Consider solving $Ax = b$ where $A \in \mathbb{R}^{n\times n}$ and $b \in \mathbb{R}^n$. Given a perturbation $\hat A = A + \Delta A$, suppose the solution to the system $\hat A\hat x = b$ is $\hat x = x + \Delta x$. Can I prove the following unequality?
$$\lVert{\Delta x}\rVert / \lVert \hat x \rVert \leq \lVert A^{-1}\rVert\lVert \Delta A \rVert$$
I'm not sure where to start, so even just hints would be appreciated.
I will use $\tilde A$ and $\tilde x$ for pertubations. I hate that triangle ;)
You can start by using $Ax=b$ and $(A+\tilde A)(x+\tilde x)=b$ Expanding the second equation gives you $Ax + \tilde Ax + A \tilde x + \tilde A \tilde x=b$. You use your knowledge of $Ax=b$ and arrive at $$A \tilde x = -\tilde A x - \tilde A \tilde x. $$ Since $A$ is invertible (otherwise, there would be no unique solution $x$ to $Ax=b$), you can determine $\tilde x = -A^{-1}\tilde Ax - A^{-1}\tilde A \tilde x$. By taking the norm of both sides, you arrive at $$\|\tilde x\| = \|A^{-1} \tilde{A}(x+\tilde x)\|. $$ Taking the sub-multiplicativity into account, you get exactly where you want to be!