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The matrix $M=I+A\in \mathbb{R}^{n\times n}$, where $I$ is the identity and $A=-A^T$ is skew-symmetric satisfies $$ Mx\cdot x = \|x\|^2 \quad\text{for all }x\in\mathbb{R}^n. $$ Therefore, $M=LU$ has an unique $LU$-factorization. Numerical experiments with Matlab suggest that there holds the following estimate $$ \|L\| + \|U\|\leq C(1+\|A\|^2), $$ where $\|\cdot\|$ is the spectral norm and the real constant $C>0$ does not depend on the dimension $n\in\mathbb{N}$. Does anyone know of a proof for that?

Winfried
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  • I observed $C\leq 1$. – Winfried Jul 14 '14 at 12:59
  • @NathanielJohnston You are right, I deleted my remarks. – Piotr Migdal Jul 14 '14 at 13:17
  • $C \leq 1$ isn't possible, even for $2\times 2$ matrices. Let $a = 0.379$ and consider $A = \begin{bmatrix}0 & a\ -a & 0\end{bmatrix}$. Then $|L| + |U| = 2.4984$ but $1 + |A|^2 = 1.1436$. In fact, a direct calculation shows that for $2 \times 2$ matrices the smallest possible $C$ is approximately $2.1846$. – Nathaniel Johnston Jul 14 '14 at 13:42
  • Are you assuming that $L$ has ones on the diagonal? The claim will be false without some kind of normalisation like this, but that particular choice of normalisation feels quite arbitrary in this context. It would seem more natural to set things up so that if the chosen factorisation of $I+A$ is $LU$, then the chosen factorisation of $I-A$ is $U^TL^T$. – Neil Strickland Jul 14 '14 at 20:35
  • I should have written $C\lesssim 1$, I just wanted to say that the constant is small. However, I don't really care about the size of the constant. In my computations $L$ is normalized to have one's at the diagonal, but your normalization would also be fine. – Winfried Jul 15 '14 at 07:57
  • I would be very careful in assuming this is true based on numerical examples only. The worst-case bounds for the growth factor in LU factorization of general (non-antisymmetric) matrices are quite hard to reach in practice. – Federico Poloni Jul 15 '14 at 11:58
  • That is true. However, most of the worst-case examples I know are not uniformly positive. Nevertheless, I would be also happy with a counterexample for the estimate. – Winfried Jul 15 '14 at 12:03
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    For what it's worth, it seems like the smallest possible $C$ in the $3\times 3$ case is also approximately $2.1846$, based on parametrizing all such $3\times 3$ matrices and constructing a mesh over them. So we might (naively) guess that this $C$ works in all dimensions. – Nathaniel Johnston Jul 15 '14 at 15:36
  • The question is more interesting if you modify a bit the definition of $LU$. Usually, the diagonal of $L$ is the unit, and that of $U$ is free. But since you estimate $|L|+|U|$, it is more natural to require that the diagonal of $L$ equals that of $U$. – Denis Serre Nov 19 '14 at 16:42

1 Answers1

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EDIT 1. For $n=2$, the smallest $C$ is as follows. In all calculations we use the "Groebner" library of Maple. Let $C_n$ be the smallest $C$ (if there exists) in dimension $n$.

Let $a\approx 0.3787$ be the number satisfying $4a^8+27a^6+34a^4-117a^2+16=0$. Note that $a$ can be calculated by radicals. Then

$C_2=\dfrac{\sqrt{2a^2+4+2a\sqrt{a^2+4}}+\sqrt{2a^4+6a^2+4+2a(a^2+1)\sqrt{a^2+4}}}{2(a^2+1)}$.

Moreover $C_2\approx 2.184596$ is a root of the following polynomial: $20736x^8-122832x^6+126913x^4-62224x^2+1024$. Note that $C_2$ can be calculated by radicals.

EDIT 2. When $n=3$, let $A=\begin{pmatrix}0&-a_3&a_2\\a_3&0&-a_1\\-a_2&a_1&0\end{pmatrix}$ ; there are $2$ cases.

Case 1. $a_1a_2a_3\geq 0$. Then the bound for $C$ seems to be $C_2$.

Case 2. $a_1a_2a_3\leq 0$. Then the bound is $>C_2$ and $C_3>C_2$. Indeed, let $a_1=-0.25,a_2=0.29,a_3=0.029$ ; then $(||L||+||U||)/(1+{a_1}^2+{a_2}^2+{a_3}^2)\approx 2.18577$.

loup blanc
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