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Chapter 6: Q16E (page 331)

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

Short Answer

Expert verified

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

Step by step solution

01

\(QR\) factorization of a Matrix

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 \({Q^T}A = R\).

02

Finding the matrix \(R\)

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

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Most popular questions from this chapter

Let \(\left\{ {{{\bf{v}}_1}, \ldots ,{{\bf{v}}_p}} \right\}\) be an orthonormal set in \({\mathbb{R}^n}\). Verify the following inequality, called Bessel’s inequality, which is true for each x in \({\mathbb{R}^n}\):

\({\left\| {\bf{x}} \right\|^2} \ge {\left| {{\bf{x}} \cdot {{\bf{v}}_1}} \right|^2} + {\left| {{\bf{x}} \cdot {{\bf{v}}_2}} \right|^2} + \ldots + {\left| {{\bf{x}} \cdot {{\bf{v}}_p}} \right|^2}\)

Let \({{\bf{u}}_1},......,{{\bf{u}}_p}\) be an orthogonal basis for a subspace \(W\) of \({\mathbb{R}^n}\), and let \(T:{\mathbb{R}^n} \to {\mathbb{R}^n}\) be defined by \(T\left( x \right) = {\rm{pro}}{{\rm{j}}_W}x\). Show that \(T\) is a linear transformation.

For a matrix program, the Gram–Schmidt process worksbetter with orthonormal vectors. Starting with \({x_1},......,{x_p}\) asin Theorem 11, let \(A = \left\{ {{x_1},......,{x_p}} \right\}\) . Suppose \(Q\) is an\(n \times k\)matrix whose columns form an orthonormal basis for

the subspace \({W_k}\) spanned by the first \(k\) columns of A. Thenfor \(x\) in \({\mathbb{R}^n}\), \(Q{Q^T}x\) is the orthogonal projection of x onto \({W_k}\) (Theorem 10 in Section 6.3). If \({x_{k + 1}}\) is the next column of \(A\),then equation (2) in the proof of Theorem 11 becomes

\({v_{k + 1}} = {x_{k + 1}} - Q\left( {{Q^T}T {x_{k + 1}}} \right)\)

(The parentheses above reduce the number of arithmeticoperations.) Let \({u_{k + 1}} = \frac{{{v_{k + 1}}}}{{\left\| {{v_{k + 1}}} \right\|}}\). The new \(Q\) for thenext step is \(\left( {\begin{aligned}{{}{}}Q&{{u_{k + 1}}}\end{aligned}} \right)\). Use this procedure to compute the\(QR\)factorization of the matrix in Exercise 24. Write thekeystrokes or commands you use.

In exercises 1-6, determine which sets of vectors are orthogonal.

\(\left[ {\begin{align}{ 2}\\{ - 7}\\{-1}\end{align}} \right]\), \(\left[ {\begin{align}{ - 6}\\{ - 3}\\9\end{align}} \right]\), \(\left[ {\begin{align}{ 3}\\{ 1}\\{-1}\end{align}} \right]\)

In exercises 1-6, determine which sets of vectors are orthogonal.

\(\left[ {\begin{align} 2\\{-5}\\{-3}\end{align}} \right]\), \(\left[ {\begin{align}0\\0\\0\end{align}} \right]\), \(\left[ {\begin{align} 4\\{ - 2}\\6\end{align}} \right]\)
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