Given a symmetric operator $A$ and a starting vector $v_0$, the Lanczos recurrence builds an orthonormal basis ${v_0, v_1, \ldots, v_n}$ of the Krylov subspace $\mathrm{span}{v_0, Av_0, A^2 v_0, \ldots, A^n v_0}$ such that the matrix elements

$$\langle v_i \mid A \mid v_j \rangle$$

vanish whenever $|i - j| > 1$. The defining property: in the Krylov basis, $A$ is tridiagonal.

The diagonal entries $\alpha_j = \langle v_j \mid A \mid v_j \rangle$ and off-diagonal entries $\beta_j = \langle v_{j+1} \mid A \mid v_j \rangle$ are the Lanczos coefficients.

This is not a coincidence of the construction; it is forced by Hermitian symmetry. The orthogonalization step in the recurrence only has to subtract against the previous TWO basis vectors because $A$‘s symmetry guarantees orthogonality to everything further back.