This is inspired by a recent question about the existence of orthogonal Cauchy-like matrices. It is proved that there are indeed such matrices, i.e. there are vectors $x,y,r,s\in\mathbb R^n$ such that for the matrix $C$ defined by $$ C_{ij} = \frac{r_i s_j}{ x_i - y_j}, $$ we have $C C^T = I$.

In the pdf with the solution, the following fact is used: If $C$ is a $n \times n$ real Cauchy matrix $C$, i.e. there are vectors $x,y\in\mathbb R^n$ such that $$C_{ij} = \frac{1}{ x_i - y_j},$$ then its inverse admits the factorization $$C^{-1} = D_aC^TD_b,$$ where $D_a,D_b$ denote the diagonal matrices with the entries of $a, b\in\mathbb R^n$, which can be computed by explicit formulae obtained by Lagrangian interpolation. BTW, it is not clear to me why the RHS features $C^T=(( \frac{1}{ x_j - y_i}))_{i,j=1}^n$, as the given formula $(7)$ in Theorem 1 of the original article (accessible here with different notation, as it calls $a,b$ what we call $x,y$ here) seems to contain rather $(( { x_j - y_i}))_{i,j=1}^n$ than the reciprocals.

In any case, my question is

whether the vectors $x,y,r,s\in\mathbb R^n$ defining a Cauchy-like matrix can all have rational entries, i.e. whether there is a Cauchy matrix $C$ such that each entry of the corresponding vectors $a,b$ in the factorization of $C^{-1}$ quoted above is a (rational) square?

Note that the given construction shows that there are Cauchy matrices where $a,b$ have only positive entries and uses their square roots at some point. But for rational square roots, I wouldn't know how to go about that even for $n=2$.

  • $\begingroup$ That denominator in the inverse is expected. If $C$ satisfies $D_x C - C D_y = \text{(rank 1)}$, then by multiplying by $C^{-1}$ on both sides you get that $C^{-1}$ satisfies $C^{-1}D_x - D_y C^{-1} = \text{(rank 1)}$, and by expanding you get those denominators. $\endgroup$ – Federico Poloni Aug 6 '20 at 6:58
  • $\begingroup$ @Federico yes I agree that that makes sense. In fact, I have a hard time interpreting the formula derived from Lagrange in the article in order to put it into a form with $D_a$ and $D_b$ for some vectors a,b. $\endgroup$ – Wolfgang Aug 6 '20 at 7:08

Let us start with four rational numbers $x_1, x_2; y_1,y_2$ so that using the cross ratio $$ r=(x_1,x_2,y_1,y_2) $$ the numbers $-r$ and $1-r$ are rational squares. (Exchanging the mid components $x_2, y_1$ bring $r$ into $1-r$. Exchanging the first and/or last two components we obtain the multiplicative inverse, etc. - so we want in the following to set in evidence squares times cross ratio values known to be squares.)

For instance for $0,1;-1,8/17$ we are producing $r=-9/16$. It may be simpler to follow the construction based on this example.

Let $C$ be the associated Cauchy matrix. In the example: $$ C=\begin{bmatrix}1 & -17/8\\1/2 & 17/9\end{bmatrix}\ . $$ Let $a_1,a_2;b_1,b_2$ be the rational squares: $$ \begin{aligned} a_1 &= (x_1-y_1)^2 \; (x_1, y_1, y_2, x_2) \ ,\\ a_2 &= (x_1-y_2)^2 \; (x_1, y_2, y_1, x_2) \ ,\\[2mm] b_1 &= 1\ ,\\ b_2 &= (-1)\; \left(\frac{x_2-y_1}{x_1-y_1}\right)^2 \;(x_1, x_2, y_1, y_2) \ . \end{aligned} $$ Then we have $C^{-1}=D_a\; C^T\; D_b$.

Let now $L$ be a the Cartan-like matrix of the shape $L=D_r\; C\; D_s$. Then: $$ \begin{aligned} L &=D_r\; C\; D_s\ ,\\ L^T &=D_s\; C^T\; D_r\ ,\\ L^{-1} &=D_s^{-1}\; C^{-1}\; D_r^{-1}\\ &=D_s^{-1}\; D_a C^T D_b\; D_r^{-1} \ ,\\[3mm] &\qquad\text{ and we want $L^T=L^{-1}$, i.e.}\\[3mm] D_s\; C^T\; D_r &= D_s^{-1}\; D_a C^T D_b\; D_r^{-1}\text{ i.e.}\\ C^T &= D_s^{-2}\; D_a C^T D_b\; D_r^{-2}\ . \end{aligned} $$ Recall that $-r$ and $1-r$ both squares implies $a,b$ squares, as wanted in the OP, so we can arrange for $s,t$ with rational entries.

To make the above easy to test, here is some sage code making the computations.


def r(s,t,u,v):
    return (s-u)/(s-v)/(t-u)*(t-v)

a1 = (x1-y1)^2 * r(x1, y1, y2, x2)
a2 = (x1-y2)^2 * r(x1, y2, y1, x2)
b1 = 1
b2 = (-1) * (x2-y1)^2 / (x1-y1)^2 * r(x1, x2, y1, y2)

C = matrix([ [1/(x1-y1), 1/(x1-y2)] , [1/(x2-y1), 1/(x2-y2)] ])
Da = diagonal_matrix( [a1, a2] )
Db = diagonal_matrix( [b1, b2] )

print("Is C^-1 = Da * C^T * Db? %s"
      % bool(C^-1 == Da * C.transpose() *Db))

And we get:

Is C^-1 = Da * C^T * Db? True

We use now instead of general variables the special values:

x1, x2, y1, y2 = 0, 1, -1, 8/17

(just replace the first var line with the above, keep the next lines of the used code) and ask for the values of $a$, $b$:

sage: a1, a2, b1, b2
(16/25, 576/7225, 1, 9/4)

Now consider the matrix $L$ given by

sage: L = diagonal_matrix([sqrt(b1), sqrt(b2)]) * C * diagonal_matrix([sqrt(a1), sqrt(a2)])
sage: L
[ 4/5 -3/5]
[ 3/5  4/5]

Which is an orthogonal matrix. (I found this problem while searching for the tag elliptic-curves, but the above solution is maybe closer to K-theory.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.