Return to Answer

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.

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).

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.

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.

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.

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.