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}



What is the non-asymptotic upper bound for the leading eigenvector of the random matrix?
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.](https://mathoverflow.net/a/359646/11260)) So for $n\gg 1$ this becomes a Gaussian, $P(x)\propto e^{-nx^2/2}$, with mean zero and variance $1/n$.


![](https://i.stack.imgur.com/lt1JW.jpg)
Plot of $P(x)$ for $n=2,3,5,10,20$ (larger $n$ gives more peaked distribution).


