How to lower bound $\tau$ based on the expression of $H$?

Question

Let $A=\{a_{ij}\}_{1\le i,j\le n}$ be an $n$ by $n$ normalized symmetric Gaussian random matrix with $E[a_{ij}]=0$ and $E[a_{ij}^2]=1/n$. Ordering its eigenvalues by $\lambda_1\le \lambda_2\le \cdots \lambda_n$ with corresponding eigenvectors $v_1,\dots, v_n \in \mathbb{R}^n$. Let $u_0$ be a vector on $\mathbb{R}^n$ uniformly distributed on the unit sphere. (We also know that $v_i$ is uniformly distributed on the unit sphere for $i=1,\dots, n$.)

Define $H_j(t)=u_t\cdot v_j$ for $j=1,\dots, n$ and time $t\ge 0$, solving the following ODE with initial value $H_j(0)$: $$ \frac{1}{2}H_j'(t)=\sum_{i=1}^n[(\lambda_i-\lambda_j)H_i^2(t)]H_j(t) $$

Answer 1

Let us drop the calculations and formalism for a minute, and think about this problem geometrically.

We know the eigenvectors $v_1, ..., v_n$ are orthogonal, and correspond to $n$ points on the unit sphere, i.e. $\forall j,\ \|v_j\| = 1$. Since $u_0$ also has norm 1, the function $H_j(0)$ is the cosine of the angle $\alpha_j(0)$ between $u_0$ and $v_j$ [1]: $$ H_j(0) = u_0\cdot v_j = \|u_0\|\|v_j\|\cos\alpha_j(0) = \cos\alpha_j(0) $$

Now consider the ODE for $H_1$: $$ H_1^\prime(t) = 2H_1(t)\sum_{i=1}^n (\lambda_i - \lambda_1)H_i(t)^2 $$

For $u_0 = v_1$, we have $H_1(0) = 1$, and since the eigenvectors are orthogonal $H_i(0) = 0$ for all $i > 1$. Hence $H_1^\prime(0) = 0$, and it follows that $H_1^\prime(t) = 0$ for all $t$, i.e. $v_1$ is an invariant. Furthermore because in this case $H_1(t)=1$ is maximal, we have $\tau_\varepsilon = 0 < n^{2/3}$, and at least one counter-example to the proposition [2].

However to disprove the proposition, a single counter-example is not sufficient (the measure of a finite set is zero). Next we will show that $\tau_\varepsilon < n^{2/3}$ in a neighborhood of $v_1$ for any threshold $\varepsilon$ and dimension $n$.

Pick a threshold $0 < \varepsilon \leq 1$. Since $\lambda_i - \lambda_1 \geq 0$ for all $i$, all of the terms in the formula of the derivative $H_1^\prime$ above are non-negative, and therefore if we pick $u_0$ such that $H_1(0)\geq 0$ we have: $$ \forall t\geq 0,\ H_1(t)\geq H_1(0) = \cos\alpha_1(0) $$ Hence for any choice of $u_0$ at a small enough angle from $v_1$ such that $0\leq \alpha_1(0) \leq \cos^{-1}\varepsilon$, we have: $$ \forall t\geq 0,\quad H_1(t)\geq H_1(0) = \cos\alpha_1(0) \geq \varepsilon \quad\implies\quad \tau_\varepsilon = 0 $$ proving that $\tau_\varepsilon < n^{2/3}$ occurs with non-zero probability for any dimension $n$ and threshold $\varepsilon$, i.e. $P( \tau_\varepsilon < n^{2/3}) > 0$.

[1] Note that the only way to satisfy the assumption $\sum_{i=1}^nH_i(t)^2 = 1$ for all $t$ is to allow $\| u_t \| \leq 1$. So in general $H_j(t) = \|u_t\|\cos\alpha_j(t)$.
[2] Naturally I am assuming $\varepsilon \leq 1$.