Skip to main content

Subsection 3.5.1 Additional homework

Homework 3.5.1.1.

Consider the matrix \(\left( \begin{array}{c} A \\ \hline B \end{array} \right) \) where \(A \) has linearly independent columns. Let

  • \(A = Q_A R_A \) be the QR factorization of \(A \text{.}\)

  • \(\left( \begin{array}{c} R_A \\ \hline B \end{array} \right) = Q_B R_B\) be the QR factorization of \(\left( \begin{array}{c} R_A \\ \hline \hline B \end{array} \right) \text{.}\)

  • \(\left( \begin{array}{c} A \\ \hline \hline B \end{array} \right) = Q R\) be the QR factorization of \(\left( \begin{array}{c} A \\ \hline \hline B \end{array} \right) \text{.}\)

Assume that the diagonal entries of \(R_A \text{,}\) \(R_B \text{,}\) and \(R \) are all positive. Show that \(R = R_B \text{.}\)

Solution
\begin{equation*} \left( \begin{array}{c} A \\ \hline \hline B \end{array} \right) = \left( \begin{array}{c | c} Q_A \amp 0 \\ \hline 0 \amp I \end{array} \right) \left( \begin{array}{c} R_A \\ \hline B \end{array} \right) = \left( \begin{array}{c | c} Q_A \amp 0 \\ \hline 0 \amp I \end{array} \right) Q_B R_B \end{equation*}

Also, \(\left( \begin{array}{c} A \\ \hline B \end{array} \right) = Q R \text{.}\) By the uniqueness of the QR factorization (when the diagonal elements of the triangular matrix are restricted to be positive), \(Q = \left( \begin{array}{c | c} Q_A \amp 0 \\ \hline 0 \amp I \end{array} \right) Q_B \) and \(R = R_B \text{.}\)

Remark 3.5.1.1.

This last exercise gives a key insight that is explored in the paper

  • [14] Brian C. Gunter, Robert A. van de Geijn, Parallel out-of-core computation and updating of the QR factorization, ACM Transactions on Mathematical Software (TOMS), 2005.