# A determinantal formula

In my research, I encounter the following formula which I believe is correct (checked for $n\le3$). Is it classical ? If so, what is a reference ?

I am given a real symmetric matrix $$S:=\int Y(t)Y(t)^Td\mu(t),$$ where $\mu$ is a probability and $Y(t):\Omega\rightarrow{\mathbb R}^n$.

Let $\sigma_k(S)$ be the elementary symmetric polynomial in the eigenvalues of $S$. For instance, $\sigma_1(S)$ is the trace and $\sigma_n(S)$ the determinant. The following formula gives $\sigma_k(S)$ in terms of the Gram matrix $G_k(s_1,\ldots,s_k)$ whose entries are the scalar products $Y(s_i)\cdot Y(s_j)$.

$$\sigma_k(S)=\frac1{k!}\int^{\otimes k}\det G_k(s_1,\ldots,s_k)\,d\mu(s_1)\cdots d\mu(s_k).$$

Remark that $S$ is positive semi-definite. The integrand is non-negative, as well as $\sigma_k(S)$. The integrand vanishes identically iff $Y(t)$ takes values in a subspace of dimension $<k$, which is the condition under which $\sigma_k(S)$ vanishes. It follows that, if the formula above failed, it would be because of an inequality between strictly positive numbers.

Edit. I delete the question mark in the formula above, because Marcel's answer and my comment to it, yield a proof.

• Is it correct that the last sentence in the OP implies 'you already know for sure that if the proposed equation is false, then the it fails at most in the form (positive real number)$\neq$(some other positive real number)'? I.e., it is for sure that it cannot fail in the form $0\neq$(positive real number)? Oct 3, 2017 at 13:49
• @PeterHeinig Yes, this is what I mean. Oct 3, 2017 at 14:00
• Comments are not for extended discussion; this conversation has been moved to chat. Oct 3, 2017 at 16:36

The case $k=n$ is a consequence of the identity
$$\int \det(f_j(s_k))\det(g_j(s_k))\prod_{j=1}^N d\mu(s_j) = N!\ \det\left(\int d\mu(t) f_j(t)g_k(t)\right)$$
• @DenisSerre Proof for $k<n$ should be very similiar. Your formula is then a nice generalization of Andreief. Oct 3, 2017 at 15:01
• Now, I think I see a proof. It consists in expanding $\det(I_n+xS)$. Write $I_n+xS=e_1e_1^T+\cdots+e_ne_n^T+xS$, which is an $S'$, for a mesure $\mu'$ (which has to be divided by $n+x$ in order to qualify as a probability). Then apply the (proven) case $k=n$, and check carefully what is the coefficient of $x^k$. Oct 3, 2017 at 15:12
• The Andreief formula is just the discrete sum $\rightarrow$ continuous sum version of the Binet-Cauchy formula. The latter involves set of rows/columns instead of tuples which accounts for $N!$. Oct 3, 2017 at 19:12