Suppose I have two sequences $(A_n)$, $(B_n)$ of $n\times n$ symmetric real matrices (growing sizes) such that $||A_n-B_n||_F\to 0$ as $n \to \infty$, where $||\cdot||_F$ denotes the Frobenius norm. I want to show that
$$|\lambda_i(A_n)-\lambda_i(B_n)|\to 0 \hspace{1cm} \text{as} \hspace{1cm}n \to \infty$$ where $\lambda_i$ denotes the $i$th largest eigenvalue.
My first instinct was to invoke the continuity of eigenvalues on the space of real symmetric $n\times n$ matrices. Indeed using Gershgorin's Theorem one can show that for any two such matrices $A$,$B$ we have
$$|\lambda_i(A)-\lambda_i(B)|\leq n||A-B||_F$$
The problem is that $n$ is growing here, and so $n||A_n-B_n||_F$ may not be a null sequence.
As an alternative I thought about using Weyl's inequalities:
$$\lambda_i(B_n)+\lambda_n(A_n-B_n) \leq \lambda_i(A_n)\leq \lambda_i(B_n)+\lambda_1(A_n-B_n)$$
or
$$\lambda_n(A_n-B_n) \leq \lambda_i(A_n)-\lambda_i(B_n)\leq \lambda_1(A_n-B_n) \hspace{1cm} (*)$$
Now, because $A_n-B_n$ is real symmetric, its singular values are the absolute values of its eigenvalues. Furthermore, the largest singular value of $A_n-B_n$ (spectral norm) is dominated by the Frobenius norm, i.e. $\sigma_1(A_n-B_n)\leq ||A_n-B_n||_F$. Hence the LHS and RHS of $(*)$ both converge to zero as $n\to \infty$. The desired result follows.
Is this correct?
Thanks a lot for your help.