Let A is a $n\times n$ matrix given by \begin{align*} a_{ij} = [\Gamma(\lambda_{i}+\mu_{j})] \end{align*} where $0 < \lambda_{1} < \ldots < \lambda_{n}$ and $0 < \mu_{1} < \ldots < \mu_{n}$ are real positive numbers and $\Gamma$ denotes the Gamma function given by $\Gamma(z) = \int_0^\infty t^{z-1}e^{-t}dt$ \, for Re(z)>0.
We need to show that matrix A is non-singular.
I have no idea how to start. Any hint or solution will be appreciated.

Thanks in advance.

  • $\begingroup$ Can you give some motivation for why you expect it to be non-singular? $\endgroup$ – Alex M. Jul 19 '18 at 10:34
  • 1
    $\begingroup$ $\Gamma(2)=\Gamma(1)=1$. $\endgroup$ – Bullet51 Jul 19 '18 at 11:11
  • $\begingroup$ thanks for replying. Here what are your $ \lambda_{i}$ and $\mu_{j}$?? I have checked it by Mathematica. And I couldn't find any factorization of this matrix too. I have tried it to prove it in many ways. But no success. $\endgroup$ – prince Jul 19 '18 at 12:10
  • $\begingroup$ I don't think, this matrix can get all it's entries equal to 1. $\endgroup$ – prince Jul 19 '18 at 12:15
  • $\begingroup$ Bullet51, can you please tell me, why did you mention $\Gamma(2)=\Gamma(1)=1$? $\endgroup$ – prince Jul 19 '18 at 12:42

By Andreieff's identity: $$ {\rm det}(A)=\frac{1}{n!}\int_{(0,\infty)^{n}} {\rm det}[e^{-\frac{t_k}{2}}t_k^{\lambda_i-\frac{1}{2}}]_{1\le i,k\le n}\ \times\ {\rm det}[e^{-\frac{t_k}{2}}t_k^{\mu_j-\frac{1}{2}}]_{1\le k,j\le n}\ \ dt_1\cdots dt_n $$ $$ =\frac{1}{n!}\int_{(0,\infty)^{n}} \left(\prod_{l=1}^{n} \frac{e^{-t_l}}{t_l}\right)\times {\rm det}[t_k^{\lambda_i}]_{1\le i,k\le n}\ \times\ {\rm det}[t_k^{\mu_j}]_{1\le k,j\le n}\ \ dt_1\cdots dt_n\ . $$ By the identity, e.g., in Reference for exponential Vandermonde determinant identity , the two determinants in the integrand have the same sign (write $t_l=e^{\alpha_l}$ to see the usual expression of the Harish-Chandra-Itzykson-Zuber integral). It is then easy to conclude that ${\rm det}(A)>0$ and so $A$ is nonsingular. The same argument and the use of Sylvester's criterion shows that $A$ is in fact positive definite. It is even totally positive.

Edit: I just found this other MO question: A difficult determinant which is directly related to this one. In the case where the Vandermonde-like determinants have integer exponents then one can of course bring Schür functions into play. One can then go a long way towards the computation of ${\rm det}(A)$ (see Marcel's answer to that question).

  • $\begingroup$ Nice answer! I was about to write an answer saying use the expansion of $\det(A+B)$ and notice that $e^{xy}$ is a totally positive kernel, when I just realized that the Andreieff identity is the integral version of the $\det(A+B)$ expansion...right? $\endgroup$ – Suvrit Jul 20 '18 at 23:06
  • $\begingroup$ A very nice argument! A somewhat similar one can be found in Karlin's book; this is too long for a comment, so I wrote the details as an answer. $\endgroup$ – Mateusz Kwaśnicki Jul 21 '18 at 9:36
  • $\begingroup$ @Abdelmalek, Sir, Can you please any book or article name, where I can find Andreieff's identity? I couldn't find it anywhere online. Thank you. $\endgroup$ – prince Jul 22 '18 at 18:58
  • $\begingroup$ @Abdelmalek, Sir, I just forgot to thank you, thanks a lot for helping me with this problem, thanks a lot. $\endgroup$ – prince Jul 22 '18 at 19:50

This is merely a reformulation of Abdelmalek Abdesselam's answer, in a somewhat different language and with different references. It should be a comment to that answer, but it is unfortunately too long. Long story short: see Karlin, Total positivity, formula (2.10) in Section 1.2, with $u(t) = t$ and $\sigma(dt) = \mathbb{1}_{(0, \infty)}(t) t^{-1} e^{-t} dt$.

The kernel $K(x,y)$ is said to be totally positive on $X \times Y$, where $X, Y \subseteq \mathbb{R}$, if $$ K\pmatrix{x_1&x_2&\cdots&x_n\\y_1&y_2&\cdots&y_n} := \det \left|\matrix{K(x_1,y_1)&K(x_1,y_2)&\cdots&K(x_1,y_n)\\K(x_2,y_1)&K(x_2,y_2)&\cdots&K(x_2,y_n)\\\vdots&\vdots&&\vdots\\K(x_n,y_1)&K(x_n,y_2)&\cdots&K(x_n,y_n)\\}\right| \geqslant 0 $$ whenever $x_1 < x_2 < \ldots < x_n$ and $y_1 < y_2 < \ldots < y_n$ (and, of course, $x_1, x_2, \ldots, x_n \in X$, $y_1, y_2, \ldots, y_n \in Y$). It is strictly totally positive if strict inequality holds. A standard reference for totally positive kernels is Karlin's book Total positivity (Stanford, 1968). Our goal is thus to prove that the kernel $G(\mu,\nu) = \Gamma(\mu + \nu)$ is strictly totally positive.

It is known that the kernel $e^{x y}$ is strictly totally positive on $\mathbb{R} \times \mathbb{R}$; see, for example, Example (i) in Section 2.1 of Karlin's book. Substituting $t = e^y$, we see that $K(x, t) = t^x$ and $\check{K}(t, x) = t^x$ are strictly totally positive on $\mathbb{R} \times (0, \infty)$ and $(0, \infty) \times \mathbb{R}$, respectively.

Define $\sigma(dt) = t^{-1} e^{-t} dt$ on $(0, \infty)$. Observe that $$ G(\mu,\nu) = \Gamma(\mu + \nu) = \int_0^\infty t^{\mu + \nu - 1} e^{-t} dt = \int_0^\infty K(\mu, t) \check{K}(t, \nu) \sigma(dt) . $$ The basic composition formula (as it is called by Karlin, see (2.5) in Section 1.2 in his book; Karlin's reference for this formula is problem 68 in Pólya and Szegő, Aufgaben und Lehrsdtze aus der Analysis, vol. 1) tells us that $$ \begin{aligned} & G\pmatrix{\mu_1&\mu_2&\cdots&\mu_n\\\nu_1&\nu_2&\cdots&\nu_n} = \idotsint\limits_{0<t_1 < t_2 < \ldots < t_n} K\pmatrix{\mu_1&\mu_2&\cdots&\mu_n\\t_1&t_2&\cdots&t_n} \times \\ & \hspace{10em} \times \check{K}\pmatrix{t_1&t_2&\cdots&t_n\\\nu_1&\nu_2&\cdots&\nu_n} \sigma(dt_1) \sigma(dt_2) \ldots \sigma(dt_n) . \end{aligned} $$ The right-hand side is clearly positive, and our claim is proved.

The above argument is (essentially) contained in Karlin's book, when he proves that moment sequences generate totally positive kernels, see formula (2.10) in Section 1.2 of his book. He is only concerned with integer moments, but the argument carries over with no modifications to arbitrary moments.

  • $\begingroup$ @prince: You're welcome. But hey, you should have accepted Abdelmalek's answer! Mine was just an extended comment to his answer. If it is possible to change the accepted answer, I strongly suggest that you do this. $\endgroup$ – Mateusz Kwaśnicki Jul 22 '18 at 19:17
  • $\begingroup$ sorry if I did something wrong, actually I am new here, I had ticked both the answers, but now I understood that we can pick just one among all answers, I like both the answers by the way. Thanks for both of you. :) $\endgroup$ – prince Jul 22 '18 at 19:45
  • $\begingroup$ but can you please give me some textbook name where I can find Andrieiff's identity? $\endgroup$ – prince Jul 22 '18 at 19:47
  • $\begingroup$ @prince: As I understand, Adreev's identity is just a name for what Karlin calls the basic composition formula. I would not be surprised if it has even more names. $\endgroup$ – Mateusz Kwaśnicki Jul 22 '18 at 20:43
  • $\begingroup$ The Andreief formula is just the continuous analogue of the Binet-Cauchy formula for determinants. As to the history of it, one can't go wrong with checking the relevant entry in the book by Thomas Muir on googlebooks $\endgroup$ – Abdelmalek Abdesselam Jul 24 '18 at 15:51

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