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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.

With $A_0=A$, repeat the shifted QR algorithm:The QR shift ${\mu }_k$ is chosen from $A_k$, for example its bottom-right entry or an eigenvalue of a trailing two-by-two block. Explicitly using the orthogonality of $Q_k$ givesThus each update is a matrix similarity preserving eigenvalues. Accumulate $Q_0{Q}_1\cdots Q_k$ if eigenvectors or the full reduction are wanted. Once a subdiagonal entry is negligible, block deflation in the QR algorithm separates the corresponding blocks and iteration continues on the active block.

The initial bottom-right entry is zero, so the chosen QR shift is ${\mu }_0=0$. One valid QR factorization is $Q_0=A$, $R_0=I$, using explicitly that $A^{T}A=I$. The update isIt repeats forever and never approaches a diagonal matrix. With the standard positive-diagonal convention for $R$, this factorization is unique, so the Rayleigh-shift iteration is stationary and fails to diagonalize $A$.

The failure is independent of harmless sign conventions. The columns of $A$ are already orthonormal, so another QR factorization can only replace $Q_0$ by $AD$ and $R_0$ by $D$, where $D$ is diagonal with entries $\pm 1$. Then $A_1=DAD$ still has zero diagonal and off-diagonal magnitudes one. Subsequent bottom-right shifts remain zero and the off-diagonal magnitude never decreases. The eigenvalues are $+1,-1$, with equal absolute values, explaining the lack of separation in the unshifted QR algorithm.

Let the real QR shift be $\mu$ and put $r=\sqrt{1+{\mu }^2}$. A valid QR factorization, including the singular cases $\mu =\pm 1$, isHere $Q^{T}Q=I$, and multiplying verifies $QR=A-\mu I$. Using this orthogonality in the shifted update givesIt is diagonal exactly when ${\mu }^2-1=0$. Therefore all real shifts giving the eigenvalues in one iteration areThe resulting diagonal is $(-1,1)$ or $(1,-1)$, respectively. Both QR shifts are exact eigenvalues. Sign changes in the QR factorization conjugate $A_1$ by a diagonal sign matrix and do not alter its zero pattern. This is Exact-shift diagonalization of a two-state swap matrix.

A QR shift is a scalar ${\mu }_k$ subtracted before the QR factorization in a QR algorithm step and restored after reversing the factors:The final identity uses the orthogonality of $Q_k$, proving that the eigenvalues are preserved. Choosing a shift near an eigenvalue can accelerate block deflation in the QR algorithm. A common choice is the current bottom-right entry, but it can stall, as in Exact-shift diagonalization of a two-state swap matrix.