= Solution
A real square <orthogonal matrix> satisfies $Q^TQ=QQ^T=I$, so it preserves the <Euclidean norm>. The columns of $Z_h$ are orthonormal, hence $\|Z_hv\|_2=\|v\|_2$. The <induced matrix norm> therefore gives
$$
\|AZ_h\|_2=\sup_{\|v\|_2=1}\|AZ_hv\|_2\leq\boxed{\|A\|_2}.
$$
For the left truncation, $\|Q_\ell^TA\|_2\leq\|A\|_2$, and equality is guaranteed for arbitrary $A$ only when $\boxed{\ell=m}$. Indeed, if $\ell<m$, take $A=q_{\ell+1}e_1^T$, where $q_{\ell+1}$ is an omitted column of $Q$. Then $\|A\|_2=1$ but $Q_\ell^TA=0$. Merely taking $\ell\geq n$ does not guarantee preservation for an arbitrary $A$.
A full <QR decomposition> of a real $m\times n$ matrix is $A=QR$, with $Q$ orthogonal and $R$ upper trapezoidal; for a square matrix, $R$ is upper triangular. Fix the following sign convention for a <Givens rotation>: it is the identity except for its $(i,k)$ block
$$
\begin{pmatrix}c&s\\-s&c\end{pmatrix},\qquad c=\cos\theta,\ s=\sin\theta.
$$
The block transpose times the block is $I_2$ because $c^2+s^2=1$, proving orthogonality. Writing $r=\sqrt{x_i^2+x_k^2}$, the choice $\boxed{c=x_i/r,\ s=x_k/r}$ gives $(Jx)_k=0$ and $(Jx)_i=r$. To zero the other coordinate take $\boxed{c=x_k/r,\ s=-x_i/r}$, giving $(Jx)_i=0$. If $r=0$, both entries are already zero and the identity suffices.
For a <tridiagonal matrix>, successively rotate adjacent row pairs $(1,2),(2,3),\ldots,(n-1,n)$, each time zeroing the current entry $(k+1,k)$. Earlier columns remain triangular because both affected rows are already zero there. The result $R=J_{n-1}\cdots J_1A$ is upper triangular, so $Q=J_1^T\cdots J_{n-1}^T$ gives its <QR decomposition>. Only $n-1$ rotations are required; the resulting $R$ has at most two superdiagonals.
For a full square matrix, process columns $j=1,\ldots,n-1$. Within column $j$, for $i=n,n-1,\ldots,j+1$, rotate rows $(i-1,i)$ to eliminate the current $(i,j)$ entry. Previously cleared columns stay clear. With $J_1,\ldots,J_s$ in application order, $s=n(n-1)/2$, one has $\boxed{R=J_s\cdots J_1A,\quad Q=J_1^T\cdots J_s^T}$. This is <QR decomposition by Givens rotations>.
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