Norm Of Matrix Inverse I'm reading Heath's Scientific Computing.
The p-norm of an n-vector $x$ is defined as $\Vert x \Vert_p=\left( \sum_{i=1}^{n} \vert x_i \vert ^{p} \right) ^ {\frac{1}{p}}$  
The matrix norm of an $m$ x $n$ matrix $A$ is defined as $\Vert A \Vert = \max_{x \neq 0} \frac{\Vert Ax \Vert}{\Vert x \Vert}$
The condition number of non-singular square matrix $A$ is defined as $cond(A)=\Vert A \Vert \Vert A^{-1}\Vert$  
In one of the examples, it is stated that:  
$\Vert A \Vert \Vert A^{-1}\Vert = \left( \max_{x \neq 0} \frac{\Vert Ax \Vert}{\Vert x \Vert} \right) \left( \min_{x \neq 0} \frac{\Vert Ax \Vert}{\Vert x \Vert} \right)^{-1}$   
By definition, $\Vert A^{-1}\Vert=\left( \max_{x \neq 0} \frac{\Vert A^{-1}x \Vert}{\Vert x \Vert}\right)$  
I'm looking for clarification on how you get from $\left( \max_{x \neq 0} \frac{\Vert A^{-1}x \Vert}{\Vert x \Vert}\right)$  to $\left( \min_{x \neq 0} \frac{\Vert Ax \Vert}{\Vert x \Vert} \right)^{-1}$
 A: Let 
$y=A^{-1}x$
and
$x=Ay$.
Note that $x$ is 0 if and only if $y$ is 0.  
Then 
$\frac{\| A^{-1}x \|}{\| x \|}=\frac{\| y \|}{\| Ay \|}$.
Since all of these norms are greater than 0 as long as $x \neq 0$, maximizing $\| A^{-1}x \| / \| x \|$ is equivalent to maximizing 
$\| y \| / \| Ay \|$, or minimizing 
$\| Ay \| / \| y \|$ (and taking the reciprocal.)
A: \begin{align*}
\Vert A^{-1} \Vert
& = \max_{x \neq 0} \frac{\Vert A^{-1} x\Vert}{\Vert x \Vert} \\
& = \max_{Ax \neq 0} \frac{\Vert A^{-1} Ax\Vert}{\Vert Ax \Vert} \\
& = \max_{Ax \neq 0} \frac{\Vert x\Vert}{\Vert Ax \Vert} \\
& = \max_{x \neq 0} \frac{\Vert x\Vert}{\Vert Ax \Vert} \\
& = \left( \left( \max_{x \neq 0} \frac{\Vert x\Vert}{\Vert Ax \Vert} \right)^{-1} \right)^{-1} \\
& = \left( \min_{x \neq 0} \frac{\Vert Ax\Vert}{\Vert x \Vert} \right)^{-1} \\
\end{align*}
The second equality follows from the fact that $x \neq 0$ if and only if $Ax \neq 0$ since $A$ is nonsingular (and hence has a kernel which only includes the zero vector).
The same trick is used a few lines later.
