= Solution
Multiplying the <Arnoldi decomposition> by $Q_k^T$ shows that the leading square block is
$$
H_k=Q_k^TAQ_k.
$$
It is an <upper Hessenberg matrix> by the <Arnoldi iteration>. Because $A=A^T$, it is also a <symmetric matrix>. Entries below the first subdiagonal vanish by the Hessenberg structure, and their reflected entries above the first superdiagonal vanish by symmetry. Consequently
$$
\boxed{T_k=Q_k^TAQ_k\text{ is symmetric tridiagonal}},\qquad
\boxed{AQ_k=Q_{k+1}\widetilde T_k}.
$$
The extra last row has only its final entry possibly nonzero. With the usual choice of positive normalization factors, the individual column relations are the three-term <Lanczos algorithm> recurrence
$$
Aq_j=\beta_{j-1}q_{j-1}+\alpha_jq_j+\beta_jq_{j+1},
\qquad \alpha_j=q_j^TAq_j,\quad \beta_j>0,
$$
where the first term is omitted for $j=1$. This reduction is an exact-arithmetic statement; <round-off error> can affect the <orthogonality> in an implementation.
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