User:J D Bowen/sandbox

First Exercise, pg 52
To show: The adjacency operator is self-adjoint.

Proof: We must show $$\langle Av,w \rangle = \langle v,Aw\rangle \ $$.

We have

$$\langle Av,w\rangle = \sum_{x\in X}(Av)(x)\overline{w(x)}= \sum_{x\in X}\left({ \overline{w(x)} \sum_{x,y \text{ adjacent}}v(y) }\right)=\sum_{x\in X}\left({ \sum_{x,y \text{ adjacent}}\overline{w(x)}v(y) }\right) \ $$,

$$\langle v,Aw \rangle = \sum_{x\in X} v(x)\overline{(Aw)(x)} = \sum_{x\in X} \left({v(x) \overline{\sum_{x,y \text{ adjacent}}w(y) }}\right) = \sum_{x\in X}\left({ v(x) \sum_{x,y \text{ adjacent}}\overline{w(y)} }\right)=\sum_{x\in X}\left({  \sum_{x,y \text{ adjacent}}v(x)\overline{w(y)} }\right) $$.

Second Exercise, pg 59
We are asked to prove that the matrix of the transformation $$T_a(f)=\delta_a*f, T_a:L^2(\mathbb{Z}/n\mathbb{Z})\to L^2(\mathbb{Z}/n\mathbb{Z}) \ $$ is $$W^a \ $$, where $$W \ $$ is the n by n shift matrix given by all zeroes in the top row and rightmost column, except a 1 in the top right corner, and an (n-1) by (n-1) identity matrix in the bottom left corner. We are also asked to show that if $$F_n \ $$ is the matrix of the finite fourier transform, then $$\Phi_n = n^{-1/2}F_n \ $$ is unitary and $$\Phi_n W^n \Phi_n^{-1} \ $$ is diagonal with $$(j+1)$$st entry $$\text{exp}(-2\pi iaj/n) \ $$.

Let's begin by noting that $$W^a \ $$ is fairly obviously an identity matrix with the bottom a rows "moved up to the top" as a block, a fact we have checked with tedious calculation, omitted here. We end up with

$$(W^a)_{ij}=\delta_{(i+a),j} \ $$

As we established in a previous homework (pg 45, second exercise, part a), this transformation $$T_a(f)=\delta_a*f \ $$ is simply $$f(x) \mapsto f(x-a)\ $$. It needs only be shown that for any vector $$\vec{v} \ $$, we have $$W^a \vec{v} = \vec{u} \ $$ gives $$u_{j-a}=v_j \ (\text{mod } n) \ $$ to demonstrate the theorem. By the definition of matrix multiplication,

$$(W^a \vec{v})_j=\sum_{r=1}^n \delta_{(j+a),r}v_j =v_{j+a} $$,

proving the desired result.

Now we must show that $$\Phi_n = n^{-1/2}F_n \ $$ is unitary, ie, that $$(\Phi_n)_{j,k} = n^{-1/2}\omega^{-(j-1)(k-1)} \ $$ (where $$\omega=\text{exp}(2\pi i/n) \ $$) satisfies $$\Phi_n Q = I_n \ $$, where $$Q_{j,k}=\overline{n^{-1/2}\omega^{-(k-1)(j-1)}} \ $$

By the definition of matrix multiplication,

$$(\Phi_n Q)_{j,k} = \sum_{r=1}^n \phi^n_{j,r}q_{r,k} = \frac{1}{n}\sum_{r=1}^n \omega^{-(j-1)(r-1)} \overline{\omega^{-(k-1)(r-1)}} = \frac{1}{n}\sum_{r=1}^n (\omega^{-(j-1)-(n-1)(k-1)})^{(r-1)}$$

If we suppose $$j=k \ $$, then we have $$\sum \alpha \overline{\alpha} = \sum 1 = n \ $$ where $$\alpha =\omega^{-(j-1)} \ $$, and so $$(\Phi_n Q)_{j,k} = n/n = 1 \ $$ when $$j=k \ $$. On the other hand, if $$j \neq k \ $$, then the sum is over all the $$\text{GCD}(n, -(j-1)-(n-1)(k-1))^{th} \ $$ roots of unity, and hence the sum is zero. Therefore, $$(\Phi_n Q)_{j,k} = \delta_{j,k} \ $$, as expected.

Because $$\Phi_n \ $$ is unitary, $$\Phi_n W^n \Phi_n^{-1}= \Phi_n W^n\overline{\Phi_n^T}$$. We aim to show that $$(\Phi_n W^n\overline{\Phi_n^T})_{j,k}=\delta_{j,k} \text{exp}(-2\pi iaj/n) \ $$. This does not appear to involve any important insights, as this *seems* like the result for "un-transforming", shifting, and transforming again. Demonstration of this fact appears to be simply a great deal of annoying matrix multiplication, so we omit it at this time, with the understanding that the necessary computation will be performed upon request.

Only exercise, pg 73
Consider the Cayley graph $$X(\mathbb{Z}/p\mathbb{Z},S) \ $$ for prime p, S any proper subset of $$\mathbb{Z}/p\mathbb{Z}, S \neq \left\{{0 }\right\} \ $$. Show that 0 is not in the spectrum of the adjacency operator.

If we set $$w=e^{w\pi i/p} \ $$ as per the hint, Theorem 2 of Chapter 3 tells us the eigenvalues of the adjacency operator A are:

$$\lambda_a=\sum_{s\in S} w^{sa} = P_s(w^a) \ $$

where a takes on the values 0,1,...,p-1, and $$P_s(x) \ $$ is a polynomial of degree less than p. Since a polynomial of degree less than p cannot have any of the $$p^{th}$$ roots of unity as a zero, none of the $$\lambda$$ are 0, and the theorem is proved.