Suppose that $x^{*}$ is an approximate numerical solution to $Ax=b.$ Show that $$\frac{||x-x^*||}{||{x}||} \leq k(A) \frac{||r||}{||b||}$$ where $k(A)$ is the condition number of $A$ and $r=b-Ax^*$ is the residual vector.
Here are my workings, are they correct? looking for some help clarifying.
the error vector $e$ is defined as $ e = x - x^*$.
Using the residual to determine, or estimate, the error in the calculated solution, the first step is to realize that these two vectors are related through the formula $r = Ae$ or equivalently $e = A^{-1}r$
and we know that $k(A)$is the condition number of $A$ and is defined as $k(A) = ||A||\cdot ||A^{-1}||$.
Since $b$ is nonzero, then $x$ is nonzero.
First since $Ax=b$ then $||b|| = ||Ax||\leq ||A|| \cdot ||x||$. From this we have the first inequality.
$$\frac{1}{||x||} \leq \frac{||A||}{||b||}$$. As for the second inequality, since $e = A^{-1}r,$ then $||e|| \leq ||A^{-1}|| \cdot ||r||.$ Then the theorem follows by combining the two inequalities.