Prove $k(AB) \leq k(A)k(B)$ where $k(\cdot)$ denotes the condition number

Question

Given a $n \times n$ matrix $A$ and $B$, we need to prove $k(AB) \leq k(A)k(B)$ where $k(\cdot)$ denotes the condition number of a matrix.

Is there any thing wrong in the below proof?

$$k(AB) = \|AB\| \cdot \|(AB)^{-1}\| \leq \|A\| \cdot \|B\| \cdot \|B^{-1}\| \cdot \|A^{-1}\| =k(A)k(B) .$$

Answer 1

You proof is correct. Both $A, B$ should be invertible, so $\|(AB)^{-1}\|=\|B^{-1}A^{-1}\|\le \|B^{-1}\|\|A^{-1}\| $.