# 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:

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

For $t \le 0$: We want to show that for any $a_i,a_j \in \mathbb{R}$, $$\sum \frac{a_ia_j}{\lambda_i^2 + t \lambda_i \lambda_j + \lambda_j^2} \ge 0$$ $$\sum a_i a_j \int_0^{\infty} e^{-(\lambda_i^2 + t \lambda_i \lambda_j + \lambda_j^2)u} du \ge 0$$ $$\int_0^{\infty} \sum a_i a_j e^{-(\lambda_i^2 + t \lambda_i \lambda_j + \lambda_j^2)u} du \ge 0$$ So it suffices to show that for all $u \ge 0$, $$\sum a_i a_j e^{-(\lambda_i^2 + t \lambda_i \lambda_j + \lambda_j^2)u} \ge 0$$ Let $u = v^2$, $v \ge 0$. By replacing $\lambda_i v$ with $\lambda_i$, we really need to show: $$\sum a_i a_j e^{-(\lambda_i^2 + t \lambda_i \lambda_j + \lambda_j^2)} \ge 0$$ By replacing $a_ie^{-\lambda_i^2}$ by $a_i$ , we really need to show: $$\sum a_i a_j e^{-t \lambda_i \lambda_j} \ge 0$$ To prove the last inequality, expand the exponential function in Taylor series, then $$e^{-t\lambda_i \lambda_j} = \sum_{k=0}^{\infty} \frac{(-t)^k(\lambda_i \lambda_j)^k}{k!}$$ Therefore $$\sum a_i a_j e^{-t \lambda_i \lambda_j} = \sum_k \frac{(-t)^k}{k!} \sum_{i,j} a_ia_j (\lambda_i \lambda_j)^k = \sum_k \frac{(-t)^k}{k!} (\sum_i a_i \lambda_i^k)^2 \ge 0$$
• Thank you,But for $t\in (0,2)$ we can prove it? Nov 28, 2013 at 16:40
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.