I was trying to prove this inequality, by taking $K(A) = ||A|| ||A^{-1}||$ and also the error $A(x -\hat{x}) =e$, I am thinking how to get those terms, estimates? any help in ideas to proceed.
2026-03-25 03:34:07.1774409647
Proving an inequality based on condition number.
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It is easy to get from $Ax=b$ and $\hat{A}\hat{x}=b$ to $$ \frac{\|x-\hat{x}\|}{\|x\|}\leq\|I-\hat{A}^{-1}A\|. $$
Now the rest is a bit more technical. Let $\hat{A}:=A+E$. We have $$\tag{1} I-\hat{A}^{-1}A=\hat{A}^{-1}(\hat{A}-A)=(A+E)^{-1}E. $$ From $A(A+E)^{-1}=(A+E-E)(A+E)^{-1}=I-E(A+E)^{-1}$, we have $$ (A+E)^{-1}=A^{-1}-A^{-1}E(A+E)^{-1}, $$ so with $\|A^{-1}\|\|E\|<1$, we get $$ \|(A+E)^{-1}\|\leq \|A^{-1}\|-\|A^{-1}\|\|E\|\|(A+E)^{-1}\|\implies \|(A+E)^{-1}\|\leq\frac{\|A^{-1}\|}{1-\|A^{-1}\|\|E\|}. $$ Putting this back to (1) gives $$ \|I-\hat{A}^{-1}A\|\leq\frac{\|A^{-1}\|\|E\|}{1-\|A^{-1}\|\|E\|} =\frac{\|A\|\|A^{-1}\|}{1-\|A^{-1}\|\|E\|}\frac{\|E\|}{\|A\|}. $$ Now just plug back $E=\hat{A}-A$.