Timeline for Matrix eigenvalues inequality (2)

7 events

when toggle format	what		by	license	comment
Feb 2, 2019 at 20:31	history	edited	Xiaopai Song	CC BY-SA 4.0	added 1 character in body
Feb 2, 2019 at 15:08	comment	added	Xiaopai Song		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.
Feb 2, 2019 at 14:57	comment	added	Xiaopai Song		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 $X'(A+YBY')^{-1}X$.
Feb 2, 2019 at 14:28	history	edited	Xiaopai Song	CC BY-SA 4.0	deleted 27 characters in body
Feb 2, 2019 at 3:24	history	edited	Xiaopai Song	CC BY-SA 4.0	edited title
Feb 2, 2019 at 2:51	history	edited	Martin Sleziak	CC BY-SA 4.0	minor typos
Feb 2, 2019 at 2:39	history	asked	Xiaopai Song	CC BY-SA 4.0