0
$\begingroup$

We have $a_1,a_2,...,a_n\in (0,1)$ and matrix M= \begin{bmatrix}2a_1&a_2&a_3&.&.\\a_2&2a_2&a_3&.&.\\a_3&a_3&2a_3&.&.\\.&.&.&.&.\end{bmatrix}

We need to check if M is positive definite.

I am trying to evaluate it's determinant as a polynomial in $a_i$ as principal minor are of the same type. And using that frame a condition for positive definiteness of M.

$\endgroup$
  • $\begingroup$ What is the question? "Is there a better way to do that?" $\endgroup$ – Federico Poloni May 22 at 7:55
  • $\begingroup$ @FedericoPoloni no, I am unable to find the determinant $\endgroup$ – mayank May 22 at 7:56
  • 1
    $\begingroup$ Try the case $n = 2$ and you'll find that the answer depends on whether $a_2$ is smaller than $4a_1$ or not. I haven't tried $n = 3$ but please try it first and see if you can observe a pattern. If yes, then try to prove it; otherwise I don't know what kind of answer you should expect for this question - in other words, what does "we need to check" mean in your post. $\endgroup$ – WhatsUp May 22 at 13:08
  • 1
    $\begingroup$ A right forum for such type questions is math.stackexchange.com $\endgroup$ – user64494 May 22 at 16:13
  • $\begingroup$ @user64494 Could you expand on your reasoning? $\endgroup$ – Yemon Choi May 22 at 22:03
2
$\begingroup$

If $D_n$ is the leading principal minor of order $n$, then it seems to me you should have $$D_n = 2 a_n D_{n-1} - a_n^2 D_{n-2}$$

$\endgroup$
  • $\begingroup$ Can you give some hint how you get D_{n-2} term $\endgroup$ – mayank May 22 at 17:02
  • $\begingroup$ Actually I found it empirically. But it should be provable... $\endgroup$ – Robert Israel May 22 at 18:11
  • $\begingroup$ Yeah it seems correct, but wasn't visible to me initially. Thanks a lot. $\endgroup$ – mayank May 23 at 0:24

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.