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.



convergence of 2nd eigenvalue
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 $c/m\_{11}(\infty)$.


