Show that $\lVert Ax \rVert \geq \lVert x \rVert \lVert A^{-1} \rVert^{-1}$ where $\lVert \cdot \rVert$ is a vector induced norm

Question

I want to prove that $$\lVert Ax \rVert \geq \lVert x \rVert \lVert A^{-1} \rVert^{-1}$$ for all $x \in \mathbb{R}^n$, where $A \in \mathbb{R}^{n\times n}$ is invertible, and $\lVert \cdot \rVert$ is a vector induced matrix norm, defined as $$\lVert A \rVert = \sup_{x\neq0}{\lVert Ax \rVert \over \lVert x \rVert}$$ (that obeys $\lVert A B \rVert \leq \lVert A \rVert\lVert B \rVert$).

I know that $\lVert Ax \rVert \leq \lVert x \rVert \lVert A \rVert$ (which follows from the definition of $\lVert A \rVert$) and that $1 \leq \lVert A \rVert \lVert A^{-1} \rVert$. I have tried manipulating the latter and even trying to apply the definition of $\lVert A \rVert$ in proving this inequality, but to no success.

Any help?

Answer 1

Note the following : $$ \|x\|_{2}=\|A^{-1}Ax\|_{2} \implies \|x\|_{2}=\frac{\|A^{-1}Ax\|_{2}}{\|Ax\|_{2}}\|Ax\|_{2} $$ The latter expression implies that since $Ax\in\operatorname{Im}(A)$ then there shall be a vector $y\neq0$ such that $$ \|x\|_{2}\leq\max_{y\neq0}\frac{\|A^{-1}y\|_{2}}{\|y\|_{2}}\|Ax\|_{2}=\|A^{-1}\|_{2}\|Ax\|_{2} $$ Therefore you end up with : $$ \|A^{-1}\|^{-1}_{2}\|x\|_{2}\leq\|Ax\|_{2} $$ Note that I am assuming you are using the following definition of $\|A\|_{p}$ $$ \|A\|_{p}=\max_{x\neq 0}\frac{\|Ax\|_{p}}{\|x\|_{p}}=\max_{\|x\|_{p}=1}\|Ax\|_{p} $$ A matrix norm sub-ordinate or induced by a vector norm.

Answer 2

Note that for any matrix $A$ and vector $x$, $\|Ax\| \leq \|A\|\|x\|$ (this should be straight forward to prove).

Now using $A^{-1}$ instead of $A$ and $Ax$ instead of $x$ we have have $\|A^{-1}(Ax)\| \leq \ \|A^{-1}\|\|Ax\|$ or in other words $\|x\| \leq \|A^{-1}\| \|Ax\|$. Multiplying by $\|A^{-1}\|^{-1}$ yields your inequality.