Assume that $A$ is a $n\times n$ positive matrix, whose eigenvalues are $a_1\ge a_2\ldots \ge a_n>0;$ $B$ is a $p \times p$ positive matrix, whose eigenvalues are $b_1\ge b_2\ldots \ge b_p>0.$ For $n\times p$ full rank matrix $X$ with $n\ge p,$

How to prove $$\det(X'AX+B)\ge c(X)\prod_{i=1}^p(a_{n-p+i}+b_{i}),$$

where $c(X)$ is a postive constant that depends only on $X$.

  • $\begingroup$ Maybe you have something more precise in mind. For instance, $c(X)=0$ always works. $\endgroup$ – Suvrit Jan 28 '19 at 19:18
  • $\begingroup$ @Suvrit Thanks. $C(X)>0.$ $\endgroup$ – Xiaopai Song Jan 28 '19 at 20:41
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    $\begingroup$ What do we get if $X=0$? $\endgroup$ – Fedor Petrov Jan 30 '19 at 21:28
  • $\begingroup$ @Fedor Petrov Thank you again. It's full rank. $\endgroup$ – Xiaopai Song Jan 30 '19 at 21:57
  • $\begingroup$ @Denis Serre Thanks for your edition. Are you interesting in another inequality mathoverflow.net/q/322262/134602? $\endgroup$ – Xiaopai Song Feb 2 '19 at 15:29

Your both conjectural inequalities are equivalent to each other and false even for $n=p$, $A=B$, $X=I$ (and $a_i=\lambda^i$ for large $\lambda$, for example).

What is true that $$|X'AX+B|\geqslant c(X)\prod_{i=1}^p (a_{n-p+i}+b_i).$$

Proof. Let $\lambda_1\geqslant \lambda_{2}\geqslant \ldots \geqslant \lambda_p$ be eigenvalues of $X'AX+B$. It suffices to prove that $$\lambda_{i}\geqslant \rho(b_i+a_{n-p+i})\quad (*)$$ for all $i=1,\dots,p$, where $\rho=\rho(X)$ depends only on $X$. Consider two cases.

1) $b_i\geqslant a_{n-p+i}$. Then $(*)$ holds with $\rho=1/2$, since $X'AX+B\geqslant B$, thus $\lambda_i\geqslant b_i$.

2) $a_{n-p+i}\geqslant b_i$. Let $L$ be the image of the operator (identified with the matrix) $X$, $\dim L=p$ and $X$ is a linear isomorphism onto $L$. Denote by $4\rho^2$, $\rho>0$, the norm of the inverse map $X^{-1}:L\mapsto X$. The intersection $L_i$ of $L$ and the space generated by $u_1,\dots,u_{n-p+i}$, where $Au_i=a_i u_i$ and $u_i$ are orthogonal eigenvectors of $A$, has dimension at least $i$. We have $$((X'AX+B) x,x)\geqslant (X'AX x,x)=(AXx,Xx)\geqslant a_{n-p+i}(Xx,Xx)\geqslant 2\rho a_{n-p+i}(x,x)$$ whenever $x\in X^{-1}L_i.$ Therefore by the variational (Courant) principle $\lambda_i\geqslant 2\rho(X)a_{n-p+i}\geqslant \rho(a_{n-p+i}+b_i)$.

  • $\begingroup$ Thank you again. Your answer is right. $\endgroup$ – Xiaopai Song Jan 31 '19 at 17:12
  • $\begingroup$ Are you interested in other similar inequalities? mathoverflow.net/q/322262/134602 and mathoverflow.net/q/322265/134602. For now, I can't prove it. $\endgroup$ – Xiaopai Song Feb 2 '19 at 3:36
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    $\begingroup$ Again, estimate consecutive eigenvalues of all involved matrices using variational principle $\endgroup$ – Fedor Petrov Feb 2 '19 at 9:24
  • $\begingroup$ Thanks for your constructive comment. I try to handle the inequality (2) mathoverflow.net/q/322262/134602 and makes comments in it. I am not familiar with the variational principle. So I can't follow you overall. If you have time, could you give some comments or answer it? Your comments are very important to me. $\endgroup$ – Xiaopai Song Feb 2 '19 at 15:17
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    $\begingroup$ I mean en.wikipedia.org/wiki/Min-max_theorem In other words, in order to estimate the $i$-th eigenvalue from below you should find a subspace of dimension $i$ onto which you have a lower bound for the quadratic form $\endgroup$ – Fedor Petrov Feb 2 '19 at 19:23

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