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

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    What other information would you like? That's all that I can think of except that $\mathbf{x}\neq\mathbf{0}$. – 2017-01-19

2 Answers 2

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

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

This gives

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

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    Exactly what I was looking for. Thank you, Fred. – 2017-01-19
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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 $$ \frac{1}{\Vert S^{-1} \Vert} \le \Vert S-T \Vert. $$

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    Nice! That is simpler. Thanks! – 2017-01-19
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    Your result is not true, since $\Vert x \Vert = \Vert S^{-1} Sx - S^{-1} Tx \Vert = \Vert S^{-1}(S-T)x \Vert \le \Vert S^{-1} \Vert \Vert S-T \Vert \Vert x \Vert$ only holds if $Tx=0$, and not for all $x \ne 0$ – 2017-01-19
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    @Fred Yes it is. The first liine in OP's post says "and suppose that $T(x)=0$". My argument is not for general $x$, but rather the specific $x$ mentioned in the hypotheses. – 2017-01-19
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    @Glitch: yes you are right. I apologize – 2017-01-19