User:J D Bowen/sandbox

=Homework Ch. 6=

pg 103
To show: that for a nonbipartite graph, the mth power of the Markov chain operator approaches a matrix with 1/n in every entry.

We rely on the fact that if two matrices A and B have row sums of 1, then their product will as well, ie, if

$$\sum_{i=1}^n a_{ij} = \sum_{i=1}^n b_{ij}=1 \ $$,

then

$$\sum_{i=1}^n (AB)_{ij} = \sum_{i=1}^n \left({ \sum_{k=1}^n a_{ik}b_{kj} }\right)= \sum_{i,k=1}^n a_{ik}b_{kj} = \sum_{k=1}^n \left({ b_{kj} \sum_{i=1}^n a_{ik} }\right) = \sum_{k=1}^n \left({b_{kj}*1}\right) = 1 \ $$.

It is clear that $$T = (1/k)A \ $$ has row sums equal to 1, since as an adjacency matrix of a k-regular graph, T has k entries of 1 in each row. A simple induction proof then demonstrates that since $$T \ $$ has all row sums equal to 1, so does $$T^m \ $$ for all m.

Turning now to the eigenvectors of $$T^m \ $$, we turn to equation (4) on page 102 to see that the spectrum of the adjacency matrix of nonbipartite graphs satisfies

$$\text{Spec}(A) = \left\{{ \lambda_1=k, \lambda_2, \dots, \lambda_n: \lambda_i\geq \lambda_j \ \text{for} \ i\geq j, \lambda_n > -k }\right\} \ $$

So as $$m\to\infty \ $$, we have

$$\text{Spec}(T^m) \to \left\{{ \lambda_1=1, \lambda_{x\neq 1}=0 }\right\} \ $$

This single non-zero eigenvector demonstrates the matrix has constant entries, and since the row sum is 1 and the row is n entries long, the entries must be $$1/n$$.

We are asked to demonstrate that this does not imply theorem 1 for even n. This is clear enough for the circle graph $$\mathbb{Z}_{2n} \ $$, for which the bipartiteness of the graph (construct sets $$\left\{{1,3,...}\right\}, \left\{{0,2,...}\right\} \ (\text{mod} \ 2n)$$ allows for the eigenvalue $$\lambda=-1 \ $$. Since then we have two non-zero eigenvalues as $$m\to\infty \ $$, with the second alternating, we see the matrix is not required to be constant.

pg 104
We are asked to show three inequalities:
 * $$\left\|f\right\|_2 \leq \left\|f\right\|_1 \leq |X|^{1/2} \left\|f\right\|_2 \ $$,
 * $$\left\|f\right\|_\infty \leq \left\|f\right\|_2 \leq |X|^{1/2} \left\|f\right\|_\infty$$,
 * $$\left\|f\right\|_\infty\leq\left\|f\right\|_1\leq |X|\left\|f\right\|_\infty \ $$.

pg 106
We are asked to show that for the circle graph $$\mathbb{Z}_n \ $$ where n is odd, with a probability function $$p:\mathbb{Z}_n\to [0,1], \Sigma p = 1 \ $$, we have

$$\left\|{T^m p - u}\right\| \leq \sqrt{n} \ \text{exp} \left({ \frac{-m\pi^2}{2n^2} }\right) \ $$,

where u is the uniform probability $$u(x)=1/n \ $$.

We are given the hint that the eigenvalues of T are $$\cos(2\pi a/n), 0 \leq a<n, \ $$ and for odd n, this function has a maximum absolute value at $$a=(n-1)/2 \ $$.

By Theorem 2, we have $$\left\|{T^m p - u}\right\| \leq \sqrt{n} \beta^m \ $$, where $$\beta = \ \text{max} |\lambda|, \lambda\neq1 \ $$. From the hint, we have

$$\beta = |\cos(\frac{2\pi\frac{n-1}{2}}{n})| = |\cos(\frac{\pi n - \pi}{n})| =| \cos(\pi-\frac{\pi}{n})| = |\cos(\frac{\pi}{n})|$$.

Using the suggested inequality $$\cos(\theta)\leq e^{-\theta^2/2}, 0\leq\theta\leq\tfrac{\pi}{2}\ $$, we reach

$$\beta \leq \text{exp} \ \left({ \frac{-\left({\frac{\pi}{n}}\right)^2}{2} }\right) = \text{exp} \ \left({ \frac{-\pi^2}{2n^2} }\right) \ $$

The result follows.