TITLE: convergence of 2nd eigenvalue
QUESTION [9 upvotes]: Fix $0<h_1<h_2<h_3<1$ reals. All matrices below are $3\times3$ real.
Suppose the sequence of matrices $M(n)$ are symmetric positive definite and these converge (point-wise) to a symmetric positive definite matrix $M$ (point-wise). Assume that the eigenvalues of $M(n)$ converge to that of $M$. 

QUESTION. Is it true that the 2nd eigenvalues of the product
  $$A(n):=\begin{pmatrix} n^{h_3-h_1}&0&0 \\ 0&1&0 \\ 0&0& n^{h_1-h_2} \end{pmatrix}\,M(n)\,
\begin{pmatrix} n^{h_3-h_1}&0&0 \\ 0&1&0 \\ 0&0& n^{h_1-h_2} \end{pmatrix}$$
  converge (to a finite number)?

The 1st eigenvalues of $A(n)$ diverge to $\infty$ while the 3rd eigenvalues of $A(n)$ converge to $0$, as $n\rightarrow\infty$. It seems that the 2nd eigenvalues can be shown to be bounded.

REPLY [5 votes]: Let $a = n^{h_3 - h_1} \to + \infty$ and $b = n^{h_1 - h_2} \to 0$. 
Note that $ab = n^{h_3 - h_2} \to +\infty$, but slower than $a$.
If $$M(n) = \pmatrix{m_{11}(n) & m_{12}(n) & m_{13}(n)\cr m_{12}(n) & m_{22}(n) & m_{23}(n)\cr
m_{13}(n) & m_{23}(n) & m_{33}(n)}$$ the characteristic polynomial of $A(n)$ is
$$ \eqalign{P_n(\lambda) = &a^2 (-m_{11}(n) \lambda^2 + (m_{11}(n) m_{22}(n) - m_{12}(n)^2) \lambda)\cr 
&+ a^2 b^2 (m_{11}(n) m_{33}(n) - m_{13}(n)^2) \lambda - \det(M(n)))\cr
&+ (\lambda^3 - m_{22}(n) \lambda^2) - b^2 (m_{33}(n) \lambda^2 - (m_{22}(n)m_{33}(n)-
m_{23}(n)^2)\lambda \cr
&= a^2 (-m_{11}(\infty) \lambda^2 + (m_{11}(\infty) m_{22}(\infty) - m_{12}(\infty)^2) \lambda) + o(a^2)}$$
as $n \to \infty$ with $\lambda$ fixed.
Note that $m_{11}(\infty)$ and $c = m_{11}(\infty) m_{22}(\infty) - m_{12}(\infty)^2$ are strictly positive.  If $0 < \lambda < c/m_{11}(\infty)$, then
$P_n(\lambda) > 0$ for sufficiently large $n$, while if $\lambda > c/m_{11}(\infty)$, $P_n(\lambda) < 0$ for sufficiently large $n$.  Thus  any interval
$c/m_{11}(\infty)-\epsilon, c/m_{11}(\infty)+\epsilon)$ will contain an eigenvalue for sufficiently large $n$.  We conclude that the second eigenvalue does converge, to $c/m_{11}(\infty)$.