Proving that the norm of a Matrix is bigger or equal to it's smallest singular value multiplied by a vector.

Question

I need to prove the following:

Let $A \in \mathbf R^{n*n}$ be a real matrix and $x \in \mathbf R^n$ a vector.show that: $$\Vert Ax \Vert_{2} \geq s_{\min}\Vert x \Vert_{2}$$ where $s_{\min}$ is the smallest singular value of A.

I proceeded like so:

per def: $\Vert A \Vert_{2} = \max\left\{ \sqrt w;\text{w is the EV of $A^TA$} \right\}$

so we see that $$\Vert A \Vert_{2} \geq s_{\min}$$ multiply each side by $\Vert x\Vert_{2}$ $$\Vert A\Vert_{2} \Vert x \Vert_{2} \geq s_{ \min}\Vert x \Vert_{2}$$ can we now conclude from this that $$\Vert Ax \Vert_{2} \geq s_{\min}\Vert x \Vert_{2}$$??

since $$\Vert A\Vert_{2} \Vert x \Vert_{2} \geq \Vert Ax \Vert_{2}$$ this doesn't seem quite right to me! please help.

Thanks a lot in advance

Answer 1

Note that $\|Ux\| = \|x\|$ for an orthogonal $U$.

Hence $\|Ax\| = \|U \Sigma V^T x \| = \|\Sigma V^T x \| = \sqrt{ \sum_l \sigma_k^2 [V^T x]_k^2 } \ge \sigma_n \|V^T x\| = \sigma_n \|x\|$.

Answer 2

$$\sigma_\text{min}\|x\|_2=\frac{\|x\|_2}{\|A^{-1}\|_2}=\frac{\|A^{-1}Ax\|_2}{\|A^{-1}\|_2}\leq\frac{\|A^{-1}\|_2\|Ax\|_2}{\|A^{-1}\|_2}=\|Ax\|_2$$

When $A$ is singular its smallest singular value is zero and thus the inequality is trivial this case.