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Suppose 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 $m \times m$ positive matrix, whose eigenvalues are $b_1\ge b_2\ldots \ge b_m>0;$ $C$ is a $p \times p$ positive matrix, whose eigenvalues are $c_1\ge c_2\ldots \ge c_p>0.$ For $n\times p$ full rank matrix $X$ with $n\ge p,$ a $n\times m$ full rank matrix $Y$ with $n\ge m,$, and $m\ge p,$

Question: How to prove it? $$det\Big(X'(A+YBY')^{-1}X+C\Big)\ge l(X,Y)\prod_{i=1}^p\Big(\frac{1}{a_i+b_{i}}+c_{p-i+1}\Big),$$ where $l(X,Y)$ is a positive constant that only depends on $X,Y.$

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  • $\begingroup$ Let $\lambda_1\geqslant \lambda_{2}\geqslant \ldots \geqslant \lambda_p$ be eigenvalues of $X('A+YBY')X+C$. It suffices to prove that $$\lambda_{p-i+1}\geqslant l(X,Y)(\frac{1}{a_i+b_i}+c_{p-i+1})\quad (*)$$ for all $i=1,\dots,p.$ Consider two cases. 1) $c_{p-i+1}\geqslant \frac{1}{a_i+b_i}$. It is easy to handle it. 2) $\frac{1}{a_i+b_i}> c_{p-i+1}.$ It suffices to prove $\lambda_{p-i+1}\ge l(X,Y) \frac{1}{a_i+b_i}.$ Let $\lambda_1\geqslant v_{2}\geqslant \ldots \geqslant v_p$ be eigenvalues of $X'(A+YBY')^{-1}X$. $\endgroup$
    – Xiaopai Song
    Commented Feb 2, 2019 at 14:57
  • $\begingroup$ Then $\lambda_{p-i+1}\ge v_{p-i+1}.$ By the variational principle, I can prove $v_{p-i+1}\ge l(X,Y) t_i^{-1},$ where $t_1\geqslant t_{2}\geqslant \ldots \geqslant t_n$ be eigenvalues of $A+YBY'$. I don't know how to do next. Because for $i=1,\ldots,p$ $t_i\le l(X,Y) (a_{i}+b_i)$ does not always hold. $\endgroup$
    – Xiaopai Song
    Commented Feb 2, 2019 at 15:08

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