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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.

$$\lambda_{j}((I+X^{*}X)(I+Y^{*}Y))\geq\sigma_j^2(I+XY^*)$$

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  • 2
    $\begingroup$ I've done some numerical tests in n=4 with no counterexamples yet. I prefer the following form of the question: $\lambda_j((I + X^*X)(I+ Y^*Y)) \geq \lambda_j((I+YX^*)(I+XY^*))$? $\endgroup$ – Chris Ramsey Nov 21 '17 at 20:41
  • $\begingroup$ Thanks. Let us see if that form has any advantage in a possible "proof ". $\endgroup$ – M. Lin Nov 22 '17 at 0:55
  • $\begingroup$ Is it known that $\lambda_j(X^*X Y^*Y) \geq \lambda_j(YX^* XY^*)$ ? $\endgroup$ – jjcale Nov 22 '17 at 20:44
  • $\begingroup$ @jjcale Yes, by similarity when $Y$ is invertible and then the general case by continuity. $\endgroup$ – Chris Ramsey Nov 22 '17 at 20:54
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    $\begingroup$ Equivalent : $\lambda_j(A B^{-1} + B + A) \geq \lambda_j(2A^{1/2} + A)$ for hermitian positive definite $A$ and $B$ : Choose $Z = XY^*$, $A=Z^* Z$, $B=YY^*$ . $\endgroup$ – jjcale Nov 25 '17 at 21:09
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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 $x^* x = 1\}$ $\geq \lambda_j(A) + 2 \lambda_j(A)^{1/2} + 1 = \lambda_j(A + 2A^{1/2} + I)$ .

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  • $\begingroup$ Nice proof! Thanks. I think this inequality is of independent interest. Are you interested in exploring the possible use of this result with me? As I don't know your real name, you are more than welcome to contact me... m_lin@i.shu.edu.cn $\endgroup$ – M. Lin Dec 3 '17 at 2:33

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