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Define a symmetric Toeplitz matrix by

H = \begin{bmatrix} 2k(0) & k(s) & k(2s) & \dots & k(NS) \\ k(s) & 2k(0) & k(s) & \dots & k((N-1)S) \\ k(2s) & k(s) & 2k(0) & \dots & k((N-2)S) \\ \vdots & \vdots & \vdots & \ddots \\ k(Ns) & k((N-1)s) & k((N-2)s) & \dots & 2k(0) \end{bmatrix}

k(.) is non-increasing function. e.g k(x) = $e^{-x}$

s, N are known constant.

How to prove that this matrix is positive-definite? Thanks a lot.

Jin
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  • seethis recent question and answer: https://math.stackexchange.com/questions/2762141/decompose-a-symmetric-matrix/2762217#2762217 – Will Jagy May 05 '18 at 19:00
  • Thanks very much! But I don't think they are similar. As k(.) could be some piecewise constant function here. – Jin May 05 '18 at 20:13
  • the $s$ does not matter at all. Is $N$ the integer one smaller than the size of the matrix? Oh, and try the 2 by 2 and 3 by 3 sizes to see what happens – Will Jagy May 05 '18 at 20:57
  • Yes, N is smaller than the size of the matrix. I saw that question, the thing I don't get is 'With 0<t<1 this means that both M and the original A are positive definite.'. I understand M is positive definite since it is the diagonal matrix with all diagonal entries being positive. But why A is also positive definite? – Jin May 05 '18 at 21:57
  • https://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia – Will Jagy May 05 '18 at 23:06
  • Thank you very much! Since I don't know the exact expression of k(.), e.g. k(x) = 1 if x <= 5. k(x) = 0.5 if 5 < x <= 10 and so on. In this way, I am not sure if I can find P so that P^T H P = D while D is a diagonal matrix with all diagonal entries being positive. Any suggestions? – Jin May 05 '18 at 23:46
  • again, you should change the names to something without functions and do the 2 by 2 case, then 3 by 3, then 4 by 4. The top row of the 4 by 4 could be $2a,b,c,d$ such that $a \geq b \geq c \geq d \geq 0.$ As far as methods: http://math.stackexchange.com/questions/1388421/reference-for-linear-algebra-books-that-teach-reverse-hermite-method-for-symmetr – Will Jagy May 05 '18 at 23:51
  • Thanks for help! I understand this and I can get D from your reference link, but I cannot prove all the diagonal elements of D are positive, right? In this sense, I cannot prove H is positive definite. – Jin May 06 '18 at 00:18
  • well, except for some denominators involved, the diagonal terms appeard to come out exactly as the principal minor terms in https://en.wikipedia.org/wiki/Sylvester%27s_criterion I have a feeling the problem can be finished with this, as each pricincipal minor is simply a smaller matrix of the same type. This should help in giving information about the determinants. – Will Jagy May 06 '18 at 00:27

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This is not true. Consider the $n\times n$ penta-band symmetric Toeplitz matrix $$ \pmatrix{ 2&1&1\\ 1&\ddots&\ddots&\ddots\\ 1&\ddots&\ddots&\ddots&\ddots\\ &\ddots&\ddots&\ddots&\ddots&1\\ & &\ddots&\ddots&\ddots&1\\ & & &1&1&2}. $$ It has positive, negative and zero eigenvalues when $n=11$.

An $n\times n$ matrix $A$ is guaranteed to be positive definite if $a_{ij}=\varphi\left(\|x_i-x_j\|^2\right)$ for some completely monotone function $\varphi:[0,\infty)\to\mathbb R$ (i.e. a function such that $(-1)^m\frac{d^m\varphi}{dx^m}\ge0$ on $(0,\infty)$ for every $m\ge0$) and some set of distinct points $\{x_1,\ldots,x_n\}$ in some Euclidean space $\mathbb R^s$. This is known as Schoenberg interpolation theorem, which is a consequence of Bochner's characterisation of positive definite functions. See chapter 15 of A Course in Approximation Theory by Cheney and Light or sec. 2.5 of this book chapter for more details.

In your case, $\varphi(x)=e^{-\sqrt{x}}$ is known to be completely monotone. So, with this function $\varphi$, $A=H-k(0)I$ is positive definite and so is $H$.

user1551
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  • Thank you very much! Question: it must be || xi - xj ||^2 inside k(.) ? Could it be just || xi - xj ||? – Jin May 06 '18 at 16:28
  • @Jin Yes, we need to use squared norm, but luckily, $\varphi(x)=e^{-\sqrt{x}}$ is completely monotone, so that $\varphi(x^2)=k(x)$ on $[0,\infty)$. – user1551 May 06 '18 at 16:41