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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?

Ondrej Draganov
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    When $A$ and $B$ are Hermitian, the statement that $|Ax|\le|Bx|$ for all $x$ is equivalent to $A^2\preceq B^2$. This is known to be false even when $A$ and $B$ are positive semidefinite. It should be easy to generate a random counterexample by computer. – user1551 Dec 16 '21 at 21:48
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    @OnDragi Did you mean that $A,B$ are symmetric and positive semidefinite? If not, then the related properties are not true. – Ben Grossmann Dec 16 '21 at 23:00

2 Answers2

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

Trevor Gunn
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  • This is a fair answer to the original question -- I edited the question and added more complete answer so it is more useful for anyone finding my question in the future. – Ondrej Draganov Dec 17 '21 at 13:41
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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): $$ A = \begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix} \hspace{1.5cm} B=\begin{pmatrix} 2 & 1 \\ 1 & 7 \end{pmatrix} $$ $$ B^2-A^2 = \begin{pmatrix} 3 & 12 \\ 12 & 45 \end{pmatrix} \hspace{1.5cm} \begin{pmatrix} -4 & 1 \end{pmatrix}\cdot (B^2-A^2) \cdot \begin{pmatrix} -4 \\ 1 \end{pmatrix} = -3$$

Ondrej Draganov
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