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Fix a Gaussian random matrix $A$ with $E[A_{ij}]=0$ for $i, j=1,\dots n$ and $E[A_{ij}^2]=\frac{1}{n}$. Let $v_1$ be the leading eigenvector of $A$. What is the non-asymptotic upper bound for $v_1$, that is something like $$ P(v_1\cdot u\ge t)\le e^{-\alpha t} $$ where $u$ is distributed uniformly on the unit sphere.

Is there any reference for this tail probability? Thank you!

Let $\{v_1,v_2,\dots, v_n\}$ be the eigenvectors corresponding to the eigenvalues $\lambda_1,\dots, \lambda_n$ of a matrix $A$ from GOE. Each of the eigenvectors $v_1,\dots, v_n$ is distributed uniformly on \begin{equation} S_+^{n-1}:=\{x=(x_1,\dots, x_n): x_i\in R, \|x\|_2=1, x_1>0\}. \end{equation}

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  • $\begingroup$ There are some results on bounds like this for singular values, see chapters 4-6 of this if this is relevant in your setting. $\endgroup$ Commented Oct 11, 2022 at 20:45
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    $\begingroup$ @Mark Thanks. But I am looking for the bounds on the eigenvectors but not eigenvalues. Here is a fact eigenvectors are distributed uniformly on the sphere. $\endgroup$ Commented Oct 11, 2022 at 21:09
  • $\begingroup$ @CarloBeenakker Yes! I find this result and I put it in my question. So they are sub-Gaussian? Because they are distributed uniformly on the sphere? $\endgroup$ Commented Oct 11, 2022 at 21:15
  • $\begingroup$ @CarloBeenakker Sorry, I mean the inner product of the $v_1$ and one fix unit vector. $\endgroup$ Commented Oct 11, 2022 at 21:59
  • $\begingroup$ @CarloBeenakker So is it concentrated about $1/n$? $\endgroup$ Commented Oct 11, 2022 at 23:42

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We know the distribution of $x=v_1\cdot u$, with $v_1$ of length $n$ uniformly distributed on the unit $n$-sphere and $u$ an arbitrary unit vector. This distribution is given by $$P(x)=\frac{\Gamma(n/2)}{\sqrt{\pi}\,\Gamma(n/2-1/2)}(1-x^2)^{n/2-3/2}\,\theta(1-x^2),\;\;n>1,$$ with $\theta(x)$ the unit step function. (Here is one derivation.) So for $n\gg 1$ this becomes a Gaussian, $P(x)\propto e^{-nx^2/2}$, with mean zero and variance $1/n$.

Plot of $P(x)$ for $n=2,3,5,10,20$ (larger $n$ gives more peaked distribution).

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  • $\begingroup$ $P(x)dx$ is the probability that the random variable $v_1\cdot u$ is in the interval $(x,x+dx)$; so this is the probability density, not the cumulative distribution function $\endgroup$ Commented Oct 12, 2022 at 18:58
  • $\begingroup$ that paper addresses a completely different problem, where the eigenvectors are not uniformly distributed on the unit sphere. $\endgroup$ Commented Oct 12, 2022 at 19:44
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    $\begingroup$ section 2 is for the GOE, theorem 4.5 in section 4 is not for the GOE. $\endgroup$ Commented Oct 12, 2022 at 20:42
  • $\begingroup$ for large $n$ this probability saturates at 1, since it's a Gaussian of width $1/\sqrt n$ --- so you cannot make it any smaller. $\endgroup$ Commented Oct 14, 2022 at 20:43

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