Setup. Consider the tridiagonal matrix $T_k$ with bottom-right $2 \times 2$ block $\begin{pmatrix} \alpha_{n-1} & \beta \\ \beta & \alpha_n \end{pmatrix}$. The Wilkinson shift $\sigma$ is the eigenvalue of this block closer to $\alpha_n$.

Key estimate. Write $\delta = (\alpha_{n-1} - \alpha_n)/2$ and $\mu = \alpha_n - \sigma = -\delta + \mathrm{sign}(\delta)\sqrt{\delta^2 + \beta^2}$. Then $|\mu| \leq \beta^2 / (2|\delta|)$ when $|\delta| \gg |\beta|$, and $|\mu| \leq |\beta|$ always.

One QR step. After the shifted QR step $T_k - \sigma I = QR$, $T_{k+1} = RQ + \sigma I$, the new subdiagonal entry satisfies $|\beta'| = |\beta| \prod_{j=1}^{n-2} \frac{|\lambda_j - \sigma|}{|d_j|}$ where $d_j$ are related to the QR factorization. The Wilkinson shift guarantees that $(T_k - \sigma I)$ has its smallest singular value at the bottom, causing $|\beta'|/|\beta|$ to be small.

Cubic convergence. For symmetric tridiagonal matrices, Wilkinson proved that $|\beta_{k+1}| = O(|\beta_k|^3)$ using the identity relating the shift quality to the subdiagonal entry. The key insight is that $|\alpha_n - \sigma_k| = O(\beta_k^2)$ (the shift approximates the eigenvalue quadratically well), and this quadratic approximation combined with the inverse-iteration-like nature of shifted QR yields cubic convergence of $\beta_k \to 0$.

Global convergence. The monotonicity $\sum_i \beta_i^2$ is non-increasing under QR steps (related to the Wilkinson monotonicity theorem) ensures global convergence. Once $|\beta_k|$ is small enough for the cubic estimate to dominate, convergence is rapid. $\square$