Minimum image norm outside a Euclidean unit ball (${\min }_{\Vert x\Vert \ge 1}\Vert Ax\Vert =\sqrt{{\lambda }_{\min }(A^{\mathsf{T}}A)}$)

For a real matrix $A$, let $B=A^{\mathsf{T}}A$. It is positive semidefinite, and the spectral theorem for real symmetric matrices gives $x^{\mathsf{T}}Bx\ge {\lambda }_{\min }\Vert x{\Vert }^2$. If ${\lambda }_{\min }>0$, rescaling any longer vector to unit length decreases its image norm, so minimizers lie on the unit sphere. Lagrange multipliers applied to $x^{\mathsf{T}}Bx$ with $\Vert x\Vert =1$ give $Bx=\lambda x$. Thus the minimizers are exactly the unit vectors in the least eigenspace.

If ${\lambda }_{\min }=0$, the minimum is zero and the full argmin consists of all $x\in \ker A$ with $\Vert x\Vert \ge 1$. Indeed, $\ker (A^{\mathsf{T}}A)=\ker A$ because $x^{\mathsf{T}}{A}^{\mathsf{T}}Ax=\Vert Ax{\Vert }^2$. Restricting the minimizers to the sphere would omit longer kernel vectors in this case. For a ball constraint $\Vert x\Vert \le 1$, the argmin is always $\ker A$ intersected with the ball.