Question

In Exercises 13 and 14, the columns of Q were obtained by applying the Gram-Schmidt process to the columns of A. Find an upper triangular matrix R such that \(A = QR\). Check your work.

13. \(A = \left( {\begin{aligned}{{}{}}5&9\\1&7\\{ - 3}&{ - 5}\\1&5\end{aligned}} \right),{\rm{ }}Q = \left( {\begin{aligned}{{}{}}{\frac{5}{6}}&{ - \frac{1}{6}}\\{\frac{1}{6}}&{\frac{5}{6}}\\{ - \frac{3}{6}}&{\frac{1}{6}}\\{\frac{1}{6}}&{\frac{3}{6}}\end{aligned}} \right)\)

Short answer

The upper triangular matrix is \(R = \left( {\begin{aligned}{{}{}}6&{12}\\0&6\end{aligned}} \right)\).

Step-by-step solution

The QR Factorization

When \(A\) is an\(m \times n\) matrix that haslinearly independent columns, then \(A\) may be factored as \(A = QR\), with \(Q\) is an\(m \times n\) matrix wherein, columns provide an orthonormal basisfor \({\mathop{\rm Col}\nolimits} A\), and \(R\) is an \(n \times n\) upper triangular invertible matrix which has positive entries on its diagonal.

Find an upper triangular matrix R

It is given that \(A = \left( {\begin{aligned}{{}{}}5&9\\1&7\\{ - 3}&{ - 5}\\1&5\end{aligned}} \right),{\rm{ }}Q = \left( {\begin{aligned}{{}{}}{\frac{5}{6}}&{ - \frac{1}{6}}\\{\frac{1}{6}}&{\frac{5}{6}}\\{ - \frac{3}{6}}&{\frac{1}{6}}\\{\frac{1}{6}}&{\frac{3}{6}}\end{aligned}} \right)\).

Obtain the upper triangular matrix \(R\) as shown below:

\(\begin{aligned}{}R & = {Q^T}A\\ & = \left( {\begin{aligned}{{}{}}{\frac{5}{6}}&{\frac{1}{6}}&{ - \frac{3}{6}}&{\frac{1}{6}}\\{ - \frac{1}{6}}&{\frac{5}{6}}&{\frac{1}{6}}&{\frac{3}{6}}\end{aligned}} \right)\left( {\begin{aligned}{{}{}}5&9\\1&7\\{ - 3}&{ - 5}\\1&5\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}{}}{\frac{{25}}{6} + \frac{1}{6} + \frac{3}{2} + \frac{1}{6}}&{\frac{{15}}{2} + \frac{7}{6} + \frac{5}{2} + \frac{5}{6}}\\{ - \frac{5}{6} + \frac{5}{6} - \frac{1}{2} + \frac{1}{2}}&{ - \frac{3}{2} + \frac{{35}}{6} - \frac{5}{6} + \frac{5}{2}}\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}{}}6&{12}\\0&6\end{aligned}} \right)\end{aligned}\)

Thus, the upper triangular matrix is \(R = \left( {\begin{aligned}{{}{}}6&{12}\\0& 6\end{aligned}} \right)\).