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I am currently reading the book of Akhiezer http://www.maths.ed.ac.uk/~aar/papers/akhiezer.pdf and I saw on page 1 that a sequence $\{s_n\}_{n=0}^\infty$ of real numbers is positive definite if $$ \sum_{i,j=0}^{n}s_{i+j}c_ic_j \ge 0,$$ for any $n\ge 0$ and $c_0, c_1,\cdots,c_n \in \mathbb{R}.$

He now further said that the the positive property of this sequence $\{s_n\}_{n=0}^\infty$ is equivalent to the fact that the all determinants $D_n$ of each associated Hankel matrix $$H_n=(s_{i+j})_{0\le i,j \le n},$$ $D_n= \det(H_n)> 0$ for all $ n\ge 0$.

My question is why is this determinant positive for $n> 0$.

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    Your example is irrelevant, as the matrix is not a Hankel matrix. – 2017-02-19
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    Thank you for the comment. I have corrected it – 2017-02-19
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    What if $s_n=0$ for all $n$? – 2017-02-19
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    Although I can't open the pdf in the link, looking at the cached version I think that the definition in this post should read a little differently. The quadratic hankel form that Jaynot gives is supposed to be positive definite, so it should read as **strictly** greater than zero for arbitrary **nonzero** $c$. Otherwise none of this makes sense! – 2017-02-19
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    I just uploaded the page that shows it. – 2017-02-19

1 Answers 1

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Let $H_k$ be the $k$th Hankel matrix of the sequence.

For the forward direction of the equivalence, we use the fact that positive definiteness of each $H_k$ individually implies positive determinant of $H_k$.
In more detail: Akhiezer defines a positive sequence as one in which every quadratic (Hankel) form is positive definite. This translates to $c^T H_k c > 0$ for all nonzero $c$. For example in the case $k=2$, the quadratic form is $s_0c_0^2 + 2s_1c_0c_1 + s_2c_1^2 = c^T H_2 c$.

Now let $c$ be any eigenvector of $H_k$ with eigenvalue $\lambda$ ($c$ is nonzero by definition of eigenvector). Then $0 < c^T H_k c = c^T \lambda c = \lambda |c|^2,$ which shows $\lambda > 0.$ so all eigenvalues of $H_k$ are greater than 0, and the determinant of $H_k$, which is the product of its eigenvalues, is positive. Note that this argument is not specific to Hankel or even symmetric matrices, in general, positive definiteness implies positive determinant.

The other direction follows immediately from Sylvester's Criterion (https://en.wikipedia.org/wiki/Sylvester%27s_criterion) , which characterizes positive definite matrices as having all their leading minors positive (necessarily and sufficiently).

For $n \geq k$ we find $H_k$ embedded in $H_n$ as the $k$th leading minor of $H_n$. Assuming that $H_i$ for $i \leq n$ has positive determinant, Sylvester's Criterion implies that $H_n$ is positive definite. Since $H_k$ by assumption has positive determinant for all $k$, this shows that all Hankel matrices of the sequence are positive-definite.

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    Thank you Badam. I am still a bit confused why this determinant is positive. Could you please explain the forward implication? – 2017-02-19
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    For example the case $n=2$, why is $Det (H_2) = s_0s_2 - s_1^2 > 0$ – 2017-02-19
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    edited answer, hope that helps. – 2017-02-19
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    Yes, thank you Badam. I read through your answer and it made a lot of sense to me. Thanks a lot. – 2017-02-19