# Some questions about Hilbert matrix

This Exercise $12$ page $27$ from Hoffman and Kunze's book Linear Algebra.

The result o Example $16$ suggests that perhaps the matrix

$$A = \begin{bmatrix} 1 & \frac{1}{2} & \ldots & \frac{1}{n}\\ \frac{1}{2} & \frac{1}{3} & \ldots & \frac{1}{n+1}\\ \vdots & \vdots & & \vdots \\ \frac{1}{n} &\frac{1}{n+1} &\ldots &\frac{1}{2n-1} \end{bmatrix}$$

is invertible and $A^{-1}$ has integer entries. Can you prove that?

I will copy their words before that example:

Some people find it less awkward to carry along two sequences of matrices, one describing the reduction of $A$ to the identity and the other recording the effect of the same sequence of operations starting from the identity.

In this post, Adrián Barquero says:

If your linear algebra textbook is Kenneth Hoffmann and Ray Kunze's Linear Algebra book then I was in the same situation you're right now a few years ago, but I was adviced at the moment that this particular exercise wasn't as easy as one might expect at first.

The matrix you have is called the Hilbert matrix and the question you have was already asked a couple of times in math overflow here and here. They have excellent answers so I will just point you to them.

My question: Is it possible answer the question made by the authors without using high techniques? I am not allowed to use determinants here.

PS: Those answers in MO don't answer the question above. I apologise if this question is unappropriated or if I was unable to understand their answers.

Yes, I believe a student at the level of this book is able to answer the question. But this is not easy. I think it would be an issue at the Mathematical Olympiad.

I suggest you do this proof by induction on the size $$n$$ of the $$n \times n$$ Hilbert matrix $$A_n.$$ Start the induction step with $$n = 2$$ . The statement is essentially the following:

$$A_n = \left(\frac{1}{i + j-1}\right)_{ij} \mbox{ is a Hilbert matrix }\Rightarrow A_n^{-1} \in M_{n \times n}(\mathbb{Z}).$$

Step 1. Verify that by inspection for $$n = 2$$.

Step 2. Verify that if $$A_n$$ is a Hilbert matrix, then $$A_{n+1}= \begin{pmatrix} A_n & v^{T} \\ v & \frac{1}{2(n+1)-1} \end{pmatrix} \quad$$ is a Hilbert matrix with $$v=\left(\frac{1}{n+1},\dots,\frac{1}{2(n+1)-2}\right)$$. Now apply matrix inversion in block form, special case 1. And in this special case in which the blocks are matrices that commute, there is an exercise in the Hoffman book (do not remember which page).

Edit 1. And in this particular case, this formula can be obtained using the definition of the inverse of a matrix. Just partition the inverse matrix into blocks of the same size and make the product and solve a matrix system. I believe that a student at the Mathematical Olympiad may have this idea, and an average student can understand it.

Edit 2. Done what was said in Edit 1, according to the notation of the link, just to verify that for $$k =\frac{1}{2(n+1)-1} - vA_n^{-1}v^T,$$ we have that $$\frac{1}{k}A_n^{-1}v^T \in\mathbb{Z}$$. But this can be checked again by induction.

• Unfortunately I can't follow your link because the server isn't reachable. However, if I do the calculation myself, I get your $k$ as the reciprocal of the bottom right entry of the inverse, so it seems we need not just $\frac{1}{k}A_n^{-1}v^T \in\mathbb{Z}$ but $\frac{1}{k}\in\mathbb{Z}$? I don't see how to show either of them by induction. Feb 25, 2012 at 10:12
• Feb 25, 2012 at 13:42
• That's a long article. What part of it do you think answers my comment? Feb 25, 2012 at 13:53
• @Joriki. It makes sense that you pointed out. But if not by induction we have to explicitly calculate the entries of the Shur complement as a function of $1, \dots, n$ which is the sum of a series of integers. Feb 25, 2012 at 14:05
• See en.wikipedia.org/wiki/Schur_complement View the block inversion formula for matrices and Shur complement formula. Feb 25, 2012 at 14:07