Singular value or eigenvalue problems lie at the center of matrix analysis. One classical result is


$$\lambda\_{j}(X^{\*}X+Y^{\*}Y)\geq 2\sigma\_j(XY^\*)$$


for $j \in \{1, \ldots, n\}$, where $\lambda\_j(\cdot)$ and $\sigma\_j(\cdot)$ denote the $j$th largest eigenvalue and singular value, respectively, and $X$ and $Y$ are $n \times n$ complex matrices. I conjecture that the following does hold.


$$\lambda\_{j}((I+X^{\*}X)(I+Y^{\*}Y))\geq\sigma\_j^2(I+XY^\*)$$



A singular value-eigenvalue inequality
The conjecture is true.


**Lemma 1** : For every matrix $Z$ holds $\lambda\_j(Z^\* Z + Z^\* + Z ) \leq \lambda\_j(Z^\* Z + 2 (Z^\* Z)^{1/2})$ .


For a proof see the proof of $\lambda\_j(Z^\* + Z ) \leq \lambda\_j(2 (Z^\* Z)^{1/2})$ in Bhatia, Matrix Analysis, Proposition III.5.1 (Fan-Hoffman).


**Lemma 2** : For $a \geq 0 , b > 0$ holds $(a+b)(1+b^{-1}) \geq a + 2 a^{1/2} + 1$ .


Proof left to the reader.


**Proof of the conjecture :**


Choose $Z = XY^\*$, $A=Z^\* Z$, $B=YY^\*$ .


We may assume that $B$ is invertable.


Then we have to show :


$\lambda\_j((I+B^{-1})^{1/2}(A+B)(I+B^{-1})^{1/2}) \geq \lambda\_j(A + 2A^{1/2} + I)$ .


Let $V\_j$ be the span of the eigenvectors to the j largest eigenvalues of $A$ and $W\_j = (I+B^{-1})^{-1/2}(V\_j)$ .


Then


$\lambda\_j((I+B^{-1})^{1/2}(A+B)(I+B^{-1})^{1/2})$
$ \geq min \{x^\* ((I+B^{-1})^{1/2}(A+B)(I+B^{-1})^{1/2})x : x \in W\_j$ and $x^\* x = 1\}$
$\geq min \{x^\* ((I+B^{-1})^{1/2}(\lambda\_j(A)I+B)(I+B^{-1})^{1/2})x : x \in W\_j$ and $x^\* x = 1\}$
$\geq \lambda\_j(A) + 2 \lambda\_j(A)^{1/2} + 1 = \lambda\_j(A + 2A^{1/2} + I)$ .


