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Chapter 6: Q13E (page 331)

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

Expert verified

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

Step by step solution

01

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.

02

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 \(R = \left( {\begin{aligned}{{}{}}6&{12}\\0& 6\end{aligned}} \right)\).

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Most popular questions from this chapter

24. Question: In Exercises 23 and 24, all vectors are in \({\mathbb{R}^n}\). Mark each statement True or False. Justify each answer.

  1. Not every orthogonal set in \({\mathbb{R}^n}\) is linearly independent.
  2. If a set \(S = \left\{ {{{\mathop{\rm u}\nolimits} _1}, \ldots ,{{\mathop{\rm u}\nolimits} _p}} \right\}\) has the property that \({{\mathop{\rm u}\nolimits} _i} \cdot {{\mathop{\rm u}\nolimits} _j} = 0\) whenever \(i \ne j\), then \(S\) is an orthonormal set.
  3. If the columns of a \(m \times n\) matrix A are orthonormal, then the linear mapping \({\mathop{\rm x}\nolimits} \mapsto A{\mathop{\rm x}\nolimits} \) preserves lengths.
  4. The orthogonal projection of y onto v is the same as the orthogonal projection of y onto \(c{\mathop{\rm v}\nolimits} \) whenever \(c \ne 0\).
  5. An orthogonal matrix is invertible.

In Exercises 7鈥�10, let\[W\]be the subspace spanned by the\[{\bf{u}}\]鈥檚, and write y as the sum of a vector in\[W\]and a vector orthogonal to\[W\].

7.\[y = \left[ {\begin{aligned}1\\3\\5\end{aligned}} \right]\],\[{{\bf{u}}_1} = \left[ {\begin{aligned}1\\3\\{ - 2}\end{aligned}} \right]\],\[{{\bf{u}}_2} = \left[ {\begin{aligned}5\\1\\4\end{aligned}} \right]\]

Use the inner product axioms and other results of this section to verify the statements in Exercises 15鈥�18.

15.\(\left\langle {{\rm{u,}}\,c{\rm{v}}} \right\rangle = c\left\langle {{\rm{u,}}\,{\rm{v}}} \right\rangle \) for all scalars \(c\).

Let \({\mathbb{R}^{\bf{2}}}\) have the inner product of Example 1, and let \({\bf{x}} = \left( {{\bf{1}},{\bf{1}}} \right)\) and \({\bf{y}} = \left( {{\bf{5}}, - {\bf{1}}} \right)\).

a. Find\(\left\| {\bf{x}} \right\|\),\(\left\| {\bf{y}} \right\|\), and\({\left| {\left\langle {{\bf{x}},{\bf{y}}} \right\rangle } \right|^{\bf{2}}}\).

b. Describe all vectors\(\left( {{z_{\bf{1}}},{z_{\bf{2}}}} \right)\), that are orthogonal to y.

Suppose \(A = QR\), where \(R\) is an invertible matrix. Showthat \(A\) and \(Q\) have the same column space.

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