Question

Suppose \(A = QR\) is a \(QR\) factorization of an \(m \times n\) matrix

A (with linearly independent columns). Partition \(A\) as \(\left[ {\begin{aligned}{{}{}}{{A_1}}&{{A_2}}\end{aligned}} \right]\), where \({A_1}\) has \(p\) columns. Show how to obtain a \(QR\) factorization of \({A_1}\), and explain why your factorization has the appropriate properties.

Short answer

It is shown that how to obtain a \(QR\) factorization of \({A_1}\).

Step-by-step solution

\(QR\) factorization of a Matrix

A matrix with order \(m \times n\) can be written as the multiplication of an upper triangular matrix \(R\) and a matrix \(Q\) which is formed by applying the Gram–Schmidt orthogonalization processto the \({\rm{col}}\left( A \right)\).

The matrix \(R\) can be found by the formula \({Q^T}A = R\).

Obtain \(QR\) factorization of  \(A\)

Given that \(A = QR\).

Since \(\left[ {\begin{aligned}{{}{}}{{A_1}}&{{A_2}}\end{aligned}} \right]\) is the partition of\(A\), we can also obtain \(Q\) as portioned such as \(Q = \left[ {\begin{aligned}{{}{}}{{Q_1}}&{{Q_2}}\end{aligned}} \right]\), where both \({A_1}\) and \({Q_1}\) have \(p\) columns.

Now we can partition \(R\) as

\(R = \left[ {\begin{aligned}{{}{}}B&C\\0&D\end{aligned}} \right]\)

Here the matrix \(B\) is \(p \times p\).

Then, find \(A = QR\).

\(\begin{aligned}{}\left[ {\begin{aligned}{{}{}}{{A_1}}&{{A_2}}\end{aligned}} \right] = \left[ {\begin{aligned}{{}{}}{{Q_1}}&{{Q_2}}\end{aligned}} \right]\left[ {\begin{aligned}{{}{}}B&C\\0&D\end{aligned}} \right]\\\left[ {\begin{aligned}{{}{}}{{A_1}}&{{A_2}}\end{aligned}} \right] = \left[ {\begin{aligned}{{}{}}{{Q_1}B}&{{Q_1}}\end{aligned}C + {Q_2}D} \right]\end{aligned}\)

On comparing both sides, we get\({A_1} = {Q_1}B\), this the \(QR\) factorization of matrix\({A_1}\).

Since,\({Q_1}\) is obtained from \(Q\), thus the columns of \({Q_1}\) are orthonormal.

Also \(B\) is an upper triangular since it is in the 1^st entry of the partition. Also, the diagonal entries of the matrix \(B\) are positive since these entries are from the diagonal entries of \(R\).