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.



Prove that the matrix $[\Gamma(\lambda_{i}+\mu_{j})]$ is nonsingular
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](https://mathoverflow.net/questions/277655/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](https://en.wikipedia.org/wiki/Sylvester%27s_criterion) shows that $A$ is in fact positive definite. It is even [totally positive](https://en.wikipedia.org/wiki/Totally_positive_matrix).




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**Edit:**
I just found this other MO question: [A difficult determinant](https://mathoverflow.net/questions/297665/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).


