User:Sujaykazi/sandbox

The Alon-Boppana bound provides a lower bound on the second-largest eigenvalue of the adjacency matrix of a graph.

Theorem statement
Let $$G$$ be a $$d$$-regular graph on $$n$$ vertices, and let $$A$$ be its adjacency matrix. Let $$\lambda_1 \ge \lambda_2 \ge ... \ge \lambda_n$$ be its eigenvalues. Then $$\lambda_2 \ge 2\sqrt{d-1} - o(1).$$

More precisely, the $$o(1)$$ term refers to the asymptotic behavior as $$n$$ grows without bound while $$d$$ remains fixed. Also, the fact that $$G$$ is $$d$$-regular implies that $$\lambda_1 = d,$$ with the all-ones vector $$\mathbf{1}$$ being the corresponding eigenvector.

First proof
Let the vertex set be $$V.$$ By the min-max theorem, it suffices to construct a nonzero vector $$z\in\mathbb{R}^{|V|}$$ such that $$z^{\text{T}}\mathbf{1} = 0$$ and $$\frac{z^{\text{T}}Az}{z^{\text{T}}z} \ge 2\sqrt{d-1} - o(1).$$

Pick some value $$r\in\mathbb{N}.$$ For each vertex in $$V,$$ define a vector $$f(v)\in\mathbb{R}^{|V|}$$ as follows. Each component will be indexed by a vertex $$u$$ in the graph. For each $$u,$$ if the distance between $$u$$ and $$v$$ is $$k,$$ then the $$u$$-component of $$f(v)$$ is $$f(v)_u = w_k = (d-1)^{-k/2}$$ if $$k\le r-1$$ and $$0$$ if $$k\ge r.$$ We claim that any such vector $$x = f(v)$$ satisfies


 * $$\frac{x^{\text{T}}Ax}{x^{\text{T}}x} \ge 2\sqrt{d-1}\left(1 - \frac{1}{2r}\right).$$

To prove this, let $$V_k$$ denote the set of all vertices that have a distance of exactly $$k$$ from $$v.$$ First, note that


 * $$x^{\text{T}}x = \sum_{k=0}^{r-1}|V_k|w^2_k.$$

Second, note that


 * $$x^{\text{T}}Ax = \sum_{u\in V}x_u \sum_{u'\in N(u)}x_{u'} \ge \sum_{k=0}^{r-1}|V_k|w_k\left[w_{k-1} + (d-1)w_{k+1}\right] - (d-1)|V_{r-1}|w_{r-1}w_r,$$

where the last term on the right comes from a possible overcounting of terms in the initial expression. The above then implies


 * $$x^{\text{T}}Ax \ge 2\sqrt{d-1}\left(\sum_{k=0}^{r-1}|V_k|w^2_k - \frac{1}{2}|V_{r-1}|w^2_r\right),$$

which, when combined with the fact that $$|V_{k+1}| \le (d-1)|V_k|$$ for any $$k,$$ yields


 * $$x^{\text{T}}Ax \ge 2\sqrt{d-1}\left(1 - \frac{1}{2r}\right)\sum_{k=0}^{r-1}|V_k|w^2_k.$$

The combination of the above results proves the desired inequality.

For convenience, define the $$(r-1)$$-ball of a vertex $$v$$ to be the set of vertices with a distance of at most $$r-1$$ from $$v.$$ Notice that the entry of $$f(v)$$ corresponding to a vertex $$u$$ is nonzero if and only if $$u$$ lies in the $$(r-1)$$-ball of $$x.$$

The number of vertices within distance $$k$$ of a given vertex is at most $$1 + d + d(d-1) + d(d-1)^2 + ... + d(d-1)^{k-1} = d^k + 1.$$ Therefore, if $$n \ge d^{2r-1} + 2,$$ then there exist vertices $$u, v$$ with distance at least $$2r.$$

Let $$x = f(v)$$ and $$y = f(u).$$ It then follows that $$x^{\text{T}}y = 0,$$ because there is no vertex that lies in the $$(r-1)$$-balls of both $$x$$ and $$y.$$ It is also true that $$x^{\text{T}}Ay = 0,$$ because no vertex in the $$(r-1)$$-ball of $$x$$ can be adjacent to a vertex in the $$(r-1)$$-ball of $$y.$$

Now, there exists some constant $$c$$ such that $$z = x - cy$$ satisfies $$z^{\text{T}}\mathbf{1} = 0.$$ Then, since $$x^{\text{T}}y = x^{\text{T}}Ay = 0,$$


 * $$z^{\text{T}}Az = x^{\text{T}}Ax + c^2y^{\text{T}}Ay \ge 2\sqrt{d-1}\left(1 - \frac{1}{2r}\right)(x^{\text{T}}x + c^2y^{\text{T}}y) = 2\sqrt{d-1}\left(1 - \frac{1}{2r}\right)z^{\text{T}}z.$$

Finally, letting $$r$$ grow without bound while ensuring that $$n \ge d^{2r-1} + 2$$ (this can be done by letting $$r$$ grow sublogarithmically as a function of $$n$$) makes the error term $$o(1)$$ in $$n.$$

Second proof (slightly weaker statement)
This proof will demonstrate a slightly weaker result, but it provides better intuition for the source of the number $$2\sqrt{d-1}.$$ Rather than showing that $$\lambda_2 \ge 2\sqrt{d-1} - o(1),$$ we will show that $$\lambda = \max(|\lambda_2|, |\lambda_n|) \ge 2\sqrt{d-1} - o(1).$$

First, pick some value $$k\in\mathbb{N}.$$ Notice that the number of closed walks of length $$2k$$ is


 * $$\text{tr}A^{2k} = \sum_{i=1}^{n}\lambda^{2k}_i\le d^{2k} + n\lambda^{2k}.$$

However, it is also true that the number of closed walks of length $$2k$$ starting at a fixed vertex $$v$$ in a $$d$$-regular graph is at least the number of such walks in an infinite $$d$$-regular tree, because an infinite $$d$$-regular tree can be used to cover the graph. By the definition of the Catalan numbers, this number is at least $$C_k(d-1)^k,$$ where $$C_k = \frac{1}{k+1}\binom{2k}{k}$$ is the $$k^{\text{th}}$$ Catalan number.

It follows that


 * $$\text{tr}A^{2k} \ge n\frac{1}{k+1}\binom{2k}{k}(d-1)^k$$
 * $$\implies \lambda^{2k} \ge \frac{1}{k+1}\binom{2k}{k}(d-1)^k - \frac{d^{2k}}{n}.$$

Letting $$n$$ grow without bound and letting $$k$$ grow without bound but sublogarithmically in $$n$$ yields $$\lambda \ge 2\sqrt{d-1} - o(1).$$



The intuition for the number $$2\sqrt{d-1}$$ comes from considering the infinite $$d$$-regular tree. This graph is a universal cover of $$d$$-regular graphs, and it has spectral radius $$2\sqrt{d-1}.$$

Saturation of the bound
A graph that essentially saturates the Alon-Boppana bound is called a Ramanujan graph. More precisely, a Ramanujan graph is a $$d$$-regular graph such that $$|\lambda_2|, |\lambda_n| \le 2\sqrt{d-1}.$$