5

While working on a quantum information project, I encountered the following two Hankel matrices

$$ a_{i,j} = (i+j)!(2n-(i+j))! ,\qquad b_{i,j} = (i+j+1)!(2n-(i+j))! $$

where $0 \le i,j \le n$ and $!$ denotes the factorial. I would like to know if the matrices are positive semidefinite for any integer $n \geq 2$. Their positive semidefiniteness is related to the quantum entanglement of some systems.

Computing the eigenvalues via Mathematica up to $n \approx 30$ seems to affirm that these matrices are indeed positive definite. How would I prove it for all $n$?

Rodrigo de Azevedo
  • 23,223
saikohage
  • 53
  • 4
  • This is how I would proceed: realize $A$ as the Gram matrix of an appropriate inner product. In general, given an inner product space $V$ with inner product $\langle \cdot, \cdot \rangle$, and given elements $v_1, \dots, v_n \in V$, one can form the Gram matrix $A$ whose entries are $a_{ij} = \langle v_i, v_j \rangle$. Then $A$ is positive definite if and only if $v_1, \dots, v_n$ are linearly independent. – nowhere Jun 02 '23 at 04:28
  • 5
    For your matrix, I would choose inner product $\langle f, g \rangle = \int_0^1 f(x) g(x) (1 - x)^{2n} dx$, and $v_j = (\frac{x}{1 - x})^j$, $j = 1, \dots, n$. Then $\langle v_i, v_j \rangle = \int_0^1 x^{i + j} (1 - x)^{2n - (i + j)} dx$ is a beta integral. I think it just evaluates to $\frac{(i + j)! (2n - (i + j))!}{(2n + 1)!}$, which is just $\frac{a_{ij}}{(2n + 1)!}$. – nowhere Jun 02 '23 at 04:29
  • @nowhere Thank you so much! I couldn't come up with the inner product by myself. I really appreciate it! – saikohage Jun 02 '23 at 06:06
  • No problem. I just happened to have seen Gram matrix when studying linear algebra, and beta integral when learning probability. For the sake of completeness, I will write a full answer tomorrow. – nowhere Jun 02 '23 at 07:20
  • @RodrigodeAzevedo They are the inverse of binomial coefficient; I originally had $a_{i,j} = \binom{2n}{i+j}^{-1}$. I encountered these matrices while working on a quantum information project -- the positivity is related to the entanglement of certain qubit systems. – saikohage Jun 02 '23 at 11:24
  • @RodrigodeAzevedo Yeah I agree with your edits. Indices start at 0. Do I need to do something like approval to reflect someone's edit? – saikohage Jun 02 '23 at 12:10
  • @saikohage Not that I know of. If you have some time, please consider appending further background, say, how you did arrive at these matrices. – Rodrigo de Azevedo Jun 02 '23 at 12:15

1 Answers1

2

Let me write a full answer based on my comment, including the thought process that went into it.

Your observation based on Mathematica is correct. The matrices are indeed positive semidefinite. In fact, they are positive definite.

Let's focus on the first matrix $a_{ij} = (i + j)!(2n - (i + j))!$. Our main idea is to express $A$ as the Gram matrix associated with a suitably chosen inner product. Here is the relevant result that we need:

Lemma: Let $V$ be a vector space equipped with inner product $\langle\cdot, \cdot \rangle$, and let $v_1, \dots, v_n \in V$. Let $A$ be the associated Gram matrix, that is, $a_{ij} = \langle v_i, v_j \rangle$. Then $A$ is positive semidefinite. Moreover, $A$ is positive definite if and only if $v_1, \dots, v_n$ are linearly independent.

If you need a proof of the lemma, it can be found in Olver-Shakiban: Applied Linear Algebra, 2e, p.161.

Now we just need to choose an appropriate $V$, $\langle \cdot, \cdot\rangle$, and $v_0, v_1, \dots, v_n$ so that $\langle v_i, v_j \rangle = (i + j)! (2n - (i + j)!)$. After some trial and error (details explained later), we take $V = L^2_\rho(0, 1)$ - square integrable (real-valued) functions on open interval $(0, 1)$ with respect to weight $\rho(x) = (1 - x)^{2n}$, $\langle f, g \rangle = \int_0^1 f(x) g(x) (1 - x)^{2n} dx$, and $v_j = \big(\frac{x}{1 - x}\big)^j$. Let's compute $\langle v_i, v_j \rangle$: \begin{align*} \langle v_i, v_j \rangle & = \int_0^1 \Big(\frac{x}{1 - x}\Big)^i \Big(\frac{x}{1 - x}\Big)^j (1 - x)^{2n} dx\\ & = \int_0^1 x^{i + j}(1 - x)^{2n - (i + j)} dx\\ & = \frac{(i + j)! (2n - (i + j))!}{(2n + 1)!}. \end{align*} The closed form expression can be obtained using the beta integral.

We have shown that $a_{ij} = \frac{1}{(2n + 1)!} \langle v_i, v_j \rangle$, so $A$ is positive semidefinite. (The positive constant factor $1/(2n + 1)!$ doesn't matter.) In addition, $v_0, v_1, \dots, v_n$ are linearly independent. Indeed, if we set $c_0 + c_1 v_1 + \dotsb + c_n v_n = 0$, then $c_0 + c_1 \big(\frac{x}{1 - x}\big) + \dotsb + c_n \big(\frac{x}{1 - x}\big)^n = 0$ for all $x \in (0, 1)$. This is a polynomial expression in $\big(\frac{x}{1 - x}\big)$. For it to be identically zero, we must have $c_0 = c_1 = \dots = c_n = 0$. Having established linear independence of $v_0, v_1, \dots, v_n$, we conclude that $a_{ij} = \frac{1}{(2n + 1)!}\langle v_i, v_j \rangle$ is positive definite.

For the other matrix $b_{ij} = (i + j + 1)! (2n - (i + j))!$, the same method works. Just use the inner product $\langle f, g \rangle = \int_0^1 f(x) g(x) x(1 - x)^{2n} dx$ instead.

Let me explain how I came up with the inner product above. The thought process might be instructive.

First, I recall a similar technique being used to show that the Hilbert matrix $a_{ij} = \frac{1}{i + j - 1}$ is positive definite. In this case, one can realize $a_{ij} = \int_0^1 x^{i - 1} x^{j - 1} dx$.

Second, our $a_{ij}$ only depends on $i + j$, just like the Hilbert matrix. So I should probably pick $v_1, \dots, v_n$ so that $v_j(x) = \text{expression}(x)^j$ is the $j$-th power of something. That way, I get $\langle v_i, v_j \rangle = \int \text{expression}(x)^{i + j}$.

Third, what kind of integral evaluates to a product of factorials? That made me think the beta integral $B(m, n) = \int_0^1 x^{m - 1} (1 - x)^{n - 1} dx = \frac{(m - 1)!(n - 1)!}{(m + n - 1)!}$ is somehow related. To get $(i + j)!(2n - (i + j))!$, I should set $m = i + j + 1$, $n = 2n - (i + j) + 1$, and the integral should be $\int_0^1 x^{i + j} (1 - x)^{2n - (i + j)}dx$.

Piecing these clues together, I found my answer.

nowhere
  • 1,201