I got stuck on this problem. Suppose $Y_{k\times d}$ is a matrix, and $Y_1$ is the submatrix of $Y$ consisting of the first $m-$rows of $Y$. I am wondering wether the following inequality is true?
$$tr(Y_1^\mathrm{T}Y_1(Y^\mathrm{T}Y)^{-1})\ge \frac{tr(Y_1^\mathrm{T}Y_1)}{tr(Y^\mathrm{T}Y)}$$ I have done a lot of numerical experiments in matlab, and did not find any counter-example, but I do not know how to prove it.
Moreover, what I really want to know is that is there any equality holds between the LHS and RHS or if there is a constant $C$ such that
$$tr(Y_1^\mathrm{T}Y_1(Y^\mathrm{T}Y)^{-1})\le C\frac{tr(Y_1^\mathrm{T}Y_1)}{tr(Y^\mathrm{T}Y)}$$.
Any comments would be highly appreciated.
We have $B:=Y^TY\succeq A:=Y_1^TY_1\succeq0$. Your first inequality is equivalent to $\operatorname{tr}(AB^{-1})\ge\frac{\operatorname{tr}(A)}{\operatorname{tr}(B)}$, which follows from Cauchy-Schwarz inequality $$ \langle A^{1/2}B^{-1/2},A^{1/2}B^{-1/2}\rangle \langle B^{1/2},B^{1/2}\rangle \ge\langle B^{1/2},B^{-1/2}A^{1/2}\rangle^2 $$ where $\langle X,Y\rangle$ denotes the Frobenius inner product $\operatorname{tr}(X^TY)$.
The desired constant $C$ in your second inequality does not exist. Intuitively, that's because $B^{1/2}$ can be almost orthogonal to $B^{-1/2}A^{1/2}$. For instance, let $$ Y=\pmatrix{t&0\\ 0&t\\ 0&1},\,Y_1=\pmatrix{t&0\\ 0&t},\,Y^TY=\pmatrix{t^2&0\\ 0&t^2+1},\,Y_1^TY_1=\pmatrix{t^2&0\\ 0&t^2} $$ for some nonzero $t$. Then $$ \operatorname{tr}\left(Y_1^TY_1(Y^TY)^{-1}\right)=1+\frac{t^2}{t^2+1}, \quad\frac{\operatorname{tr}(Y_1^TY_1)}{\operatorname{tr}(Y^TY)}=\frac{2t^2}{2t^2+1}. $$ Clearly, there does not exist any constant $C$ such that $1+\frac{t^2}{t^2+1}\le\frac{Ct^2}{1+t^2}$ for all $t$, because $\lim\limits_{t\to0}\left(1+\frac{t^2}{t^2+1}\right)=1>0=\lim\limits_{t\to0}\frac{2Ct^2}{1+t^2}$.