= Solution
Because $r<n$, choose a <unit vector> $v$ in the <null space> of $X^T$, so that $v^TX=0$. 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 $\alpha>1$. Set
$$
\boxed{A=J\bigl(I_n+(\alpha-1)vv^T\bigr)}.
$$
Its <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 $\operatorname{range}(JX)$, which is <orthogonal> to $Jv$. Hence $PJv=0$ for $P=QQ^T$, and
$$
\|(I_m-P)A\|_2\geq\|(I_m-P)Av\|_2=\|\alpha Jv\|_2=\alpha.
$$
Conversely, an <orthogonal projection> is a contraction in the <Euclidean norm>, so $\|(I_m-P)A\|_2\leq\|A\|_2=\alpha$. Therefore
$$
\boxed{\|A-\widehat A\|_2=\sigma_1(A)=\alpha>\sigma_2(A)=1}.
$$
If $X$ is rank deficient, complete the <QR decomposition> within $J(v^\perp)$: this space has dimension $n-1\geq r$ 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 $\operatorname{range}(AX)$, the conclusion holds without any completion. A fixed sampling <matrix> can thus miss the unique largest <singular value> completely.
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