# Norm of the upper triangular part of symmetric matrix

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 $D$?

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

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.
The $\log n$ result mentioned by Christian Remling is a special case of the results in
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.
• 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. Aug 16 '14 at 18:43
• @Yemon, Yes, you are right. If the transpose $^T$ is replaced with the conjugate transpose $^*$, then my comment is valid. Aug 17 '14 at 3:48