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Of course no. Remember that the operator norm of $A$ wrto the Eucliedan norms is the attained at an eigenvector of $S:=A^TA. $ Try a suitable simple binary $2\times 2$ matrix and compare the values of $\|Ax\|$ on the eigenvectors of $S$ and in the three nonzero binary vectors $(01), (10), (11).$

However, if instead you take in the domain $\mathbb{R}^n$ either the $l^1$ norm $\| \cdot \|\cdot \|1 $ |_1$ or the $l^\infty$ norm $\| \cdot\|\infty $ \|\cdot\|_\infty$ then, whatever norm you have in the target space $\mathbb{R}^m$, the operator norm of $A$ is attained in an extremal point of the unit ball of the domain, which is in both cases a binary vector.
show/hide this revision's text 1

Of course no. Remember that the operator norm of $A$ wrto the Eucliedan norms is the attained at an eigenvector of $S:=A^TA. $ Try a suitable simple binary $2\times 2$ matrix and compare the values of $\|Ax\|$ on the eigenvectors of $S$ and in the three nonzero binary vectors $(01), (10), (11).$

However, if instead you take in the domain $\mathbb{R}^n$ either the $l^1$ norm $\| \cdot \|1 $ or the $l^\infty$ norm $\| \cdot\|\infty $ then, whatever norm you have in the target space $\mathbb{R}^m$, the operator norm of $A$ is attained in an extremal point of the unit ball of the domain, which is in both cases a binary vector.