= Solution
Put $C=Q_1A$ and write its second column as $v$. For $n\geq3$, take
$$
\boxed{Q_2=J(2,n)J(2,n-1)\cdots J(2,3),}
$$
where each angle is computed from the current second column, in the right-to-left application order. These <Givens rotations> eliminate its entries in rows $3,\ldots,n$ by accumulating their <Euclidean norm> in row two.
Each rotation acts only on rows with indices at least two. The first column of $C$ is $\|a\|_2e_1$, whose entries in all those rows are zero. Mixing two zeros leaves them zero, and row one is untouched. Thus the first column is unchanged. The resulting <matrix> has the form
$$
\boxed{Q_2Q_1A=
\begin{pmatrix}
\|a\|_2&v_1\\
0&\sqrt{v_2^2+\cdots+v_n^2}\\
0&0\\
\vdots&\vdots\\
0&0
\end{pmatrix}\quad(n\geq3).}
$$
This verifies both required properties, including the case of zero second-column tail. When $n=2$ there are no rows to eliminate: take $Q_2=I_2$, the empty product. No sign change of $v_2$ is required in that case. Each <Givens row-pair update> preserves previously eliminated columns, which is the central invariant of <QR decomposition by Givens rotations>.
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