= Solution
Proceed column by column. At stage $j=1,\ldots,n-2$, consider the entries below the diagonal in column $j$, indexed $j+1,\ldots,n$. If this tail is nonzero, part (c) provides a <Householder matrix> sending it to a multiple of its first coordinate vector. Embed this <matrix> as an <orthogonal> <matrix> acting identically on coordinates $1,\ldots,j$, and apply the similarity transformation $A\mapsto Q_j^TAQ_j$. The tail's entries below row $j+1$ become zero. Symmetry is preserved, so the corresponding entries in row $j$ vanish too. If the tail is already zero, use the <identity element>[identity] transformation.
Earlier reduced columns remain unchanged in their required zero positions: all their entries in the newly mixed tail coordinates were already zero, and the first $j$ coordinates are fixed. Thus each stage preserves the previous zeros while creating the next ones. After $n-2$ stages, all entries at distance greater than one from the diagonal vanish. With the product $Q=Q_1\cdots Q_{n-2}$ one has $Q^TQ=I$ and
$$
\boxed{Q^TAQ\text{ is symmetric tridiagonal}.}
$$
This constructive reduction uses only reflections on successive trailing subspaces.
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