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Let $a_1, ...,a_n , b_1,...b_n$ be $2n$ distinct elements of a field and define $$h_{ij}:=\dfrac1{a_i-b_j} , \forall i,j=1,2 ,\dots,n. $$ Is the $n \times n$ matrix $H:=(h_{ij})$ non-singular ?

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  • $\begingroup$ For $n=2,$ it is non singular. $\endgroup$ – Bumblebee Nov 2 '14 at 14:56
  • $\begingroup$ Remark: the matrix in question is an instance of a Cauchy matrix. $\endgroup$ – user1551 May 2 '15 at 10:20
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Theorem: \begin{align} \det H_n= (-1)^{\frac{n(n-1)}{2}} \dfrac{\prod_{i< j}{(a_i - a_j)(b_i - b_j)}}{\prod_{1\leq i, j \leq n}(a_i - b_j)} \end{align} where $H_n = \left(\dfrac{1}{a_i - b_j}\right)_{n \times n}$


Proof:

Firstly, we prove $$\det H_n =C_n \dfrac{\prod_{i< j}{(a_i - a_j)(b_i - b_j)}}{\prod_{1\leq i, j \leq n}(a_i - b_j)} $$ for some constant $C_n$

If we compute the determinant of $H_n$ by definition as the sum corresponding to all the permutations, we see that

\begin{align} \det H_n &= \sum_{\sigma \in S_n}sgn(\sigma)\prod_{i=1}^n\dfrac{1}{a_i - b_{\sigma(i)}}\\ &= \dfrac{\sum_{\sigma \in S_n}sgn(\sigma)\prod_{j \neq \sigma(i)}(a_i - b_j)}{\displaystyle\prod_{1\leq i,j \leq n}(a_i - b_j)} \end{align} which is in form of $$\dfrac{P(a_i, b_j, i, j = 1,2,\cdots, n)}{\prod_{1\leq i, j \leq n}(a_i - b_j)}$$ where $P$ is a polynomial of degree $n^2 - n$ in $a_i$'s and $b_j$'s

If we make $a_i = a_j$, the determinant is zero and the same thing happens when we make $b_i=b_j$, so in the factorization of $P$ there should be the term $\prod_{i< j}{(a_i - a_j)(b_i - b_j)}$, which is already of degree $n(n-1)$. So $P$ is equal to $\prod_{i< j}{(a_i - a_j)(b_i - b_j)}$ multiplied by a constant.

In summary

$$\det H_n =C_n \dfrac{\prod_{i< j}{(a_i - a_j)(b_i - b_j)}}{\prod_{1\leq i, j \leq n}(a_i - b_j)} $$ with $C_n {\prod_{i< j}{(a_i - a_j)(b_i - b_j)}} = \sum_{\sigma \in S_n}sgn(\sigma)\prod_{j \neq \sigma(i)}(a_i - b_j)$

Then we prove the value of the constant $C_n = (-1)^{\frac{n(n-1)}{2}}$ by induction:

I will suppose all $a_i, b_j$ are non zero, if one of them is zero, make a translation of all the $a_i$ and $b_j$ of the same size, this does not change the determinant.

Firstly, it's easy to verify $C_2=-1$.

Suppose now that we already know $C_{n-1}=(-1)^{\frac{(n-1)(n-2)}{2}}$.

We are going to figure out what's the term in $C_n {\prod_{i< j}{(a_i - a_j)(b_i - b_j)}}$ that contains neither $a_1$ nor $b_1$. We can see that the term is $$C_n (\prod_{j=2}^n a_jb_j)\prod_{2\leq i< j}(a_i - a_j)(b_i - b_j) \tag 1$$.

Then what is the term in $\sum_{\sigma \in S_n}sgn(\sigma)\prod_{j \neq \sigma(i)}(a_i - b_j)$ that contains neither $a_1$ nor $b_i$? We see that this term is(only $\sigma(1) = 1$ gives terms containing neither $a_1$ nor $b_1$) $$\left(\sum_{\sigma \in S_n, \sigma(1)= 1} sgn(\sigma)\prod_{i\geq 2, j \neq \sigma(i), j\geq 2}(a_i - b_j)\right)\prod_{j=2}^n(-a_jb_j) \tag 2$$.

Therefore we have $(1) = (2)$, i.e.

\begin{align} C_n &= \dfrac{\left(\sum_{\sigma \in S_n, \sigma(1)= 1} sgn(\sigma)\prod_{i\geq 2, j \neq \sigma(i), j \geq 2}(a_i - b_j)\right)\prod_{j=2}^n(-a_jb_j)}{(\prod_{j=2}^n a_jb_j)\prod_{2\leq i< j}(a_i - a_j)(b_i - b_j)} \\ &= (-1)^{n-1}\dfrac{\left(\sum_{\sigma \in S_n, \sigma(1)= 1} sgn(\sigma)\prod_{i\geq 2, j \neq \sigma(i), j \geq 2}(a_i - b_j)\right)}{\prod_{2\leq i< j}(a_i - a_j)(b_i - b_j)} \end{align}

We take matrix $H_{n-1} = (\dfrac{1}{a_i -b_j})_{2\leq i, j \leq n}$, then we see that

\begin{align} \det H_{n-1} &= \dfrac{\left(\sum_{\sigma \in S_n, \sigma(1)= 1} sgn(\sigma)\prod_{i\geq 2, j \neq \sigma(i), j \geq 2}(a_i - b_j)\right)}{\displaystyle\prod_{2\leq i,j \leq n}(a_i - b_j)}\\ = & C_{n-1} \dfrac{\prod_{2\leq i< j}{(a_i - a_j)(b_i - b_j)}}{\prod_{2\leq i, j \leq n}(a_i - b_j)} \end{align}

i.e.

$$C_{n-1}{\prod_{2\leq i< j}{(a_i - a_j)(b_i - b_j)}} = \sum_{\sigma \in S_n, \sigma(1)= 1} sgn(\sigma)\prod_{i\geq 2, j \neq \sigma(i), j \geq 2}(a_i - b_j)$$

Plug the above equation in the expression for $C_n$, we get $C_n = (-1)^{n-1}C_{n-1} = (-1)^{\frac{n(n-1)}{2}}$

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