Does this sequence of eigenvalues converge? Let $(A_n) $ be a sequence of $k\times k$ real positive-definite matrices ($k$ fixed) and suppose that $A_n\to A$ for some positive-definite matrix $A$, where convergence is element wise.
Let $e_i$ denote the $k\times 1$ vector having 1 in the $i$th position and zeros elsewhere, and define
$$B_n:=A^{1/2}_n ( ne_1e'_1+e_2e_2')A^{1/2}_n$$
Question: Can we show the convergence of $\lambda_2(B_n)$? ($\lambda_2(C)$ denotes the second largest eigenvalue of symmetric matrix $C$).
From Weyl's inequalities we have $0\leq \lambda_2(B_n)\leq e_2'A_ne_2$ so the sequence $\lambda_2(B_n)$ is bounded. But I don't see how to go further.
Any counterexample?
 A: Yes, $\lambda_2(B_n)$ exists and is equal to $\frac{\det A[1:2,\,1:2]}{a_{11}}$. More generally, suppose $k>1$ and $P_n=A_n^{1/2}\left(\sum_{i=1}^{k-1}ne_ie_i^T+e_ke_k^T\right)A_n^{1/2}$. We will prove in the below that
$$
\lim_{n\to\infty}\lambda_k(P_n)
=\frac{\det A[1:k,\,1:k]}{\det A[1:(k-1),\,1:(k-1)]}.
$$
Let $X_n\in\mathbb R^{n\times k}$ be the submatrix taken from the first $k$ columns of $A_n^{1/2}$ and $D_n=\operatorname{diag}(n,\ldots,n,1)\in\mathbb R^{k\times k}$. Then $P_n=(X_nD_n^{1/2})(D_n^{1/2}X_n^T)$. By Sylvester's eigenvalue theorem, $P_n$ share the same multi-set of nonzero eigenvalues with the $k\times k$ matrix $(D_n^{1/2}X_n^T)(X_nD_n^{1/2})$. Therefore
\begin{aligned}
\lambda_k(P_n)
&=\lambda_\min\left(D_n^{1/2}X_n^TX_nD_n^{1/2}\right)\\
&=\frac{1}{\lambda_\max\left(\left(D_n^{1/2}X_n^TX_nD_n^{1/2}\right)^{-1}\right)}\\
&=\left\|D_n^{-1/2}(X_n^TX_n)^{-1}D_n^{-1/2}\right\|_2^{-1}.
\end{aligned}
So, if we denote by $X\in\mathbb R^{n\times k}$ the submatrix taken from the first $k$ columns of $A^{1/2}$, then $\lim_{n\to\infty}\lambda_k(P_n)=y_{kk}^{-1}$, where $Y=(X^TX)^{-1}$. As $A^{1/2}$ is symmetric, $X^T$ is precisely the submatrix taken from the first $k$ rows of $A^{1/2}$. Therefore $X^TX=A[1:k,\,1:k]$ and by Laplace expansion along the last row or column of $A[1:k,\,1:k]$, we obtain
$$
y_{kk}=\frac{\det A[1:(k-1),\,1:(k-1)]}{\det A[1:k,\,1:k]}.
$$
Now the result follows.
