Prove that $1/||S^{-1}||\leq ||S-T||$

Question

Let $||\cdot||$ denote the operator norm, with $T\in L(\mathbb{R}^n)$ and $S\in GL(n)$, and suppose that $T(\mathbf{x})=\mathbf{0}$. On one hand, we have that $$||S(\mathbf{x})||=||{S(\mathbf{x})-T(\mathbf{x})}||=||{(S-T)(\mathbf{x})}||\leq ||{S-T}||||{\mathbf{x}}||.$$ So, I need something like $$||\mathbf{x}||/||S^{-1}||\leq ||S(\mathbf{x})||$$ to obtain the result I want. I know that $1/||S^{-1}||\leq||S||$, but I don't see exactly how to use it.

EDIT: $\mathbf{x}\neq\mathbf{0}.$

Answer 1

For all $x \in \mathbb R^n$ we have

$$||x||=||S^{-1}(S(x))|| \le ||S^{-1}||*||S(x)||.$$

This gives

$$\frac{||x||}{||S^{-1}||} \le ||S(x)||.$$

Answer 2

Since $x \neq 0$ we can estimate $$ \Vert x \Vert = \Vert S^{-1} Sx - S^{-1} Tx \Vert = \Vert S^{-1}(S-T)x \Vert \le \Vert S^{-1} \Vert \Vert S-T \Vert \Vert x \Vert. $$ Divide both sides by $\Vert S^{-1} \Vert \Vert x \Vert$ and we get $$ \frac{1}{\Vert S^{-1} \Vert} \le \Vert S-T \Vert. $$