Find a \(QR\) factorization of the matrix in Exercise 12.

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

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

From exercise 11 we have,

\(A = \left( {\begin{aligned}{{}{r}}1&3&5\\{ - 1}&{ - 3}&1\\0&2&3\\1&5&2\\1&5&8\end{aligned}} \right)\)

Again, with help of exercise 12 where we have calculated the orthogonal basis for columns of \(A\) we have

\(\left\{ {\left( {\begin{aligned}{{}{r}}1\\{ - 1}\\0\\1\\1\end{aligned}} \right),\left( {\begin{aligned}{{}{r}}{ - 1}\\1\\2\\1\\1\end{aligned}} \right),\left( {\begin{aligned}{{}{r}}3\\3\\0\\{ - 3}\\3\end{aligned}} \right)} \right\}\)

Now normalizing them we get,

\(\begin{aligned}{}\left\{ {\frac{{{v_1}}}{{\left\| {{v_1}} \right\|}},\frac{{{v_2}}}{{\left\| {{v_2}} \right\|}},\frac{{{v_3}}}{{\left\| {{v_3}} \right\|}}} \right\} & = \left\{ {\frac{{\left( {\begin{aligned}{{}{r}}1\\{ - 1}\\0\\1\\1\end{aligned}} \right)}}{{\sqrt 4 }},\frac{{\left( {\begin{aligned}{{}{r}}{ - 1}\\1\\2\\1\\1\end{aligned}} \right)}}{{\sqrt 8 }},\frac{{\left( {\begin{aligned}{{}{r}}3\\3\\0\\{ - 3}\\3\end{aligned}} \right)}}{{\sqrt {4 \cdot 9} }}} \right\}\\\left\{ {\frac{{{v_1}}}{{\left\| {{v_1}} \right\|}},\frac{{{v_2}}}{{\left\| {{v_2}} \right\|}},\frac{{{v_3}}}{{\left\| {{v_3}} \right\|}}} \right\} & = \left\{ {\left( {\begin{aligned}{{}{r}}{\frac{1}{2}}\\{\frac{{ - 1}}{2}}\\0\\{\frac{1}{2}}\\{\frac{1}{2}}\end{aligned}} \right),\left( {\begin{aligned}{{}{r}}{ - \frac{1}{{2\sqrt 2 }}}\\{\frac{1}{{2\sqrt 2 }}}\\{\frac{1}{{\sqrt 2 }}}\\{\frac{1}{{2\sqrt 2 }}}\\{\frac{1}{{2\sqrt 2 }}}\end{aligned}} \right),\left( {\begin{aligned}{{}{r}}{\frac{1}{2}}\\{\frac{1}{2}}\\0\\{ - \frac{1}{2}}\\{\frac{1}{2}}\end{aligned}} \right)} \right\}\end{aligned}\)

Hence the matrix \(Q\) will be:

\(Q = \left( {\begin{aligned}{{}{r}}{\frac{1}{2}}&{ - \frac{1}{{2\sqrt 2 }}}&{\frac{1}{2}}\\{ - \frac{1}{2}}&{\frac{1}{{2\sqrt 2 }}}&{\frac{1}{2}}\\0&{\frac{1}{{\sqrt 2 }}}&0\\{\frac{1}{2}}&{\frac{1}{{2\sqrt 2 }}}&{ - \frac{1}{2}}\\{\frac{1}{2}}&{\frac{1}{{2\sqrt 2 }}}&{\frac{1}{2}}\end{aligned}} \right)\)

Now, calculate \({Q^T}A = R\) by using \(A\) and \(Q\).

\(\begin{aligned}{}R & = {Q^T}A\\ & = \left( {\begin{aligned}{{}{r}}{\frac{1}{2}}&{ - \frac{1}{2}}&0&{\frac{1}{2}}&{\frac{1}{2}}\\{ - \frac{1}{{2\sqrt 2 }}}&{\frac{1}{{2\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}}&{\frac{1}{{2\sqrt 2 }}}&{\frac{1}{{2\sqrt 2 }}}\\{\frac{1}{2}}&{\frac{1}{2}}&0&{\frac{{ - 1}}{2}}&{\frac{1}{2}}\end{aligned}} \right)\left( {\begin{aligned}{{}{r}}1&3&5\\{ - 1}&{ - 3}&1\\0&2&3\\1&5&2\\1&5&8\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}{}}2&8&7\\0&{2\sqrt 2 }&{3\sqrt 2 }\\0&0&6\end{aligned}} \right)\end{aligned}\)

Hence, the required factorization is,\(A = \left( {\begin{aligned}{{}{r}}{\frac{1}{2}}&{ - \frac{1}{{2\sqrt 2 }}}&{\frac{1}{2}}\\{ - \frac{1}{2}}&{\frac{1}{{2\sqrt 2 }}}&{\frac{1}{2}}\\0&{\frac{1}{{\sqrt 2 }}}&0\\{\frac{1}{2}}&{\frac{1}{{2\sqrt 2 }}}&{ - \frac{1}{2}}\\{\frac{1}{2}}&{\frac{1}{{2\sqrt 2 }}}&{\frac{1}{2}}\end{aligned}} \right)\left( {\begin{aligned}{{}{}}2&8&7\\0&{2\sqrt 2 }&{3\sqrt 2 }\\0&0&6\end{aligned}} \right)\).