The matrix condition number of a matrix $A$ is defined as $\kappa(A) = \Vert A\Vert \Vert A^{-1}\Vert$. I want to show that $$\frac{\Vert x\Vert}{\Vert Ax\Vert}\Vert A\Vert = \kappa(A).$$
Edit There was a mistake in my original statement. It is clear that $$\frac{\Vert x\Vert}{\Vert Ax\Vert}\Vert A\Vert = \frac{\Vert A^{-1}Ax\Vert}{\Vert Ax\Vert}\Vert A\Vert\leq\Vert A^{-1}\Vert\Vert A\Vert = \kappa(A).$$ On the other hand we have $$\frac{\Vert x\Vert}{\Vert Ax\Vert}\Vert A\Vert \geq \frac{\Vert x\Vert}{\Vert A\Vert\Vert x\Vert}\Vert A\Vert =1.$$ Now my question was whether finding $A$ and $x$ such that the bound is attained suffices to prove the desired direction of the inequality.
(Disclaimer: I asked the same question a couple of months ago. However, the given proof was incorrect, so I came up with the above idea)
The equality $\frac{\|x\|}{\|Ax\|} = \kappa(A)$ is false, as well as both inequalities. For $A = 2I$ ($I$ being the identity), you have $\kappa(A) = 1$ and $\|x\|/\|Ax\| = 1/2$, while for $A= I/2$, you have $\kappa(A)= 1$ and $\|x\|/\|Ax\|=2$.