The question is from the book 'Statistical Models and Methods for Financial Markets', Page 43.
Suppose $\boldsymbol{x_1,..., x_n}$ are $n$ independent observations from a multivariate population with mean $\boldsymbol{μ}$ and covariance matrix $\boldsymbol{V}$
The mean vector μ can be estimated by $$\bar{\boldsymbol{x}} = \sum_{i=1}^n \boldsymbol{x}_n$$
And the covariance matrix can be estimated by
$$
\hat{\boldsymbol{V}} = \sum_{i=1}^n (\boldsymbol{x}_i - \bar{\boldsymbol{x}}) (\boldsymbol{x}_i - \bar{\boldsymbol{x}})^T/(n-1)
$$
which is the sample covariance matrix.
Let $\hat{\boldsymbol{a}}_j = (\hat{a}_{1j} ,..., \hat{a}_{pj} )^T$ be the eigenvector corresponding to the $j$th largest eigenvalue $\hat{\lambda_j}$ of the sample covariance
matrix $\hat{\boldsymbol{V}}$.
For fixed $p$, the following asymptotic results have been established
under the assumption $λ1 > ··· > λp > 0$ as $n\to \infty$:
(i) $\sqrt{n} (\hat{\lambda}_i -\lambda_i)$ has a limiting $N(0, 2\lambda_i^2 )$ distribution.
Moreover, for $i \neq j$, the
limiting distribution of $\sqrt{n} (\hat{\lambda}_i -\lambda_i,\hat{\lambda}_j -\lambda_j)$ is that of two independent normals.
(ii) $\sqrt{n} (\hat{\boldsymbol{a}}_i -\boldsymbol{a}_i)$ has a limiting normal distribution with mean $\boldsymbol{0}$ and covariance
matrix
$$
\sum_{h\neq i} \frac{\lambda_i \lambda_h}{(\lambda_i - \lambda_h)^2} \boldsymbol{a}_h \boldsymbol{a}_h^T
$$
It seems like it's a direct application of multivariate CLT,but I cannot derive these two statements.