Assume $Q$ is a positive definite random matrix such that $0 < \lambda_{\min}(Q)....\leq \lambda_{\max}(Q) \leq 1$ holds. I want to show that \begin{align} E\left[\frac{\lambda_{\min}(Q)}{\lambda_{\min}(Q)+\lambda_{\max}(Q)}\right] \geq \frac{\lambda_{\min}(E[Q])}{\lambda_{\min}(E[Q])+\lambda_{\max}(E[Q])} \end{align}

Or I want to show the following Kantorovich type inequality for expected values: \begin{align} E\left[\frac{(y^Ty)^2}{(y^TQy)(y^TQ^{-1}y)}\right] \geq \frac{4\lambda_{\min}(E[Q])\lambda_{\max}(E[Q])}{\left[\lambda_{\min}(E[Q])+\lambda_{\max}(E[Q])\right]^2} \end{align}


1 Answer 1


$\newcommand\lmax{\lambda_{\max}(P)}\newcommand\lmin{\lambda_{\min}(P)}$Your first displayed inequality for all positive definite random matrices $Q$ means exactly that the function $$P\mapsto\frac\lmin{\lmin+\lmax}$$ on the set of all positive definite matrices is convex. Looking at just the diagonal positive definite matrices, we see that this convexity implies the convexity of $\frac x{x+y}$ in $x,y>0$ such that $x\le y$. However, $\frac x{x+y}$ is not convex in $x\in(0,y]$ for any $y>0$.

So, your first displayed inequality does not hold for some positive definite random matrices $Q$.

The second inequality is also false in general. E.g., let $P(Q=D(1,a))=1/2=P(Q=D(a,1))$, where $0<a<1$ and $D(s,t)$ stands for the diagonal $2\times2$ matrix with the diagonal entries $s,t$. Then $EQ=D(\frac{1+a}2,\frac{1+a}2)$ and hence the right-hand side of your second inequality is $1$. On the other hand, for any $2\times1$ matrix $y$ with both entries nonzero, the left-hand side of your second inequality goes to $0$ as $a\downarrow0$. So, your second inequality does not hold.

  • $\begingroup$ Thanks. Actually, my hunch was that the first inequality is not correct. In my case, I just need to show the second inequality. $\endgroup$
    – Ripon
    Commented Sep 16, 2020 at 21:51
  • $\begingroup$ @SarowarMorshedRipon : Your second inequality also fails to hold in general. $\endgroup$ Commented Sep 16, 2020 at 23:50

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