 Solution

Apply part (b) to the first column of $A$, using an orthogonal reflection to zero all entries below the first. Next apply a Householder reflection to the trailing subvector of the second column, embedded as $\mathrm{diag}(1,H_2)$ so it leaves the first row and the zeros already created intact. Continue similarly through the columns. If a trailing column subvector is zero, skip that step.

The product of these embedded orthogonal matrices sends $A$ to an upper triangular matrix $R$. Write that product as $Q^T$; its transpose $Q$ is also orthogonal. HenceThe argument uses no inverse of $A$ and therefore includes singular matrices.