= Minimum image norm outside a Euclidean unit ball
{title2=$\min_{\|x\|\ge1}\|Ax\|=\sqrt{\lambda_{\min}(A^{\mathsf T}A)}$}
For a real <matrix> $A$, let $B=A^{\mathsf T}A$. It is <positive semidefinite>, and the <spectral theorem for real symmetric matrices> gives $x^{\mathsf T}Bx\ge\lambda_{\min}\|x\|^2$. 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 $\|x\|=1$ 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 $\|x\|\ge1$. Indeed, $\ker(A^{\mathsf T}A)=\ker A$ because $x^{\mathsf T}A^{\mathsf T}Ax=\|Ax\|^2$. Restricting the minimizers to the sphere would omit longer <kernel> vectors in this case. For a ball constraint $\|x\|\le1$, the <argmin> is always $\ker A$ intersected with the ball.
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