Suppose $0<a_1<a_2<\cdots <a_n$,$\lambda_1<\lambda_2<\cdots<\lambda_n$and \begin{align*} A=\left({a_i}^{\lambda_j}\right)= \begin{pmatrix} {a_1}^{\lambda_1}&{a_1}^{\lambda_2}&\ldots &{a_1}^{\lambda_n}\\ {a_2}^{\lambda_1}&{a_2}^{\lambda_2}&\ldots &{a_2}^{\lambda_n}\\ \vdots &\vdots&\ldots&\vdots\\ {a_n}^{\lambda_1}&{a_n}^{\lambda_2}&\ldots &{a_n}^{\lambda_n} \end{pmatrix} \end{align*} Prove that the determinant of A is greater than zero,$\mathrm{i,e.}$ $\det A>0$.
It is easy to check that the assertion is true when $n=2$,for when $n=2$, \begin{align*} \det A&=\begin{vmatrix} {a_1}^{\lambda_1}&{a_1}^{\lambda_2}\\ {a_2}^{\lambda_1}&{a_2}^{\lambda_2} \end{vmatrix} ={a_1}^{\lambda_1}{a_2}^{\lambda_2}-{a_1}^{\lambda_2}{a_2}^{\lambda_1}\\ &=\exp\{\lambda_1\ln a_1+\lambda_2 \ln a_2\}-\exp\{\lambda_2\ln a_1+\lambda_1 \ln a_2\} \end{align*} Since $(\lambda_1\ln a_1+\lambda_2 \ln a_2)-(\lambda_2\ln a_1+\lambda_1 \ln a_2)=(\lambda_2-\lambda_1) (\ln a_2-\ln a_1)>0$,then the case for $n=2$holds.While about the general case for $n>2$,I have no idea. Can anyone help me?It's a homework.