 Solution

At step $k$, choose a real shift ${\mu }_k$, factor $A_k-{\mu }_{k}I=Q_k{R}_k$, and setThe 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 $e=(1,\dots ,1{)}^{\mathsf{T}}$. Then $A=I-2e{e}^{\mathsf{T}}/n$ and $e^{\mathsf{T}}e=n$, 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: 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.