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Given a $n \times n$ real Cauchy like matrix $C$, i.e., for real vectors $r$, $s$, $x$, $y$

$$ C_{ij} = \frac{r_i s_j}{ x_i - y_j} $$

Can a Cauchy-like $C$ be orthogonal, i.e., $C C^T = I$ for $n > 2$?

There exists such an orthogonal $C$ for $n = 2$ , $x = [1,0.4]$, $y = [6.25,0.625]$, $r = [-1.8114, 1.4811]$, and $s = [2.3367, -0.1225]$ with

$$ C = \begin{bmatrix} 0.8062 & 0.5916 \\ -0.5916 & 0.8062 \end{bmatrix} $$

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  • $\begingroup$ Your example for n = 2 is incorrect. With those vectors, C = [ 0.806228262857143 -0.042266000000000; -15.381717200000002 0.806376666666667]. And for that $C$, $CC^T \ne I$. I don't think there is an example for n = 2, let alone n > 2. $\endgroup$ Jun 6, 2020 at 15:06
  • $\begingroup$ @MarkL.Stone I have double check again and it works for me. I get $r_i s_j = [ -4.2327, 0.2219; 3.4609, -0.1814]$ and for $x_i - y_j = [ -5.2500, 0.3750; -5.8500, -0.2250]$. $\endgroup$
    – Jiro
    Jun 6, 2020 at 15:13
  • $\begingroup$ Ahh, you have a typoi in the formula for $C_{ij}$. Should be $y_j$, not $y_i$. $\endgroup$ Jun 6, 2020 at 15:27
  • $\begingroup$ @MarkL.Stone yes, Sorry, Federico was kind enough to fix the typo $\endgroup$
    – Jiro
    Jun 6, 2020 at 15:29

2 Answers 2

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Please give a look at this short note.

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  • $\begingroup$ Dear Dario, thank you very much for this excellent and very general solution. I'm currently still reviewing it. $\endgroup$
    – Jiro
    Jun 6, 2020 at 18:33
  • $\begingroup$ Can there be solutions where all 4 vectors have rational entries? I.e. how to find a Cauchy matrix such that all the entries of the vectors a, b are squares? $\endgroup$
    – Wolfgang
    Jun 6, 2020 at 21:05
  • $\begingroup$ @Wolfgang Are you referring to $a$ and $b$ in the note? I believe that should directly follow from the given reference (Schechter, 1959). $\endgroup$
    – Jiro
    Jun 6, 2020 at 22:20
  • $\begingroup$ Yes I mean a,b from the note. Now, having only squares as entries is rather a number theoretical problem, which may not be straightforward! That's why I would love to see an explicit solution with not-too-big numbers. $\endgroup$
    – Wolfgang
    Jun 7, 2020 at 11:23
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Yes, here is an example for n = 3.

r' =  [-1.085216443606418   4.526028779191116  -0.111128247133696]
s' =  [0.552760089055250    0.079464242975571  -0.006871962674798]
x' =  [-2.748286685551109   1.237373231951619  -1.153274317177488]
y' =  [ 3.750107821687254  -2.661680052752817  -1.152510574989135]
C =
   [0.092309621606373   0.995719385103790  -0.004673316541826
   -0.995651542853765   0.092242250711209  -0.013014315066420
    0.012527528559379  -0.005854341324361  -0.999904389287486]

C*C' =  
   [1.000000000000081   0.000000000000002   0.000000000000001
    0.000000000000002   1.000000000000000   0.000000000000003
    0.000000000000001   0.000000000000003   1.000000000000528]
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  • $\begingroup$ Thanks, that works for me. Ok, I will dig deeper into why this is feasible. $\endgroup$
    – Jiro
    Jun 6, 2020 at 17:09
  • $\begingroup$ This answer would be much more helpful if you told us how you found that counterexample. $\endgroup$ Jun 6, 2020 at 18:34
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    $\begingroup$ @Federico Poloni I used the BARON global optimization solver (with YALMIP as front end) and vector variables r,s,x,y to solve a feasibility problem. n=3; r=sdpvar(n,1); s=sdpvar(n,1); x=sdpvar(n,1); y=sdpvar(n,1);C=kron(r,s')./(repmat(x,1,n)-repmat(y,1,n)');optimize([-1000<=[x;y;r;s]<=1000;C*C'==eye(n)],[],sdpsettings('solver','baron'')) . I could have played around with the elment-wise bounds of +/-1000, but it wasn't necessary. YALMIP doesn't support implicit expansion Introduced in MATLAB 2016b),, so I explicitly expanded. $\endgroup$ Jun 6, 2020 at 19:08

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