= Solution
A generic lower triangular matrix has a dense first column. The first left <Householder transformation> in its <QR decomposition> mixes those rows, usually creating entries throughout the rows to the right. Subsequent eliminations impose upper triangularity but no further systematic zeros; thus $R$ is generically a full upper triangular matrix.
For <Householder bidiagonalization> use two-sided operations instead. At stage $j$, a right <Householder transformation> acting on columns $j,\ldots,n$ clears the current row $j$ beyond its diagonal. Then a left <Householder transformation> acting on rows $j+1,\ldots,n$ clears column $j$ below its first subdiagonal. The left operation leaves row $j$ unchanged; both operations preserve the already completed lower-bidiagonal rows and columns. Continue for $j=1,\ldots,n-1$, omitting a final unnecessary left operation. The result is
$$
\boxed{B=PAQ,\qquad P,Q\text{ orthogonal},\quad B\text{ lower bidiagonal}.}
$$
It follows that $B^TB=Q^TA^TAQ$. Hence the two symmetric matrices are <orthogonally similar> and have the same <eigenvalues>. By part (c), compute the <eigenvalues> of the symmetric <tridiagonal matrix> $B^TB$ and take their nonnegative square roots to obtain all <singular values> of $A$.
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