Hilbert matrices are positive definite

(Argument stolen from Yahoo Answers)

It can be seen by using its integral form:

\displaystyle H_{ij} = \int_0^1 dx\,x^{i+j-2}

Then we can express the positive definite condition as

\displaystyle \sum_{i,j} x_i H_{ij} x_j > 0

for \mathbf{x} \ne 0.

Replacing and operating:

\displaystyle \sum_{i,j} x_i \int_0^1 du\,u^{i+j-2} x_j = \int_0^1 du \sum_{i,j} x_i u^{i+j-2} x_j

\displaystyle = \int_0^1 du \sum_{i,j} x_i u^{i-1} x_j u^{j-1}

\displaystyle = \int_0^1 du \sum_i x_i u^{i-1} \sum_j x_j u^{j-1}

\displaystyle = \int_0^1 du \left( \sum_i x_i u^{i-1} \right) \left( \sum_j x_j u^{j-1} \right)

\displaystyle = \int_0^1 du \left( \sum_i x_i u^{i-1} \right)^2

The term inside the parentheses is just a (polynomial) function of u:

\displaystyle = \int_0^1 du\,p(u)^2

The only way the integral can be zero is if p(u) is identically zero:

\displaystyle \sum_i x_i u^{i-1} = 0 \implies \forall i: x_i = 0.

QED

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One thought on “Hilbert matrices are positive definite

  1. Axel says:

    Great demonstration!

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