 Solution

A real square orthogonal matrix satisfies $Q^{T}Q=Q{Q}^T=I$, so it preserves the Euclidean norm. The columns of $Z_h$ are orthonormal, hence $\Vert Z_{h}v{\Vert }_2=\Vert v{\Vert }_2$. The induced matrix norm therefore givesFor the left truncation, $\Vert Q_{\ell }^{T}A{\Vert }_2\le \Vert A{\Vert }_2$, 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 $\Vert A{\Vert }_2=1$ but $Q_{\ell }^{T}A=0$. Merely taking $\ell \ge n$ 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 $(i,k)$ blockThe 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 $(1,2),(2,3),\dots ,(n-1,n)$, each time zeroing the current entry $(k+1,k)$. Earlier columns remain triangular because both affected rows are already zero there. The result $R=J_{n-1}\cdots J_{1}A$ 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 $j=1,\dots ,n-1$. Within column $j$, for $i=n,n-1,\dots ,j+1$, rotate rows $(i-1,i)$ to eliminate the current $(i,j)$ entry. Previously cleared columns stay clear. With $J_1,\dots ,J_s$ in application order, $s=n(n-1)/2$, one has $\boxed{R=J_s\cdots J_{1}A,Q=J_1^T\cdots J_s^T}$. This is QR decomposition by Givens rotations.