The theorem I am trying to prove is that the condition number of matrix A is:
$$\operatorname{cond}(A) = \|A\|\cdot\|A^{-1}\|$$
where $\|A\|$ signifies the operator norm.
From the properties of the operator norm it is know that:
$\|Ax\|\leq \|A\|\|x\|$
Then using the following equalities:
$$A(x -x_a) = r \text{ and } Ax = b$$
hence
$$\|(x -x_a)\| \leq ||A^{-1}\|\|r\|$$
and
$$\frac{1}{\|b\|} \geq \frac{1}{\|A\|\|x\|}$$
Combining the two inequalities gives:
$$\frac{\|x - x_a\|}{\|x\|} \leq \frac{\|A\|}{\|b\|}\|A^{-1}\|\|r\|$$
This shows that $\|A\|\|A^{-1}\|$ is an upper bound for this expression.
I had thought that this is true due to the fact that $b = Ax$ and $r = A(x-x_a)$ so r is always less than b but this does not take into account the possibility that $x_a$ is much bigger than $x$.
Q1.) How do you show that $\|A\|\|A^{-1}\|$ is an upper bound for this expression?
It then says to show that this quantity is always obtainable.
To do so they set
$x$ such that $\|A\|=\frac{\|Ax\|}{\|x\|}$
$r$ such that $\|A^{-1}\|= \frac{\|A^{-1}r\|}{\|r\|}$
$x_a$ such that $x - x_a = A^{-1}r$
Then it just remains to check the equality:
$$\frac{\|x - x_a \|}{\|x\|}=\frac{\|A^{-1}r\|}{\|x\|}=\frac{\|A^{-1}\|\|r \|\|A \|}{\|Ax\|}$$
Now I understand how you get from the first to second equality (that's straight forward) but I don't see how the last equality shows that the $||A||.||A^{-1}||$ is achievable. I've tried several things but I just can't see how the judicious choice of variables above can produce the $\frac{\| r\|}{\| b\|} = 1$ required?
Q2.) How can I prove that $\frac{\| r\|}{\| b\|} = 1$ as required?
Thanks
Baz