Exact-shift diagonalization of a two-state swap matrix ($\mu =\pm 1$)

For $A=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, one shifted QR algorithm step with real QR shift $\mu$ gives, with a consistent choice of factor signs,To verify this, factor $A-\mu I=QR$ with $Q=(1+{\mu }^2{)}^{-1/2}\left(\begin{array}{cc}-\mu & -1\\ 1& -\mu \end{array}\right)$. Its columns are orthonormal, and $R=Q^T(A-\mu I)$ is upper triangular. Then $A_1=Q^{T}AQ$. Thus $A_1$ is diagonal exactly when $\mu =\pm 1$, the two eigenvalues. The bottom-right-entry shift is zero and leaves the matrix unchanged under the positive-diagonal factor convention. This illustrates why a poorly chosen QR shift can stall even a two-by-two symmetric eigenvalue computation.