Proving a symmetric Cauchy matrix is positive semidefinite Let $x_1, x_2,\dots, x_n$ be positive real numbers. Let $A$ be the $n\times n$ matrix whose $i,j^\text{th}$ entry is $$a_{ij}=\frac{1}{x_i+x_j}.$$
This is a Cauchy matrix. I am trying to show that this matrix is positive semi-definite.
I have been given the following hint: Consider the matrix $T=(t^{x_i+x_j})$ where $t>0$. Use the fact that $T$ is positive semi-definite and that $$\frac{1}{x_i+x_j}=\int_0^1t^{x_i+x_j-1}dt.$$
I have managed to show that $T$ is positive semi-definite but I don't understand where to go from there or how to use the rest of the hint.
I would like another way to do this, preferably without involving integrals
Thank you.
 A: Hint (without integrals): Let $X=\operatorname{diag}(x_1,x_2,\ldots,x_n)$ and $e$ be the vector of all ones. 


*

*Prove that the Cauchy matrix $C$ satisfies the equation
$$
XC+CX=ee^T.
$$

*For any eigenvalue $\lambda$ of $C$ with $Cv=\lambda v$, pre-multiply the equation by $v^T$ and post-multiply by $v$. Conclude that $\lambda\ge 0$.

A: Regarding the hint: just as a sum of positive semidefinite operators is positive semidefinite, so is an integral of positive semidefinite operators positive semidefinite.
So, since $T(t)$ is positive semidefinite for all $t \geq 0$, it follows that
$$
A = \int_0^1 T(t)\,dt
$$
is positive semidefinite.

This paper hints at an interesting proof (without integrals) that when the numbers $x_i$ are distinct, $A$ is necessarily positive definite. We note that the determinant of a Cauchy matrix $x_i$ is given by
$$
\det(A) = \frac{\prod_{i,k,i>k}(x_i-x_k)^2}{\prod_{i,j = 1}^n(x_i+x_j)}.
$$
Because all terms being multiplied are positive, $\det(A) > 0$.  Since every principal submatrix of $A$ is itself a Cauchy matrix for some set of distinct $x_i$, we can conclude that the principal submatrices of $A$ also have positive determinant. By Sylvester's criterion, we can conclude that $A$ is positive definite.
For the more general statement with non-distinct $x_i$, it suffices to note that the limit of a sequence positive semidefinite matrices must itself be positive semidefinite.
A: There is also a proof that is similar in spirit to the hint you've mentioned. See exercise 1.6.3 (pp.24-25) of Bhatia, Positive Definite Matrices. The idea is that, instead of writing the Cauchy matrix as an integral of Gramians, we write it as an infinite sum of Gramians. More specifically, let $0<t<\min_ix_i$. Then
$$
\frac{1}{x_i+x_j-t}
=\frac{t}{x_ix_j}\left(\frac{1}{1-\frac{(x_i-t)(x_j-t)}{x_ix_j}}\right)
=\frac{t}{x_ix_j}\sum_{k=0}^\infty\left(\frac{(x_i-t)(x_j-t)}{x_ix_j}\right)^k.
$$
Therefore, by setting $\mathbf v_k=\left(\frac{(x_1-t)^k}{x_1^{k+1}},\,\frac{(x_2-t)^k}{x_2^{k+1}},\,\ldots,\,\frac{(x_n-t)^k}{x_n^{k+1}}\right)^\top$, we see that $$
\left(\frac{1}{x_i+x_j-t}\right)_{i,j\in\{1,\ldots,n\}}
=t\sum_{k=0}^\infty \mathbf v_k\mathbf v_k^\top
$$
is positive semidefinite. Now the result follows by passing the matrix on the LHS to the limit $t\to0$.
A big merit of the above proof is that it can be easily extended to prove the positive semidefiniteness of the power Cauchy matrix $\left(\frac{1}{(x_i+x_j)^p}\right)_{i,j\in\{1,\ldots,n\}}$ for any $p>0$.
A: Complementing A.Γ.'s answer, and rephrasing a bit:

Given $a_1, a_2, \dots, a_n > 0$, we build the $n \times n$ symmetric Cauchy matrix $\rm C$ whose entries are $$c_{ij} = \frac{1}{a_i + a_j}$$ Show that matrix $\rm C$ is positive semidefinite.

Henceforth, we shall assume that the given positive numbers are distinct, i.e., $$|\{a_1, a_2, \dots, a_n\}| = n$$

Let ${\rm A} := \mbox{diag} (a_1, a_2, \dots, a_n)$. Note that $\mathrm A \succ \mathrm O_n$. Consider the following matrix equation
$${\rm A X + X A} =  1_n 1_n^\top$$
Multiplying both sides by $-1$, we obtain a Lyapunov matrix equation
$${\rm (-A) X + X (-A)} = - 1_n 1_n^\top$$
where matrix $-\rm A$ is stable (or Hurwitz) and the RHS is negative semidefinite. Since the pair $(-\rm A, 1_n)$ is controllable, the Lyapunov equation has the following unique, symmetric, positive definite solution
$$\rm X = \int_0^{\infty} e^{- \tau \mathrm A} 1_n 1_n^\top e^{- \tau \mathrm A} \,{\rm d} \tau = \int_0^{\infty} \begin{bmatrix} e^{- a_1 \tau}\\ e^{- a_2 \tau}\\ \vdots\\ e^{- a_n \tau}\end{bmatrix} \begin{bmatrix} e^{- a_1 \tau}\\ e^{- a_2 \tau}\\ \vdots\\ e^{- a_n \tau} \end{bmatrix}^\top \, {\rm d} \tau = \cdots = \rm C$$
because
$$\displaystyle\int_0^{\infty} e^{-(a_i + a_j) \tau} \,{\rm d} \tau = \frac{1}{a_i + a_j}$$
Therefore, we conclude that $\rm C$ is positive definite.

Alternative solution
Vectorizing both sides of the Lyapunov equation,
$$\left( \mathrm I_n \otimes \mathrm A + \mathrm A \otimes \mathrm I_n \right) \mbox{vec} (\mathrm X) = 1_n \otimes 1_n$$
or,
$$\begin{bmatrix} \mathrm A + a_1 \mathrm I_n & & & \\ & \mathrm A + a_2 \mathrm I_n & & \\ & & \ddots & \\ & & & \mathrm A + a_n \mathrm I_n\end{bmatrix} \begin{bmatrix} \mathrm x_1\\ \mathrm x_2\\ \vdots\\ \mathrm x_n\end{bmatrix} = \begin{bmatrix} 1_n\\ 1_n\\ \vdots\\ 1_n\end{bmatrix}$$
where $\mathrm x_i$ is the $i$-th column of $\rm X$. Solving for $\mathrm x_i$,
$$\mathrm x_i = \begin{bmatrix} \frac{1}{a_1 + a_i}\\ \frac{1}{a_2 + a_i}\\ \vdots\\ \frac{1}{a_n + a_i}\end{bmatrix}$$
which is the $i$-th column of Cauchy matrix $\rm C$. Therefore, the unique, symmetric, positive definite solution of the Lyapunov equation is $\rm C$.

Addendum
The controllability matrix corresponding to the pair $(-\rm A, 1_n)$ is
$$\begin{bmatrix} | & | & & |\\ 1_n & -\mathrm A 1_n & \dots & (-1)^{n-1} \mathrm A^{n-1} 1_n\\  | & | & & |\end{bmatrix}$$
which is a square $n \times n$ Vandermonde matrix whose columns have been multiplied by $\pm 1$, which does not affect its rank. Since we assumed that the given $a_1, a_2, \dots, a_n$ are distinct, the Vandermonde matrix has full rank and, thus, the pair $(-\rm A, 1_n)$ is controllable.
