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find the value $$A_{n}=\begin{vmatrix} 1-\dfrac{1}{(n+1)^2}&\dfrac{1}{2}&\dfrac{1}{3}&\cdots&\dfrac{1}{n+1}\\ \dfrac{1}{2}&\dfrac{1}{3}&\dfrac{1}{4}&\cdots&\dfrac{1}{n+2}\\ \dfrac{1}{3}&\dfrac{1}{4}&\dfrac{1}{5}&\cdots&\dfrac{1}{n+3}\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ \dfrac{1}{n+1}&\dfrac{1}{n+2}&\dfrac{1}{n+3}&\cdots&\dfrac{1}{2n+1} \end{vmatrix}$$ show that $\det(A_{n})=0$

I have prove $$\det(A_{1})=\begin{vmatrix} \dfrac{3}{4}&\dfrac{1}{2}\\ \dfrac{1}{2}&\dfrac{1}{3} \end{vmatrix}=0$$ and $$\det(A_{2})=\begin{vmatrix} \dfrac{8}{9}&\dfrac{1}{2}&\dfrac{1}{3}\\ \dfrac{1}{2}&\dfrac{1}{3}&\dfrac{1}{4}\\ \dfrac{1}{3}&\dfrac{1}{4}&\dfrac{1}{5} \end{vmatrix}=0$$ $$\det(A_{3})=\begin{vmatrix} \dfrac{15}{16}&\dfrac{1}{2}&\dfrac{1}{3}&\dfrac{1}{4}\\ \dfrac{1}{2}&\dfrac{1}{3}&\dfrac{1}{4}&\dfrac{1}{5}\\ \dfrac{1}{3}&\dfrac{1}{4}&\dfrac{1}{5}&\dfrac{1}{6}\\ \dfrac{1}{4}&\dfrac{1}{5}&\dfrac{1}{6}&\dfrac{!}{7} \end{vmatrix}=0$$

math110
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  • It suffices to find explicitly a vector in the kernel of the matrix. With the aid of WA, it seems that the matrix has nullity $1$ and there is an eigenvector with integer coordinates, the first of which is $n$. – lhf Apr 11 '20 at 14:05

1 Answers1

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Shifting the index, we have $$A_n = H_n -(1/n^2) e_1 e_1^T$$ where $H_n$ is the Hilbert matrix and $e_1$ is the column vector $[1,0,0,\dots,0]^T$.

By the matrix determinant lemma, $$ \det(A_n)=(1-(1/n^2)e_1^T H_n^{-1} e_1) \det(H_n) =(1-(1/n^2) (H_n^{-1})_{11}) \det(H_n) = 0 $$ because the $1,1$ entry of $H_n^{-1}$ is $n^2$.

lhf
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