Condition number for the relative residual of the matrix square root

Question

Let $X$ be a square root of $A \in \mathbb{C}^{n \times n}$, i.e., such that $X^2 = A$, and let $$ \alpha(X) = \frac{\|X\|^2}{\|A\|} = \frac{\|X\|^2}{\|X^2\|} \ge 1. $$ On page 135 of Functions of Matrices: Theory and Computation (Higham, 2008) it is claimed that it is easy to show that $$ \frac{\kappa(X)}{\kappa(A)} \le \alpha(X) \le \kappa(X), $$ where $\kappa(A) = \|A\| \|A^{-1}\|$ is the condition number of $A$. I'm trying to prove this fact but am having troubles.

I think there must be some useful property I cannot see that links the norms of $A$ and $X$ and their inverses. Does anybody have a hint to give me to get me started?

Answer 1

You have that $X=AX^{-1}$ and so $|| X|| \le || A ||||X^{-1}||$ that is equivalent to second inequality. The first inequality has the same proof by using $X^{-1}=(AX^{-1})^{-1}$

Answer 2

Hint: If $A=X^2$ then $AX^{-1}=X$ and $A^{-1}X=X^{-1}$, so $$\|X\|\le\|A\|\|X^{-1}\|,\qquad\|X^{-1}\|\le\|A^{-1}\|\|X\|$$ $$\therefore\|A\|\ge\|X\|/\|X^{-1}\|,\qquad\|A^{-1}\|\ge\|X^{-1}\|/\|X\|$$