User:J D Bowen/sandbox

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

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

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

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.