Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Start with $n+1$ positive numbers in increasing order, $\rho_1 \leq \rho_2 \leq ... \leq \rho_{n+1}$. Now define the symmetric tridiagonal matrix $M$ - $$M= \begin{bmatrix} a_1 & -b_2 & 0 & 0 &... &0\\ -b_2 & a_2 & -b_3 & 0 &... &0\\ 0 & -b_3 & a_2 & 0 &... &0\\ ...\\ 0 & 0 &0 & 0 &... &a_n\\ \end{bmatrix} $$ according to $$a_k = \left(\frac{1}{\sqrt{\rho_k}} + \frac{1}{\sqrt{\rho_{k+1}}}\right)^2,$$ and $$b_k = \frac{1}{2\sqrt{\rho_k}}\left(\frac{1}{\sqrt{\rho_{k-1}}} + \frac{1}{\sqrt{\rho_{k+1}}} + \frac{2}{\sqrt{\rho_k}} \right) = \frac{1}{2\sqrt{\rho_k}}\left(\sqrt{a_{k-1}} + \sqrt{a_k}\right)$$ I suspect this matrix is always positive semi-definite, I can prove it for small values of $n$. Any ideas on how to prove something like this for general $n$?

share|improve this question
    
If you can derive a formula for your matrix's Cholesky triangle, then you have a proof of positive (semi)definiteness. –  J. M. May 7 '13 at 8:10
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.