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The set of minimizers of an objective:It is a set of inputs, whereas the minimum is an objective value. The argmin can be empty if an infimum is not attained, a singleton, or a larger set.

If $\varphi$ is strictly increasing on the range of $f$, then $f(x)\le f(y)$ if and only if $\varphi (f(x))\le \varphi (f(y))$. Quantifying over every $y$ proves that $f$ and $\varphi \circ f$ have the same argmin, whether or not a minimum is attained. Squaring therefore preserves the argmin of a nonnegative objective, while squaring a signed objective need not preserve its argmin.

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.

The map $u\mapsto u^2$ is strictly increasing on $[0,\infty )$: for $u,v\ge 0$, $u\le v$ if and only if $u^2\le v^2$. Therefore a point has the smallest $f$-value if and only if it has the smallest $f^2$-value, and the same applies to $g$. By minimizers under strictly increasing transformations, both asserted argmin equalities hold, including when the argmin sets are empty. The differentiability assumptions are unnecessary for this order argument.

The origin is feasible and has objective zero, while a Euclidean norm is nonnegative. Thus the minimum is $0$, and the complete argmin isIf $A$ is injective, this set contains only the origin. If it has a nontrivial kernel, every feasible kernel vector is also a minimizer. No Lagrange multipliers are needed.

Interpret the norms as Euclidean norms and set $B=A^{\mathsf{T}}A$. This is a real symmetric matrix with ${\mathbf{x}}^{\mathsf{T}}B\mathbf{x}=\Vert A\mathbf{x}{\Vert }^2\ge 0$, 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 $\Vert \mathbf{x}\Vert >1$ has a strictly smaller objective after rescaling to unit length. ThereforeIf ${\lambda }_{\min }=0$, the kernel of a linear map satisfies $\ker B=\ker A$, since $B\mathbf{x}=0$ implies $\Vert A\mathbf{x}{\Vert }^2=0$, and the converse is immediate. The minimum is then zero, attained precisely atIn 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.