Linear Algebra and Its Applications · Problem 7.4.16

When $A$ is symmetric, the Lanczos iteration produces a tridiagonal structure because symmetry kills off the upper Hessenberg terms. Multiplying the recurrence by $q_j^T$ isolates $a_j$ by exploiting orthonormality: the cross terms vanish, leaving $a_j = q_j^T A q_j$. The matrix $T$ in $AQ = QT$ captures this tridiagonal pattern—it's the compressed representation of $A$ in the orthonormal basis, with the $b$'s as off-diagonals and $a$'s on the main diagonal.

When $A$ is symmetric, the Lanczos iteration produces a tridiagonal structure because symmetry kills off the upper Hessenberg terms. Multiplying the recurrence by $q_j^T$ isolates $a_j$ by exploiting orthonormality: the cross terms vanish, leaving $a_j = q_j^T A q_j$. The matrix $T$ in $AQ = QT$ captures this tridiagonal pattern—it's the compressed representation of $A$ in the orthonormal basis, with the $b$'s as off-diagonals and $a$'s on the main diagonal.