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A determinant

Let n be a natural number and let \lambda_1, \; \lambda_2, \dots, \lambda_n \in \mathbb{R} be mutually distinct real numbers none of which equals 0 , -1 , \dots , -n+1. Prove that

    \[\begin{vmatrix} \frac{1}{\lambda_1} & \frac{1}{\lambda_2} & \cdots & \frac{1}{\lambda_n} \\ \frac{1}{\lambda_1+1}& \frac{1}{\lambda_2+1} & \cdots & \frac{1}{\lambda_n+1} \\ \vdots & \vdots & \ddots & \vdots \\ \frac{1}{\lambda_1 + n -1}& \frac{1}{\lambda_2+n-1} & \cdots & \frac{1}{\lambda_n+n-1} \end{vmatrix} \neq 0\]

Solution

We need to show that the rows of the above matrix are linearly idependent. Consider a linear combination with coefficients c_0 , c_1, \dots, c_{n-1} \in \mathbb{R}. Let

    \[f\left ( \lambda_i \right ) = \frac{c_0}{\lambda_i} + \frac{c_1}{\lambda_i+1} + \cdots + \frac{c_{n-1}}{\lambda_i + n -1} =0\]

for all i=1, 2, \dots, n-1. Adding these fractions together we get that

    \[f\left ( \lambda_i \right ) = \frac{P\left ( \lambda \right )}{\lambda \left ( \lambda +1 \right )\cdots \left ( \lambda +n -1 \right )} =0\]

where P(\lambda) is a polynomial in \lambda of degree at most n-1 and f(\lambda_i)=0 for i=1, 2, \dots, n. Since \lambda_i does not equal 0 , -1 , \dots , -n +1 we see that P(\lambda_i) =0 for i=1, 2, \dots, n-1. Hence P is a polynomial of degree at most n-1 with at least n roots. Therefore P=0 and f=0. We will show that this implies

    \[c_0=c_1=\cdots=c_{n-1} =0\]

Suppose , on the contrary , that c_i \neq 0. Then

    \[\lim_{\lambda \rightarrow -i } \frac{c_i}{\lambda + i} = \pm \infty\]

and hence \lim \limits_{\lambda \rightarrow -i} f(\lambda) =\pm \infty contradicting the fact that f=0.

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