= Solution
Starting with $A_0=A$, suppose columns $1,\ldots,j-1$ have already been reduced to tridiagonal form. Let $x$ be the part of column $j$ in rows $j+1,\ldots,n$. A <Householder reflection> $P_j$ on these coordinates can map $x$ to $\pm\lVert x\rVert e_1$. Extend it by the identity on the first $j$ coordinates and call the resulting orthogonal <matrix> $U_j$.
The update
$$
A_j=U_j^TA_{j-1}U_j
$$
zeros every entry below row $j+1$ in column $j$. Because it is an <orthogonal similarity>, it preserves <eigenvalues>; because the same transformation is applied on both sides, it preserves symmetry and zeros the corresponding row entries without disturbing earlier columns. After $n-2$ such steps,
$$
H=U^TAU,\qquad U=U_1U_2\cdots U_{n-2},
$$
is symmetric and tridiagonal and has the same <eigenvalues> as $A$. Every reflector is obtained from finitely many <matrix> entries using finitely many arithmetic operations and a square root, so this is a finite construction.
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