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For any $X \in \mathbb{C}^{m\times n}$, let $\Sigma(X)$ be the "middle factor" in its SVD, so that $X = U\Sigma(X) V^H$ and the diagonal of $\Sigma(X)$ is arranged in descending order.

Theorem ([Theorem 7.4.9.1, R. A. Horn and C. R. Johnson. Matrix Analysis. Cambridge University Press, Cambridge, second edition, 2012]). For any matrices $X, Y\in\mathbb{C}^{m\times n}$ and any unitarily invariant norm $\|\cdot\|$ on $\mathbb{C}^{m \times n}$, we have \begin{equation} \|X - Y\| \ge \|\Sigma(X) - \Sigma(Y)\|. \end{equation}

I am investigating the equality condition for the above inequality.

For the Frobenius norm, the inequality reduces to von Neumann's trace inequality [Theorem I, von Neumann 1937], and the equality holds if and only if there exist unitary matrices $U \in \mathbb{C}^{m\times m}$ and $V \in \mathbb{C}^{n\times n}$ such that $X = U \Sigma(X) V^H$ and $Y = U \Sigma(Y) V^H$. See the summary in Section 2.1 of Tight Error Bounds for Nonnegative Orthogonality Constraints and Exact Penalties, Chen, He, and Zhang, 2022.

Is there any known result except the above one for the Frobenius norm? Is the abovementioned condition still necessary (in some cases)? Thanks.

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  • $\begingroup$ Also, perhaps relevant, for symmetric $X,Y$, you can always find invertible $S$ such that equality holds for $X_2=S'XS$ and $Y_2=S'YS$, explicit formula for $S$ is math.stackexchange.com/a/1080923/998 $\endgroup$
    – Yaroslav Bulatov
    Commented Aug 1, 2022 at 22:59
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    $\begingroup$ Condition from Von Neumann's trace inequality should also apply to any unitarily invariant norm since its value is uniquely determined by singular values, see Theorem 1.1 (von Neuman, 1937) "The mathematics of eigenvalue optimization" paper $\endgroup$
    – Yaroslav Bulatov
    Commented Aug 1, 2022 at 23:02
  • $\begingroup$ Thank you @YaroslavBulatov for the comments. That’s also my intuition and motivation for this study. $\endgroup$
    – Nuno
    Commented Aug 2, 2022 at 0:08

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