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 for a real-valued function on a finite group $f:X \to \mathbb{R} \ $:
 * $\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 \ $.

where

$\left\|f\right\|_1 = \sum_{x\in X} |f(x)| \ $,

$\left\|f\right\|_2 = \langle f,f \rangle ^{1/2} \ $,

$\left\|f\right\|_\infty = \text{max} \ |f(x)| \ $.

Equation 1
From the definition of the inner product $\langle f,g \rangle = \sum_{x\in X} f(x)\overline{g(x)} \ $, we have

$\left\|f\right\|_2 = \sqrt{\Sigma( f(x)^2)} \ $, and we have $\left\|f\right\|_2 \leq \left\|f\right\|_1 \ $ directly from the triangle inequality.

To demonstrate the second part of equation 1, we investigate the square of the inequality

$\left({ \Sigma |f(x)| }\right)^2 \leq |X| \Sigma( f(x)^2) \ $

Expanding the LHS, we see we are attempting to show that

$\sum_{i,j=1}^n \left({|f(i)||f(j)|}\right) \leq \sum_{k=1}^n (f(k)^2n) $

Let $g=|f| \ $ for notational sake, and note that since f is real-valued, $f^2=g^2 \ $. Of course we have equality if g is constant, so let us consider only the case where the maximum value of g minus the minimum value is some positive number d.


 * Finish proof

Equation 2
For the second equation, we again square the equation to get

$(\text{max} \ |f(x)|)^2 \leq \Sigma( f(x)^2) \leq |X|( \text{max} \ |f(x)|)^2 $

Obviously the middle expression will contain as a summand at least the left expression, plus other nonnegative terms, and so the inequality holds; and while both the middle and right expressions are sums of |X| nonnegative terms, the rightmost expression is necessarily at least as large as the middle expression because it sums of the greatest of terms in the middle expression.

Equation 3
The third equation is perhaps the most trivial of all. If we look at it in the form

$\text{max} \ |f(x)| \leq \sum_{x\in X} |f(x)| \leq |X| \text{max} \ |f(x)| \ $

we see it is similar in structure to our previous inequality, and true for the same reasons: the greatest of a set of nonnegative numbers is necessarily at most the sum of those numbers, with equality only if the other numbers are all zero, and the sum of all n possibly different nonnegative numbers is naturally at most the sum of n copies of the greatest of those numbers, with equality only if all the numbers are the same.

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.