Does it hold that $\kappa(A^2) = \kappa(A)^2$?

Question

Let $A \in \mathbb{R}^{n \times n}$ be invertible and $b\in \mathbb{R}^n$. Let $x \in \mathbb{R}^n$ be the solution of $Ax=b$.

Let $\kappa(A)$ denote the condition number of matrix $A$. Does the following hold?$$\kappa(A^2) = \kappa(A)^2$$

I think that statement is correct. This is the way I have tried to prove it:

$$\kappa(A^2) = \|A^2\| \cdot \|({A^2})^{-1}\| = \|A\| \cdot \|A\| \cdot \|A^{-1}\| \cdot \|A^{-1}\| =\\ \quad\quad\quad= \|A\| \cdot \|A^{-1}\| \cdot \|A\| \cdot \|A^{-1}\| = \kappa(A) \cdot \kappa(A) = \kappa(A)^2$$

Question: Is that correct? Or do I have made a mistake?

Answer 1

Since norms are submultiplicative only, you get the inequality $$ \kappa(A^2) \le \kappa(A)^2. $$ If $A=A^{-1}$ then $A^2=I$, but $\kappa(A)$ can be larger than one.