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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.

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    $\begingroup$ Minor nitpick: what if $M(n) \equiv \mathrm{diag}(0,1,1)$? The first eigenvalue (I assume you mean the largest) of $A(n)$ remains bounded. $\endgroup$ – Willie Wong Oct 27 '16 at 13:50
  • $\begingroup$ @WillieWong: Oh, yeah. I need to fix positive definite instead of semi-definite. Thanks. $\endgroup$ – T. Amdeberhan Oct 27 '16 at 16:23
  • $\begingroup$ @ChristianRemling it looks that considering the inverse allows to get the asymptotics of the third eigenvalue, does not it? $\endgroup$ – Fedor Petrov Oct 27 '16 at 19:15
  • $\begingroup$ @FedorPetrov: I just posted that! Can we make it a joint work? $\endgroup$ – Christian Remling Oct 27 '16 at 19:17
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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 $c/m_{11}(\infty)$.

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I claim that the second eigenvalues tends to $\frac{\det(M)(M^{-1})_{33}}{M_{11}}.$ Indeed, the product of eigenvalues equals $n^{2(h_3-h_2)}\det(M(n))$, the largest eigenvalue grows as $n^{2(h_3-h_1)}M_{11}$, since $n^{-2(h_3-h_1)}A(n)=M_{11}e_{11}+o(1)$, and the largest eigenvalue of $A(n)^{-1}$ grows as $n^{2(h_2-h_1)}(M^{-1})_{33}$ by the same reason.

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Next attempt: By min-max, it's clear that the largest eigenvalue satisfies $\lambda_1=M_{11}n^{2(h_3-h_1)}(1+o(1))$ (test on $e_1$; nothing else in the matrix is as large as the $11$ entry, so this produces essentially the max of the quadratic form asymptotically).

I can now find the smallest eigenvalue by equivalently looking for the largest eigenvalue of $A^{-1}$, and then the same argument shows that $\lambda_3^{-1}=Dn^{2(h_2-h_1)}(1+o(1))$; here $D=C_{33}/\det M$, where $C_{33}$ denotes the corresponding cofactor of $M$, but in fact I only need to know that this quantity is positive, which is clear since $M^{-1}>0$ also.

Now the convergence of $\lambda_2$ follows by looking at the determinant; the limit equals $C_{33}/M_{11}$.

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