Considering pure QR algorithm (without shifts and preliminary tridiagonal reduction) are there sufficient conditions for algorithm to converge to quasi-diagonal form?


For the the following matrix 
$$
A = \left(\begin{array}{cc}
0 & 1 \\
1 & 0 
\end{array}\right).
$$
with eigenvalues $\lambda\_1 = 1$ and $\lambda\_2 = -1$ it apparently does not.
QR decomposition produces the following
$$
A = QR = \left(\begin{array}{cc}
0 & 1\\
1 & 0\end{array}\right)
\left(\begin{array}{cc}
1 & 0\\
0 & 1 
\end{array}\right),
$$
which generates stationary sequence $\{A^{(k)}\}$.



QR alogrithm for eigenvalue problem
This example is explained in section 11.5 [here](http://www.math.iit.edu/~fass/477577_Chapter_11.pdf).


