 Solution

Because $r<n$, choose a unit vector $v$ in the null space of $X^T$, so that $v^{T}X=0$. Let $J\in {\mathbb{R}}^{m\times n}$ have orthonormal columns, for example the first $n$ columns of $I_m$, and choose a real number $\alpha >1$. SetIts singular value along $v$ is $\alpha$, and its other singular values are $1$. Thus ${\sigma }_1(A)=\alpha >1={\sigma }_2(A)$. Meanwhile $AX=JX$. If $X$ has full column rank, the column space of the thin QR decomposition is exactly $\mathrm{range}(JX)$, which is orthogonal to $Jv$. Hence $PJv=0$ for $P=Q{Q}^T$, andConversely, an orthogonal projection is a contraction in the Euclidean norm, so $\Vert (I_m-P)A{\Vert }_2\le \Vert A{\Vert }_2=\alpha$. ThereforeIf $X$ is rank deficient, complete the QR decomposition within $J(v^{\perp })$: this space has dimension $n-1\ge r$ and contains all columns of $JX$, so the required orthonormal completion exists there. The same proof applies. For rank-deficient data the choice of unused columns of $Q$ matters; the construction specifies a valid choice. If $Q$ instead contains only a basis of $\mathrm{range}(AX)$, the conclusion holds without any completion. A fixed sampling matrix can thus miss the unique largest singular value completely.