Why is $\kappa(A^{-1})= \kappa(A)$? In my syllabus we have the alternative definition of the condition of a matrix:
$$\kappa(A)= \frac{\text{max}_{\| \vec{y} \| =1}\| A \vec{y} \|}{\text{min}_{\| \vec{y} \| =1}\| A \vec{y} \|}$$
In it, it also says that by definition of the condition of a matrix it follows that $\kappa(A^{-1})= \kappa(A)$. So there is no explanation for this. Therefore, my question is: Why is $\kappa(A^{-1})= \kappa(A)$?

First you must accept that if $A^{-1}$ doesn't exist then by convention $\kappa(A)=\infty$.
Once you've postulated that $A^{-1}$ exists, you want to show
$$\max_{\| x \|=1} \| A^{-1} x \|=\left ( \min_{ \| x \|=1 } \| Ax \| \right )^{-1}.$$
To see that, take a minimizer $x^*$ on the right side. Let $b=\frac{Ax^*}{\| Ax^* \|}$; then look at $A^{-1} b=\frac{x^*}{\| Ax^* \|}$. You picked $x^*$ so that among unit vectors, $\| Ax^* \|$ is as small as possible, so $\| A^{-1} b \|=\frac{1}{\| Ax^* \|}$ is as large as possible among unit vectors $b$.
Intuitively, if $A$ maps $x^*$ to some much smaller vector $b$, then $A^{-1}$ maps $b$ back to $x^*$, which is much bigger than $b$. Juggling the normalizations obfuscates this intuition a little bit, I think.
