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Let $a_1, \cdots, a_n\in\mathbb{R}^k$ be independent random vectors sampled from $N(0,\Sigma)$, where $\Sigma = \operatorname{diag}(\lambda_1, \cdots, \lambda_k)$ and $\lambda_1 \ge \cdots \ge \lambda_k$. We aim to establish a high probability bound on the largest eigenvalue or, equivalently, the spectral norm of the matrix $\sum_{i=1}^n \|a_i\|^2 a_ia_i^T$.

Can we show there exist positive constants $c_1$ and $c_2$ such that $\mathbb{P}(\|\sum_{i=1}^n \|a_i\|^2 a_ia_i^T\| \le c_1n\lambda_1) \ge c_2$?

For example, we can show the above result for the matrix $\sum_{i=1}^n a_ia_i^T$ using the bound on the norm of matrices with Gaussian entries.

Thanks!

Update: Can we get a bound of order $O(n\lambda_1)$ without $k$ if $k = O(n)$?

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    $\begingroup$ $\lVert a_i\rVert_2^2$ should concentrate around its expected value of $\text{tr}(\Sigma)$ fairly well. This is to say that $\sum_{i = 1}^n \lVert a_i\rVert^2 a_i a_i^T$ should be $\approx \text{tr}(\Sigma)$ larger than $\sum_i a_i a_i^T$. $\endgroup$
    – Mark Schultz-Wu
    Commented Jul 27 at 6:30

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