I am looking for a reference to verify the following inequality, where $X$ and $Y$ are Hermitian positive semidefinite matrices: $$ \lambda_n(X^{1/2}YX^{1/2}) = \lambda_n(XY) \leq \lambda_n(X)\lambda_1(Y). $$ This might be too trivial for MathOverflow but I am completely blanking on this.
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1$\begingroup$ Why does $XY$ have real eigenvalues? $\endgroup$– AlfCommented Nov 8, 2023 at 23:31
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2$\begingroup$ @Alf: $X=AA^*$, $Y=BB^*$, and $XY=A(A^*BB^*)$ has the same eigenvalues as $(A^*BB^*)A=(A^*B)(A^*B)^*\succeq0$. $\endgroup$– Dustin G. MixonCommented Nov 9, 2023 at 2:00
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1$\begingroup$ math.stackexchange.com/questions/3892626/… $\endgroup$– Dustin G. MixonCommented Nov 9, 2023 at 2:05
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1 Answer
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Just combining my comments to form an explicit answer. Write $X=AA^*$ and $Y=BB^*$. Then $XY$ has the same eigenvalues as $(A^*B)(A^*B)^*$, and so
$$\lambda_n(XY)=\sigma_n(A^*B)^2\leq\big(\sigma_n(A^*)\sigma_1(B)\big)^2=\lambda_n(X)\lambda_1(Y),$$
where the inequality follows from [https://math.stackexchange.com/questions/3892626/singular-values-of-product-of-matrices].
answered Nov 9, 2023 at 9:46