6
$\begingroup$

Let $D\in \mathbb{R}^{n\times n}$ denote a lower triangular matrix. With $\|\cdot\|$ denoting the spectral matrix norm, is there an estimate like

$$ \|D\| \leq C\|D+D^T\|, $$

where $C>0$ is independent of the dimension $n\in\mathbb{N}$ and $D$?

$\endgroup$
3
  • 1
    $\begingroup$ No. The best constant is $C\sim\ln n$. See this question and the link provided there for background: mathoverflow.net/questions/177198/… $\endgroup$ Aug 6 '14 at 16:32
  • $\begingroup$ @ChristianRemling I think you could post this as an answer $\endgroup$
    – Yemon Choi
    Aug 6 '14 at 17:57
  • $\begingroup$ @YemonChoi: OK, will do. $\endgroup$ Aug 6 '14 at 18:10
4
$\begingroup$

No. The best constant is $C_n\sim \ln n$. See, for example, this paper. In particular, for the lower bound, see example 3.3 of the paper.

$\endgroup$
1
4
$\begingroup$

The $\log n$ result mentioned by Christian Remling is a special case of the results in

Kwapień, S.; Pełczyński, A. The main triangle projection in matrix spaces and its applications. Studia Math. 34 1970 43–68.

$\endgroup$
0
$\begingroup$

The assumption that $D$ is triangular is not necessary (the Schur decomposition may bring a matrix to its triangular form). If the numerical range of D is contained in a sector, then Lemma 3.1 in http://www.tandfonline.com/doi/abs/10.1080/03081087.2014.933219#preview says that there is a C in terms of secant function.

$\endgroup$
3
  • $\begingroup$ Unless I have misremembered, even if a square matrix $A$ is unitarily conjugate to a triangular matrix $D$, it might not be the case that $A+A^\top$ is unitarily conjugate to $D+D^\top$. Therefore I think that the question as stated really does want to start with the assumption that $D$ is triangular. $\endgroup$
    – Yemon Choi
    Aug 16 '14 at 18:43
  • $\begingroup$ @Yemon, Yes, you are right. If the transpose $^T$ is replaced with the conjugate transpose $^*$, then my comment is valid. $\endgroup$
    – M. Lin
    Aug 17 '14 at 3:48
  • $\begingroup$ Name of the linked paper: Zhang - A matrix decomposition and its applications. $\endgroup$
    – LSpice
    Jan 20 at 3:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.