= Solution
Interpret the norms as <Euclidean norms> and set $B=A^{\mathsf T}A$. This is a <real symmetric matrix> with $\mathbf x^{\mathsf T}B\mathbf x=\|A\mathbf x\|^2\ge0$, hence a <positive semidefinite matrix>. Squaring the nonnegative objective does not change its <argmin>. Let $\lambda_{\min}$ be the smallest <eigenvalue> of $B$ and $E_{\min}$ its <eigenspace>.
First minimize on the <unit sphere>. The <Lagrange multiplier> function $L(\mathbf x,\lambda)=\mathbf x^{\mathsf T}B\mathbf x-\lambda(\mathbf x^{\mathsf T}\mathbf x-1)$ has stationarity condition $B\mathbf x=\lambda\mathbf x$. Multiplication by $\mathbf x^{\mathsf T}$ shows that the squared objective at such a point is $\lambda$. The <spectral theorem for real symmetric matrices> also gives, for every unit <vector>, $\mathbf x^{\mathsf T}B\mathbf x\ge\lambda_{\min}$, with equality exactly in $E_{\min}$. This proves global minimality, rather than merely finding stationary candidates.
If $\lambda_{\min}>0$, every point with $\|\mathbf x\|>1$ has a strictly smaller objective after rescaling to unit length. Therefore
$$
\boxed{\operatorname*{arg\,min}_{\|\mathbf x\|\ge1}\|A\mathbf x\|=\{\mathbf x\in E_{\min}:\|\mathbf x\|=1\},\quad\min=\sqrt{\lambda_{\min}}.}
$$
If $\lambda_{\min}=0$, the <kernel of a linear map> satisfies $\ker B=\ker A$, since $B\mathbf x=0$ implies $\|A\mathbf x\|^2=0$, and the converse is immediate. The <minimum> is then zero, attained precisely at
$$
\boxed{\{\mathbf x\in\ker A:\|\mathbf x\|\ge1\}.}
$$
In this case minimizers need not have unit length. This distinction is essential for a rank-deficient $A$, including any real $m\times n$ <matrix> with $m<n$. The result is the <minimum image norm outside a Euclidean unit ball>, governed by the least <singular value>.
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