Let $\Sigma_n$ be a sequence of $n\times n$ (growing size) positive definite matrices and suppose that $\lambda^{1}_{n} \to 0$, where $\lambda^{1}_{n}$ denotes the largest eigenvalue of $\Sigma_n$.
Using the eigendecomposition of $\Sigma_n$ we have that $\left\lVert \Sigma_n \right\rVert_F \leq \sqrt{n} \lambda^{1}_{n}$ so I cannot deduce that $\left\lVert \Sigma_n \right\rVert_F \to 0$, where $\left\lVert \cdot \right\rVert_F$ denotes the Frobenius norm.
But if $Q_n$ is an arbitrary bounded sequence of $n\times n$ (growing size) matrices then I can obtain the following inequality:
$$ \left\lVert \Sigma_n Q_n \right\rVert_F \leq \left\lVert \Sigma_n \right\rVert_2 \left\lVert Q_n \right\rVert_F = \lambda^{1}_{n} \left\lVert Q_n \right\rVert_F \to 0 $$ where $\left\lVert \cdot \right\rVert_2$ denotes the 2-norm.
What is going on here? Why is the matrix sequence $\Sigma_n$ not converging to zero but converges to zero when multiplied by an arbitrary bounded matrix sequence?
Since $Q_n$ is bounded in norm but its matrix size increases it seems natural to think that most of its entries must get very small as $n\to\infty$, but I believe this must happen at a rate faster than $1/\sqrt{n}$ in order to kill the growth rate $\sqrt{n}$ of $\left\lVert \Sigma_n \right\rVert_F$. But how can I show this rigorously?
Thanks a lot for your help.
Let $$\Sigma_n=\frac{1}{n^{\frac{1}{3}}}I_n$$ Then $$|\Sigma_n|_F=\sqrt{\Sigma \Sigma (a_{mn})^2}$$ $$=\sqrt{n \frac{1}{n^{2/3}}}$$ $$=n^{\frac{1}{6}}$$ although $\lambda_1=\frac{1}{n^{\frac{1}{3}}}$