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Prove that the $5 \times 5$ Hilbert matrix, $H_5$, has five positive eigenvalues.

I know that $\lambda$ is an eigenvalue of $H_5$ iff $$\det (\lambda I_n - H_5) = 0$$

I computed $\lambda I_n - H_5$. Now I have to find the determinant of this and I believe this would take a really long time and that there must be an easier way of doing this.

How can I find the eigenvalues of this matrix?

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  • $\begingroup$ Try simplifying it by row and column operations (subtracting a multiple of one column from another). $\endgroup$ – almagest Apr 7 '16 at 18:10
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Let $H_n=[h_{ij}]$; since $h_{ij}=\int_0^1x^{i-1}x^{j-1}dx=<x^{i-1},x^{j-1}>$, $H_n$ is the Gramian matrix of the basis $\{1,x,\cdots,x^{n-1}\}$ of the Euclidean space (with the above scalar product) of polynomials of degree $<n$. Consequently, $H_n$ is symmetric $>0$. Then, its eigenvalues are $>0$.

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