What is the non-asymptotic upper bound for the leading eigenvector of the random matrix?

Question

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!

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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}

Answer 1

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).