Weyl’s monotonicity theorem says that if $X \leq Y$ with $X$ and $Y$ Hermitian matrices, then $\lambda_{j}(X)\leq \lambda_{j}(Y)$ for j=1,2,...,n. I have to prove that these inequalities are equivalent to the following: There exist a unitary $U$ such that $X\leq U^*YU$.
I was able to prove the necessary condition that is if $\lambda_{j}(X)\leq \lambda_{j}(Y)$ for j=1,2,...,n then there exists a unitary matrix such that $X \leq U^*YU$ But how do i prove the sufficient condition?
Nessacerity: Suppose that$\lambda_j(X) \le \lambda_j(Y)$, for $i= 1,\ldots,n$. Let $X = V_1 D_1 V_1^*$ and $Y = V_2 D_2 V_2^*$ be spectral decomposition of $X$ and $Y$, respectively, with $D_1 \le D_2$. So $V_1^*XV_1\le V_2^*YV_2 $. Therefore $ X\le(V_1V_2^*)Y(V_2V_1^*)$.
Sufficiency: Suppose that $X \leq U^* Y U$. Then by Weyl’s monotonicity theorem, $\lambda_j(X) \le \lambda_j(U^*YU)=\lambda_j(Y)$, for $i=1,\ldots,n$.