= Solution
Set $A^{(0)}=A$. At step $k=1,\ldots,n-2$, let $u=A^{(k-1)}_{k+1:n,k}$ be the column below the diagonal. Choose a <Householder reflection> $\widehat H_k$ sending $u$ onto its first coordinate axis, and embed it as
$$
P_k=\operatorname{diag}(I_k,\widehat H_k),\qquad A^{(k)}=P_k^TA^{(k-1)}P_k.
$$
If $u=0$, use $P_k=I$. Left multiplication annihilates entries $k+2,\ldots,n$ in column $k$. Right multiplication performs the matching operation on row $k$, because the <matrix> remains <symmetric>. The first $k$ coordinates are fixed, so zeros already created in earlier columns and rows are preserved.
Here the <orthogonality> of each $P_k$ is used explicitly: $P_k^{-1}=P_k^T$, so each update is a <similarity transformation> and preserves <eigenvalues>. After $n-2$ steps,
$$
\boxed{T=Q^TAQ,\qquad Q=P_1P_2\cdots P_{n-2},}
$$
where $Q$ is <orthogonal> and $T$ is a real symmetric <tridiagonal matrix>. This is <Householder tridiagonalization>.
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