If an invertible matrix $A$ satisfies $c||x|\leq ||Ax|| \leq C||x||$ for all $x$, can one say the same about its inverse?

Question

For every $z$ with norm 1, we have that

$$ 1 = ||z|| = ||AA^{-1}z|| \leq ||A||\cdot ||A^{-1}z|| \leq C||A^{-1}z||$$ and thus, this give us that $||A^{-1}z|| \geq \frac{1}{C}$.

I thought that using the same argument with $A^{-1}A$ would yield the other inequality, but it doens't work... Can one conclude that $||A^{-1}z|| \leq \frac{1}{c}$ somehow? Or if not so, maybe that $||A^{-1}z||$ is lesser than other constant involving $c$ and $C$?

Answer 1

If $c||x||\leq ||Ax|| \leq C||x||$ and $A$ is invertible substituting $x=A^{-1}y$ to the equation we have:

$$c||A^{-1}y||\leq ||y|| \leq C||A^{-1}y||$$ Thus $||y|| \leq C||A^{-1}y||$ and so $\frac{1}{C}||y|| \leq ||A^{-1}y||$ and $c||A^{-1}y||\leq ||y||$ and so $||A^{-1}y||\leq \frac{1}{c}||y||$. Hence: $$\frac{1}{C}||y||\leq ||A^{-1}y|| \leq \frac{1}{c}||y||$$ as we wanted to prove.