Loewner order and norms of images: Does $A \preccurlyeq B$ imply $\|Ax\| \leq \|Bx\|$?

Does $A \preccurlyeq B$ imply that $\|Ax\| \leq \|Bx\|$ for all $x$?

I assume that $A,B$ are symmetric matrices and $A \preccurlyeq B$ denotes that $B-A$ is positive semi-definite. I can see that $A \preccurlyeq B$ implies several related properties, like

• $\|A\|\leq \|B\|$, if $A,B$ are positive semidefinite,
• $\|A^{\frac 1 2} x\| \leq \|B^{\frac 1 2} x\|$.

But does it also imply $\|Ax\| \leq \|Bx\|$ for all $x$? What if we assume both $A,B$ to be positive semidefinite?

$$ A = \begin{pmatrix} -1 & 0 \\ 0 & 0 \end{pmatrix}, B = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}, x = \begin{pmatrix} 1 \\ 0 \end{pmatrix} $$

No, this is not true. A very simple counter-example is $-I \preccurlyeq 0$, with $I$ being the identity matrix, as $\|-Ix\|=\|x\|>0=\|0x\|$ for $x\neq 0$.

It is not true even if we assume both $A$ and $B$ to be positive semi-definite. We have $\|Ax\|\leq\|Bx\|$ for all $x$ if and only if $A^2\preccurlyeq B^2$:

$$ \|Ax\|\leq\|Bx\| \iff x^TA^TAx\leq x^TB^TBx \iff 0\leq x^T(B^2-A^2)x, $$ and there exist $A,B$ positive semidefinite such that $A\preccurlyeq B$, but $A^2\not\preccurlyeq B^2$. An example from another question (swapped $A$ and $B$ to be consistent with the notation here): $$ A = \begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix} \hspace{1.5cm} B=\begin{pmatrix} 2 & 1 \\ 1 & 7 \end{pmatrix} $$ $$ B^2-A^2 = \begin{pmatrix} 3 & 12 \\ 12 & 45 \end{pmatrix} \hspace{1.5cm} \begin{pmatrix} -4 & 1 \end{pmatrix}\cdot (B^2-A^2) \cdot \begin{pmatrix} -4 \\ 1 \end{pmatrix} = -3$$