= Solution
At step $k$, choose a real shift $\mu_k$, factor $A_k-\mu_kI=Q_kR_k$, and set
$$
A_{k+1}=R_kQ_k+\mu_kI=Q_k^\mathsf TA_kQ_k.
$$
The last equality proves <orthogonal> similarity at every step. Thus all iterates have the same <eigenvalues>; if $A_k$ is symmetric, the last expression is symmetric, proving symmetry by <induction>.
For the specified <matrix>, let $e=(1,\ldots,1)^\mathsf T$. Then $A=I-2ee^\mathsf T/n$ and $e^\mathsf Te=n$, so $A^\mathsf T=A$ and $A^2=I$. It is a <Householder reflection>, hence <orthogonal>. With the standard positive-diagonal QR convention, part (a) gives $Q=A$, $R=I$ at the first unshifted step, and therefore $A_1=RQ=A$. The same repeats forever: \b[the unshifted <algorithm> is stationary].
If arbitrary diagonal signs in QR are allowed, $R=D$, $Q=AD$ for a sign <matrix> $D$, and the next iterate is $DAD$. Its off-diagonal magnitudes are unchanged, so even then only sign changes occur, not progress toward a diagonal form. This explains the role of the usual convention in the literal stationary-iterate statement.
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