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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.

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  • $\begingroup$ 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)? $\endgroup$
    – Peter Heinig
    Commented Oct 3, 2017 at 13:49
  • $\begingroup$ @PeterHeinig Yes, this is what I mean. $\endgroup$
    – Denis Serre
    Commented Oct 3, 2017 at 14:00
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    $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Todd Trimble
    Commented Oct 3, 2017 at 16:36

1 Answer 1

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The case $k=n$ is a consequence of the identity

$$\int \det(f_j(s_k))\det(g_j(s_k))\prod_{j=1}^N d\mu(s_j) = N!\ \det\left(\int d\mu(t) f_j(t)g_k(t)\right)$$

which I have seen under the names "Andreief identity" and also "Gram identity". The proof is elementary using the Leibniz formula for the determinant.

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    $\begingroup$ Oups! I should have found this one, since I use Andreif identity (and reprove it actually) in my last paper ... $\endgroup$
    – Denis Serre
    Commented Oct 3, 2017 at 14:51
  • $\begingroup$ @DenisSerre Proof for $k<n$ should be very similiar. Your formula is then a nice generalization of Andreief. $\endgroup$
    – Marcel
    Commented Oct 3, 2017 at 15:01
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    $\begingroup$ 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 $x^k$. $\endgroup$
    – Denis Serre
    Commented Oct 3, 2017 at 15:12
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    $\begingroup$ 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 $N!$. $\endgroup$
    – Abdelmalek Abdesselam
    Commented Oct 3, 2017 at 19:12

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