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.

  • $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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

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

  • $\begingroup$ 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

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