TITLE: How prove this $det\left(\frac{1}{\lambda^2_{i}+t\lambda_{i}\lambda_{j}+\lambda^2_{j}}\right)_{n\times n}>0,-2<t<2$
QUESTION [5 upvotes]: Question:
Show that for $t\in (-2,2)$ and $0<\lambda_1<\lambda_2<\ldots<\lambda_n$ we have
$$det(A)=det\left(\dfrac{1}{\lambda^2_{i}+t\lambda_{i}\lambda_{j}+\lambda^2_{j}}\right)_{n\times n}>0$$
My try: few days ago,I have ask this problem How prove this matrix $\det (A)=\left(\frac{1}{\ln{(a_{i}+a_{j})}}\right)_{n\times n}\neq 0$
But I want try use achille hui methods,so 
$$\dfrac{1}{\lambda^2_{i}+t\lambda_{i}\lambda_{j}+\lambda^2_{j}}=\int_{0}^{\infty}f(\lambda_{i},\lambda_{j},x)dx$$
But I can't find this $f$. Thank you very much!
Now Sanchez has prove for $t\le 0$ then $\det(A)>0$,so other case is true? Thank you

REPLY [2 votes]: When $t\in(0,2)$, you may reduce the problem to the case $t=0$ as follows. Write
\begin{align*}
\frac{1}{\lambda_i^2+t\lambda_i\lambda_j+\lambda_j^2}
&=\frac{1}{(\lambda_i+\lambda_i)^2-2(1-\frac t2)\lambda_i\lambda_j}\\
&=\frac{1}{(\lambda_i+\lambda_i)^2}
\frac{1}{1-(1-\frac t2)\frac{2\lambda_i\lambda_j}{(\lambda_i+\lambda_i)^2}}\\
&=\frac{1}{(\lambda_i+\lambda_i)^2}
\sum_{k=0}^\infty\left[\left(1-\frac t2\right)\frac{2\lambda_i\lambda_j}{(\lambda_i+\lambda_i)^2}\right]^k\\
&=\sum_{k=0}^\infty\left(1-\frac t2\right)^k\frac{2^k\lambda_i^k\lambda_j^k}{(\lambda_i+\lambda_i)^{k+2}}.
\end{align*}
So, it suffices to show that the matrix $B=\left[\frac{\lambda_i^k\lambda_j^k}{(\lambda_i+\lambda_i)^{k+2}}\right]_{i,j=1,2,\ldots,n}$ is positive definite for all integer $k\ge0$. Since $B$ is congruent to $\left[\frac1{(\lambda_i+\lambda_i)^{k+2}}\right]_{i,j=1,2,\ldots,n}$ and Hadamard products of positive definite matrices are positive definite, it suffices to prove that $C=\left[\frac1{\lambda_i+\lambda_i}\right]_{i,j=1,2,\ldots,n}$ is positive definite. But this becomes the case with $t=0$ and $\lambda_i$ replaced by $\sqrt{\lambda_i}$ in the original problem statement.