TITLE: LU factorization for $I+A$ (A skew-symmetric)
QUESTION [11 upvotes]: 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?

REPLY [4 votes]: 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 $(||L||+||U||)/(1+{a_1}^2+{a_2}^2+{a_3}^2)\approx 2.18577$.