I have a $N\times N$ sample covariance matrix of the form $S_{xx},$ which of rank $k$ such that $k\ll N.$ You can think about that $k=5$ but $N=1000$. Since $S_{xx}$ is a positive semidefinite matrix with at most $k$ nonzero eigenvalues, from the construction of $S_{xx},$ I know that the summation of of all eigenvalues of $S_{xx}$ is of order $O\left( N\right) ,$ i.e., proportional to $N.$ Now I establish that there exists a $N\times 1$ vector $% \mathbf{a}$ such that $\mathbf{a}^{\prime }S_{xx}\mathbf{a}\rightarrow 0$, as $N\rightarrow \infty,$ where $\mathbf{a}^{\prime }\mathbf{a} \neq 1$. Is there any rigorous way to find the order of $% \mathbf{a,}$ i.e., $\mathbf{a}=O\left( N^{-1}\right) $ or $\mathbf{a}% =O\left(N^{-(1+\epsilon )}\right) $ for some $\epsilon >0?$.
For simplicity, I can assume $\mathbf{a}=c \mathbf{1}_N$, where $\mathbf{1}_N$ is a $N \times 1$ vector of ones, I want to find the order of $c$ such that $ c^2\mathbf{1}_N^{\prime }S_{xx}\mathbf{1}_N\rightarrow 0$, for instance, $c$ could be proportional to $N^{-1}$. Many thanks!