 Solution

Apply part (b) to the first column of $A$, using the identity if it is already the zero vector. This gives an orthogonal matrix $H_1$ for which $H_{1}A$ has zeros below its first diagonal entry. At stage $j$, apply a Householder transformation to rows $j,\dots ,n$ of column $j$, and extend it by the identity on the first $j-1$ coordinates. If that trailing column is zero, again use the identity. The transformation preserves zeros in earlier columns, since their trailing entries already vanish.

After at most $n-1$ stages,is an upper triangular matrix. Each $H_j$ is symmetric and orthogonal, so withwe have $Q^{T}Q=I$ andNo invertibility assumption was used, so the Householder QR decomposition also exists for singular matrices.