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I have got an exercise on Hilbert matrices determinant. Let $n \in \mathbb{N}^*$ , and $H_n$ be the Hilbert matrix of size $n \times n$. Let's note $\Delta_{n} $ the determinant of $H_n$. I have to prove that :

$$ \Delta_{n+1} = \frac{(n!)^4}{(2n)!(2n+1)!} \Delta_{n} $$

And this, without the Cauchy determinant. There is a clue : "Start by substracting the last column of $H_{n+1}$ to others.".

That's what I did, and I got a matrix where the coefficient $i,j$ is (except for the last column) :

$$ \frac{1}{i+j-1} - \frac{1}{n+i} = \frac{n-j+1}{(i+j-1)(n+i)} $$

As for a specific row we can factor by $\frac{1}{n+i}$. We can do that for each line and then factorize by $$ \displaystyle \prod_{i=0}^{n+1}\left(\frac{1}{n+i}\right) = \frac{n!}{(2n+1)! } $$

Let's do something similar for the columns, and so we factorize by $$ \prod_{i=1}^{n} (1+n-i) = n! $$

To sum up I have :

$$ \Delta_{n+1} = \frac{n!^2}{(2n+1)!} X$$

Where $X$ is the determinant of $H_{n+1}$ but the last column is filled with $1$. I am stuck. I want to calculate the determinant by "developing" (I'm french I dont' know if it's the correct word) it with the last column. But i don't find anything interesting.

Could you help me please?

Axel
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Hugo Faurand
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1 Answers1

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You are almost there. Now try to do $L_i \leftarrow L_i-L_{n+1}$ for all $1\leq i \leq n$. Where $L_i$ denotes the $i$-th row. By substracting the rows you will have the opportunity to make the missing factors appear as you did with the columns, and you will end up with only $0$s in the last column except at the bottom.

Indeed, for all $1\leq i,j \leq n$ you will have,

$$\dfrac{1}{i+j-1}- \dfrac{1}{n+1+j-1} = \dfrac{n+1-i}{(i+j-1)(n+1+j-1)}$$

Now by factorizing each column: $$ \prod_{j=1}^n \dfrac{1}{n+1+j-1} =\prod_{j=1}^n \dfrac{1}{n+j} =\dfrac{n!}{(2n)!}$$

And now each row:

$$ \prod_{i=1}^n (n+1-i) = n!$$

Therefore,

$$X = \dfrac{(n!)^2}{(2n)!}\begin{vmatrix} \large{H_n} & 0 \\ * & 1 \end{vmatrix} = \dfrac{(n!)^2}{(2n)!}\Delta_n $$

Finally,

$$\boxed{\Delta_{n+1} = \dfrac{(n!)^4}{(2n)!(2n+1)!} \Delta_n }$$

Axel
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