Does $\|A\|_2\leq \|B\|_2$ imply $\|CA\|_2\leq \|CB\|_2$ or $\|AC\|_2\leq \|BC\|_2$?

Assume matrices $A$, $B$, and $C$ are of same dimensions, does $\|A\|_2\leq \|B\|_2$ imply $\|CA\|_2\leq \|CB\|_2$ or $\|AC\|_2\leq \|BC\|_2$?

$\|A\|_2$ denotes by $\lambda_{max}\sqrt{A^TA}$, and $\lambda_{max}$ is the largest eigenvalue.

No. Consider $A = \begin{bmatrix}1&0\\0&0\end{bmatrix}$, $B = \begin{bmatrix}0&0\\0&2\end{bmatrix}$, $C = \begin{bmatrix}1&0\\0&0\end{bmatrix}$. Clearly, $\|A\|_2 = 1 \le 2 - \|B\|_2$.
However, since $AC = CA = \begin{bmatrix}1&0\\0&0\end{bmatrix}$ and $BC = CB = \begin{bmatrix}0&0\\0&0\end{bmatrix}$, we have $\|AC\|_2 = 1 > 0 = \|BC\|_2$ and $\|CA\|_2 = 1 > 0 = \|CB\|_2$.
Therefore, $\|A\|_2 \le \|B\|_2$ does not imply either $\|CA\|_2 \le \|CB\|_2$ or $\|AC\|_2 \le \|BC\|_2$.
• I'm not sure if there are any reasonable set of constraints on $A,B,C$ which gives us $\|A\|_2 \le \|B\|_2 \implies \|CA\|_2 \le \|CB\|_2$. Dec 24 '15 at 9:39
Very general counterexample: let be $P$, $Q$ the projections on two orthogonal (nontrivial) subspaces. Wlog we can suppose $0<\|P\|\le\|Q\|$. Now, $$\|PP\|=\|P\|\not\le 0 =\|PQ\|,\qquad \|PP\|=\|P\|\not\le 0 =\|QP\|.$$