We have the matrix \begin{equation*}A=\begin{pmatrix}1/2 & 1/5 & 1/10 & 1/17 \\ 1/5 & 1/2 & 1/5 & 1/10 \\ 1/10 & 1/5 & 1/2 & 1/5 \\ 1/17 & 1/10 & 1/5 & 1/10\end{pmatrix}\end{equation*}
I want to find an approximation of the eigenvalues using 1 step of the QR method.
I have done the following:
First we find the QR decomposition of $A$, so we get the matrices \begin{equation*}Q=\begin{pmatrix}0.907651 & -0.415323 & -0.055244 & -0.0249738 \\ 0.363061 & 0.846984 & -0.38676 & -0.0349633 \\ 0.18153 & 0.302883 & 0.865764 & -0.354628 \\ 0.106783 & 0.135599 & 0.312758 & 0.93402\end{pmatrix}\end{equation*} and \begin{equation*}R=\begin{pmatrix}0.550872 & 0.410045 & 0.275499 & 0.136682 \\ 0 & 0.414564 & 0.306426 & 0.134404 \\ 0 & 0 & 0.412557 & 0.162503 \\ 0 & 0 & 0 & 0.017511\end{pmatrix}\end{equation*}
Then we get \begin{equation*}A_1=RQ=\begin{pmatrix}0.709622 & 0.212327 & 0.0966156 & 0.0014864 \\ 0.219891 & 0.462078 & 0.1464 & 0.0016014 \\ 0.091623 & 0.147077 & 0.408059 & 0.00481704 \\ 0.001869 & 0.00238 & 0.005474 & 0.016275\end{pmatrix}\end{equation*}
Therefore an approximation of the eigenvalues are the elements of the diagonal because the result of the QR method is approximately a diagonal matrix or not?
If the method converges, the matrices $A_k$ converge to a triangular matrix with the same eigenvalues as $A$. So the answer is yes, if you just perform one step of the $QR$ method, your estimate for the eigenvalues would be the diagonal elements of $A_1$.