Show that for sufficiently small $\delta A$, we get $||\delta x||\leq ||A^{-1}||(||\delta A||\cdot||\hat{x}||+||\delta b|| )$ is strictly an equality

Question

Proof attempt:

Given $\delta x=\hat{x}-x,\,Ax=b$. Assume $||A^{-1}\delta A||<1$

Now, $\hat{x}-x=(A+\delta A)^{-1}(b+\delta b)-A^{-1}b=((A+\delta A)^{-1}-A^{-1})b+(A+\delta A)^{-1}\delta b$

So $||\hat{x}-x||\leq ||(A+\delta A)^{-1}-A^{-1}||\cdot||b||+||(A+\delta A)^{-1}||\cdot||\delta b||$

but $||(A+\delta A)^{-1}||=||(A+\delta A)^{-1}-A^{-1}+A^{-1}||\leq||(A+\delta A)^{-1}-A^{-1}||+||A^{-1}||$

but $||(A+\delta A)^{-1}-A^{-1}||\leq||A^{-1}||\cdot||(I+A^{-1}\delta A)^{-1}(I+A^{-1}\delta A-I)||=||A^{-1}||\cdot||(I+A^{-1}\delta A)^{-1}A^{-1}\delta A||\leq\frac{||A^{-1}||\cdot||A^{-1}\delta A ||}{1-||A^{-1}\delta A||}$

Now, going back $||\hat{x}-x||\leq\frac{||A^{-1}||}{1-||A^{-1}\delta A||}(||A^{-1}\delta A||\cdot||b||+||\delta b||)\leq\frac{||A^{-1}||}{1-||A^{-1}\delta A||}(||x||\cdot||\delta A||+||\delta b||)$

$\implies ||\hat{x}-x||\leq||A^{-1}||(||\delta A||\cdot||x||+||\delta b||)$

Am I going in the right direction? Any adjustments need to be done?

Answer 1

I would start with $$ A·(\hat x-x)=(-δA·\hat x+b+δb)-b=-δA\,\hat x+δb, $$ the claimed identity follows directly.