###### Homework3.5.1.1.

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 $R = R_B \text{.}$

###### Remark3.5.1.1.

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.