1
$\begingroup$

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 $X,Y.$

Improve this question
$\endgroup$
2
  • $\begingroup$ 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
    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
    Feb 2, 2019 at 15:08

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.