Question

Find a \(QR\) factorization of the matrix in Exercise 11.

Short answer

The factorization of the matrix is,\(A = \left( {\begin{aligned}{{}{r}}{\frac{1}{{\sqrt 5 }}}&{\frac{1}{2}}&{\frac{1}{2}}\\{\frac{{ - 1}}{{\sqrt 5 }}}&0&0\\{\frac{{ - 1}}{{\sqrt 5 }}}&{\frac{1}{2}}&{\frac{1}{2}}\\{\frac{1}{{\sqrt 5 }}}&{\frac{{ - 1}}{2}}&{\frac{1}{2}}\\{\frac{1}{{\sqrt 5 }}}&{\frac{1}{2}}&{\frac{{ - 1}}{2}}\end{aligned}} \right)\left( {\begin{aligned}{{}{}}{\sqrt 5 }&{ - \sqrt 5 }&{4\sqrt 5 }\\0&6&{ - 2}\\0&0&4\end{aligned}} \right)\).

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 triangularmatrix \(R\) and a matrix \(Q\) which is formed by applying Gram–Schmidt orthogonalizationprocess to the \({\rm{col}}\left( A \right)\).

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

Finding the matrix \(R\)

From exercise 11 we have,

\(A = \left( {\begin{aligned}{{}{r}}1&2&5\\{ - 1}&1&{ - 4}\\{ - 1}&4&{ - 3}\\1&{ - 4}&7\\1&2&1\end{aligned}} \right)\)

Again, with help of exercise 11 where we have calculated the orthogonal basisfor columnsof \(A\) we have,

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

Now normalizing them we get

\(\begin{aligned}{}\left\{ {\frac{{{v_1}}}{{\left\| {{v_1}} \right\|}},\frac{{{v_2}}}{{\left\| {{v_2}} \right\|}},\frac{{{v_3}}}{{\left\| {{v_3}} \right\|}}} \right\} & = \left\{ {\frac{{\left( {\begin{aligned}{{}{r}}1\\{ - 1}\\{ - 1}\\1\\1\end{aligned}} \right)}}{{\sqrt 5 }},\frac{{\left( {\begin{aligned}{{}{r}}3\\0\\3\\{ - 3}\\3\end{aligned}} \right)}}{{\sqrt {4 \cdot 9} }},\frac{{\left( {\begin{aligned}{{}{r}}2\\0\\2\\2\\{ - 2}\end{aligned}} \right)}}{{\sqrt {16} }}} \right\}\\\left\{ {\frac{{{v_1}}}{{\left\| {{v_1}} \right\|}},\frac{{{v_2}}}{{\left\| {{v_2}} \right\|}},\frac{{{v_3}}}{{\left\| {{v_3}} \right\|}}} \right\} & = \left\{ {\left( {\begin{aligned}{{}{r}}{\frac{1}{{\sqrt 5 }}}\\{\frac{{ - 1}}{{\sqrt 5 }}}\\{\frac{{ - 1}}{{\sqrt 5 }}}\\{\frac{1}{{\sqrt 5 }}}\\{\frac{1}{{\sqrt 5 }}}\end{aligned}} \right),\left( {\begin{aligned}{{}{r}}{\frac{1}{2}}\\0\\{\frac{1}{2}}\\{ - \frac{1}{2}}\\{\frac{1}{2}}\end{aligned}} \right),\left( {\begin{aligned}{{}{r}}{\frac{1}{2}}\\0\\{\frac{1}{2}}\\{\frac{1}{2}}\\{ - \frac{1}{2}}\end{aligned}} \right)} \right\}\end{aligned}\)

Hence, the matrix \(Q\) will be:

\(Q = \left( {\begin{aligned}{{}{r}}{\frac{1}{{\sqrt 5 }}}&{\frac{1}{2}}&{\frac{1}{2}}\\{\frac{{ - 1}}{{\sqrt 5 }}}&0&0\\{\frac{{ - 1}}{{\sqrt 5 }}}&{\frac{1}{2}}&{\frac{1}{2}}\\{\frac{1}{{\sqrt 5 }}}&{\frac{{ - 1}}{2}}&{\frac{1}{2}}\\{\frac{1}{{\sqrt 5 }}}&{\frac{1}{2}}&{\frac{{ - 1}}{2}}\end{aligned}} \right)\)

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

\(\begin{aligned}{}R & = {Q^T}A\\ & = \left( {\begin{aligned}{{}{r}}{\frac{1}{{\sqrt 5 }}}&{\frac{1}{{\sqrt 5 }}}&{\frac{1}{{\sqrt 5 }}}&{\frac{1}{{\sqrt 5 }}}&{\frac{1}{{\sqrt 5 }}}\\{\frac{1}{2}}&0&{\frac{1}{2}}&{\frac{{ - 1}}{2}}&{\frac{1}{2}}\\{\frac{1}{2}}&0&{\frac{1}{2}}&{\frac{1}{2}}&{\frac{{ - 1}}{2}}\end{aligned}} \right)\left( {\begin{aligned}{{}{r}}1&2&5\\{ - 1}&1&{ - 4}\\{ - 1}&4&{ - 3}\\1&{ - 4}&7\\1&2&1\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}{r}}{\sqrt 5 }&{ - \sqrt 5 }&{4\sqrt 5 }\\0&6&{ - 2}\\0&0&4\end{aligned}} \right)\end{aligned}\)

Hence, the required factorization is,\(A = \left( {\begin{aligned}{{}{r}}{\frac{1}{{\sqrt 5 }}}&{\frac{1}{2}}&{\frac{1}{2}}\\{\frac{{ - 1}}{{\sqrt 5 }}}&0&0\\{\frac{{ - 1}}{{\sqrt 5 }}}&{\frac{1}{2}}&{\frac{1}{2}}\\{\frac{1}{{\sqrt 5 }}}&{\frac{{ - 1}}{2}}&{\frac{1}{2}}\\{\frac{1}{{\sqrt 5 }}}&{\frac{1}{2}}&{\frac{{ - 1}}{2}}\end{aligned}} \right)\left( {\begin{aligned}{{}{}}{\sqrt 5 }&{ - \sqrt 5 }&{4\sqrt 5 }\\0&6&{ - 2}\\0&0&4\end{aligned}} \right)\).