 Solution

Proceed column by column. At stage $j=1,\dots ,n-2$, consider the entries below the diagonal in column $j$, indexed $j+1,\dots ,n$. 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 $1,\dots ,j$, and apply the similarity transformation $A\mapsto Q_j^{T}A{Q}_j$. 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 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 $Q^{T}Q=I$ andThis constructive reduction uses only reflections on successive trailing subspaces.