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I am working to prove the following $$\|A^{-1}\|_\infty \ge \frac{\|U^{-1}\|_\infty}{n}$$ where $PA=LU$.

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You can LaTeX here. Just enclose it within $...$ – user2468 Mar 9 '12 at 3:27
up vote 3 down vote accepted

When you do $$PA = LU,$$ I assume you are doing a partial pivoting. If so, the elements of $L$ are less than unity with the diagonal entries being $1$. This means $\|L\|_{\infty} \leq n$. (Because the infinity norm of a matrix is the largest row sum of the absolute values and in each row of $L$ all the elements are atmost $1$.) Hence, $$U^{-1} = A^{-1}P^{-1}L$$ i.e. $$\|U^{-1}\|_{\infty} = \|A^{-1}P^{-1}L\|_{\infty} \leq \|A^{-1}\|_{\infty} \|P^{-1}\|_{\infty} \|L\|_{\infty} = \|A^{-1}\|_{\infty} \|L\|_{\infty} \leq n \|A^{-1}\|_{\infty}$$ Note that $\|P\|_{\infty} = 1$ since all the row sum is $1$. Hence, we get $$ \|A^{-1}\|_{\infty} \geq \frac{\|U^{-1}\|_{\infty}}{n}.$$

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This is great thanks a lot! I was focusing on relating the absolute row sum of inv(A) to that of inv(U) and totally missed the property of L that completes the inequality! – mathjacks Mar 9 '12 at 13:16

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