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Let $U\in\mathbb{R}^{d\times d}$ an upper triangular real matrix. Let $$S = R+R^T - R^TR,$$ which is obviously symmetric.

I'm interesting in getting an estimate of $\|S\|_F$, the Frobenius norm of $S$. Of course triangular inequality yields: $$\left|\sqrt{2}\|R\|_F-\|R\|^2_F\right| \leq\|S\|_F\leq \|R+R^T\|_F + \|R^TR\|_F \leq \sqrt{2}\|R\|_F+\|R\|^2_F.$$

Nonetheless for a diagonal matrix this estimate gets quite off for large matrices, so I'm wondering if it is possible to get a finer estimate than this one?

user70925
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1 Answers1

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Not sure if this is useful, but if you rewrite $S$ as $I-(R-I)^T(R-I)$, then \begin{align} \|S\|_F\le\|I\|_F+\|R-I\|_F^2 &= \|I\|_F+\|I\|_F^2+\|R\|_F^2-2\operatorname{tr}(R)\tag{1} \end{align} and tie occurs when $R=I$. You may further relax the $-2\operatorname{tr}(R)$ in $(1)$ to $2\|I\|_F\|R\|_F$ using Cauchy-Schwarz inequality, but then the resulting inequality is not tight because tie occurs in the Cauchy-Schwarz inequality here at $R=-I$ rather than at $R=I$.

user1551
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