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Winfried
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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?

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 $C>0$ does not depend on the dimension $n\in\mathbb{N}$. Does anyone know of a proof for that?

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?

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Winfried
Winfried
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LU factorization for $I+A$ (A skew-symmetric)

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 $C>0$ does not depend on the dimension $n\in\mathbb{N}$. Does anyone know of a proof for that?