1
$\begingroup$

Let $A$ a $n\times n$ symmetric positive-definite matrix of eigenvalues $\lambda_n \leq \ldots \leq \lambda_1$. Then we have the estimate: $$ \left(\frac{n-1}{\textrm{tr } A}\right)^{n-1} \textrm{det } A \leq \lambda_n, $$ by applying the arithmetic-geometric mean inequality on the set $\{\lambda_1,\ldots, \lambda_{n-1}\}$ of cardinality $n-1$. This is very coarse in general, but for a diagonal matrix $\ell \textrm{Id}$, it turns out to become $$ \left(1-\frac{1}{n}\right)^{n-1} \ell \leq \ell, $$ so basically we are only off from a $e^{-1}$ factor.

Is there an example where this bound is sharp? If not is possible to get rid of this constant factor by tweaking things a bit?

$\endgroup$
1
$\begingroup$

The idea is the following. First try $n=2$ and you will find out the inequality is sharp and it becomes to be an equality if $\lambda_n=0$ (though $\lambda_n>0$ but we can take $\lambda\to 0^+$). Hence our first guess is that it is sharp when $\lambda_n=0$. Now we try to argue it.

Note that your inequality is equivalent to the following one $$ \lambda_1\cdots \lambda_{n-1}\leqslant \left(\frac{\lambda_1+\cdots+\lambda_{n-1}+\lambda_{n}}{n-1}\right)^{n-1} $$ If we take $\lambda_n=0$, the inequality (just A-G inequality) is still true and the equality holds when $\lambda_1=\cdots=\lambda_{n-1}$. Now everything is clear. The inequality is sharp by taking $\lambda_1=\cdots=\lambda_{n-1}$ and $\lambda\to 0^+$.

$\endgroup$
1
  • $\begingroup$ I see. I didn't think of letting the least eigenvalue tends to zero. Thanks. $\endgroup$ – user70925 Oct 24 '17 at 22:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.