Quadratic matrix bounds

Question

Let A be a singular matrix with a simple (non-repeated) zero-eigenvalue. Dose the following inequality hold?

$$\|Ax\|^2\geq\sigma_2\|x\|^2, \qquad \forall x\notin Null(A)$$

where $\sigma_2$ is the smallest nonzero singular value of the matrix $A$. If it is true, where can I find a proof??

Answer 1

No, it's not true.

Consider the matrix $2\times2$ matrix $A = \begin{pmatrix} \sigma_2 & 0\\ 0 & 0 \end{pmatrix}$, and any vector $x = \begin{pmatrix} a \\ b \end{pmatrix}$. Then for any $b \neq 0$, we have $\left\lVert Ax\right\rVert^2 <\sigma_2\left\lVert x\right\rVert^2$

Answer 2

No. Let $A=diag(1,1,0)$. Then $A^T=A$, hence $A^TA=A^2=A$, thus $ \sigma_2=1.$

Let$x=(1,0,1)^T.$ Then $Ax=(1,0,0)^T$ This gives

$$||Ax||_2^2=1,$$

but

$$ \sigma_2 ||x||_2^2=\sqrt{2}.$$

Answer 3

It does not hold. Note in particular that for any $z \notin \operatorname{Null}(A)$, $y \in \operatorname{Null}(A)$, and $t \in \Bbb R$, taking $x = y + tz$ gives us $$ \|Ax\|^2 = \|A(y + tz)\|^2 = t^2\|Az\|^2,\\ \|x\|^2 = \|y + tz\|^2 \geq (\|y\| - \|tz\|)^2 = (\|y\| - t\|z\|)^2. $$ Thus, we have $$ \frac{\|Ax\|^2}{\|x\|^2} \leq t^2 \cdot \frac{\|Az\|^2}{(\|y\| - t\|z\|)^2}. $$ By taking a limit as $t \to 0$, we see that $\frac{\|Ax\|^2}{\|x\|^2}$ can be made arbitrarily small, and it certainly does not hold that $\frac{\|Ax\|^2}{\|x\|^2} \geq \sigma_2$.