Let $b \in \mathbb R^n$ be fixed.
Find the relative condition numbers of the following problem:
Find the solution $x \in \mathbb R^n$ of $Ax=b$ for the invertible matrix $A \in \mathbb R^{n\times n}$.
We defined the relative condition numbers as
$$\kappa_{ij}^{\mathrm{rel}}(x)=\Bigg|\frac{\partial f_i(x)}{\partial x_j}\frac{x_j}{f_i(x)}\Bigg|$$
For example if we look at the addition $y=x_1+x_2$ we have $\kappa_{\mathrm{rel}}(x_1,x_2)=|\frac{x_1}{x_1+x_2}|$ and $|\frac{x_2}{x_1+x_2}|$.
Or for the multiplication $y=x_1x_2$ we have $\kappa_{\mathrm{rel}}(x_1,x_2)=|x_2\frac{x_1}{x_1x_2}|=1$ and $|x_1\frac{x_2}{x_1x_2}|=1$
The hint was to use the implicit function theorem so maybe we need to use implicit differentiation but I don't know
Thanks!
In the case of linear systems, you can get direct estimates on the relative error... Assuming for simplicity that the error comes from an inexact right hand side you can reason as follows:
If you are solving $A \tilde x = \tilde b$ instead of $Ax = b$, by algebraic manipulation you show that $$ x -\tilde x = A^{-1}(b -\tilde b). $$
Taking norms you conclude that
$$ \|x -\tilde x\| \leq \|A^{-1}\| \|b -\tilde b\| \Rightarrow \frac{\|x -\tilde x\|}{\|x\|}\leq \|A^{-1}\| \frac{\|b -\tilde b\|}{\|x\|} $$
you can conclude that $$ \frac{\|x -\tilde x\|}{\|x\|}\leq \|A^{-1}\| \|A\| \frac{\|b -\tilde b\|}{\|b\|}. $$