From the singular value decomposition of an $m \times n$ matrix $A$, we should have $\Vert A \Vert = \sigma_1$, where $\sigma_1$ is the top singular value of $A$.
This would mean that the maximum argument (unique up to sign?) of $\Vert A \Vert = \arg\max_{\Vert x \Vert = 1} \Vert A x \Vert$ is a linear combination of the rows of $A$.
However, what is the intuition as to why $x$ must be a combination of the rows, and not anything more complex? Is this even true?
Suppose that $x$ is a unit vector that is not a linear combination of the rows of $A$. We can then write $x = u + v$, where $u$ is a linear combination of the rows of $A$ and $v$ is orthogonal to the rows of $A$ (which is to say that $Av = 0$). Note that $$ \|Ax\| = \|Au + Av\| = \|Au\|. $$ In fact, if we just got rid of the $v$ component and scaled $u$ to be a unit vector (note that $\|u\|^2 = 1 - \|v\|^2 < 1$), we'd end up with $$ \left\|A \frac{u}{\|u\|}\right\| = \frac{\|Au\|}{\|u\|} > \|Au\| = \|Ax\|. $$