 2 & 2 \\ 2 & 2 & 3 & 2 \\ 2 & 2 & 2 & 3 \end{bmatrix} = \begin{bmatrix} 21 & 20 & 20 & 20 \\ 20 & 21 & 20 & 20 \\ 20 & 20 & 21 & 20 \\ 20 & 20 & 20 & 21 \end{bmatrix}
$
Now we can see the that entry in $\displaystyle (v_1, v_1)$ is always 1 more than the entry in $\displaystyle (v_1, v_2) $
But how do I generalise this for all $\displaystyle F^{2^n}$?
Thanks!

Well, I can think of two possible approaches to this problem. One is to diagonalize F. I think you might be able to do that, since it is symmetric. That would enable you to produce a closedform expression for the required power of F.
Another approach is to reason about the problem and come up with an argument (I don't know what that would be, but it seems to me that you might be able to come up with one) showing the desired result.

I don't know what role the $\displaystyle 2^n$ part plays... but this might work for the path length being any specific number :
We consider a walk from v1 to v2. The stage at which the walk reaches v2 for the final time and loops there before terminating, we take the walk to v1 instead and loop there. This way we can construct a bijection between walks from v1 to v1 and those from v1 to v2, leaving out the case when the whole path is a succession of loops on v1 itself. Hence the result.