 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 $j,\dots ,n$ clears the current row $j$ beyond its diagonal. Then a left Householder transformation acting on rows $j+1,\dots ,n$ 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 $j=1,\dots ,n-1$, omitting a final unnecessary left operation. The result isIt follows that $B^{T}B=Q^T{A}^{T}AQ$. Hence the two symmetric matrices are orthogonally similar and have the same eigenvalues. By part (c), compute the eigenvalues of the symmetric tridiagonal matrix $B^{T}B$ and take their nonnegative square roots to obtain all singular values of $A$.