I am working to prove the following $\|A^{-1}\|_\infty \ge \frac{\|U^{-1}\|_\infty}{n}$ where $PA=LU$.
Proof involving matrix inverse of: $\|A^{-1}\|_\infty \ge \frac{\|U^{-1}\|_\infty}{n}$
4
$\begingroup$
linear-algebra
-
0You can LaTeX here. Just enclose it within `$...$` – 2012-03-09
1 Answers
3
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}.$
-
0This 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! – 2012-03-09