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Chapter 6: Q10E (page 331)

Find an orthogonal basis for the column space of each matrix in Exercises 9-12.

10. \(\left( {\begin{aligned}{{}{}}{ - 1} & 6 & 6 \\ 3 & { - 8}&3\\1&{ - 2}&6\\1&{ - 4}&{ - 3}\end{aligned}} \right)\)

Short Answer

Expert verified

An orthogonal basis for the column space is \(\left\{ {\left( {\begin{aligned}{{}{}}{ - 1}\\3\\1\\1\end{aligned}} \right),\left( {\begin{aligned}{{}{}}3\\1\\1\\{ - 1}\end{aligned}} \right),\left( {\begin{aligned}{{}{}}{ - 1}\\{ - 1}\\3\\{ - 1}\end{aligned}} \right)} \right\}\).

Step by step solution

01

The Gram-Schmidt process

With abasis\(\left\{ {{{\bf{x}}_1}, \ldots ,{{\bf{x}}_p}} \right\}\)for a nonzero subspace \(W\) of \({\mathbb{R}^n}\), the expressionis shown below:

\(\begin{aligned}{}{{\bf{v}}_1} & = {{\bf{x}}_1}\\{{\bf{v}}_2} & = {{\bf{x}}_2} - \frac{{{{\bf{x}}_2} \cdot {{\bf{v}}_1}}}{{{{\bf{v}}_1} \cdot {{\bf{v}}_1}}}{{\bf{v}}_2}\\ \vdots \\{{\bf{v}}_p} & = \frac{{{{\bf{x}}_p} \cdot {{\bf{v}}_1}}}{{{{\bf{v}}_1} \cdot {{\bf{v}}_1}}}{{\bf{v}}_p} - \frac{{{{\bf{x}}_p} \cdot {{\bf{v}}_2}}}{{{{\bf{v}}_2} \cdot {{\bf{v}}_2}}}{{\bf{v}}_p} - \ldots - \frac{{{{\bf{x}}_{p - 1}} \cdot {{\bf{v}}_{p - 1}}}}{{{{\bf{v}}_{p - 1}} \cdot {{\bf{v}}_{p - 1}}}}{{\bf{v}}_{p - 1}}\end{aligned}\)

Therefore, theorthogonal basisfor \(W\) is \(\left\{ {{{\bf{v}}_1}, \ldots ,{{\bf{v}}_p}} \right\}\). Furthermore,

\({\mathop{\rm Span}\nolimits} \left\{ {{{\bf{v}}_1}, \ldots ,{{\bf{v}}_k}} \right\} = {\mathop{\rm Span}\nolimits} \left\{ {{{\bf{x}}_1}, \ldots ,{{\bf{x}}_k}} \right\}\) for \(1 \le k \le p\).

02

Determine an orthogonal basis for the column space

Consider the columns of the matrix as \({{\bf{x}}_1},{{\bf{x}}_2}\), and \({{\bf{x}}_3}\).

Apply the Gram-Schmidt process on these vectors to obtain an orthogonal basis as shown below:

\(\begin{aligned}{}{{\bf{v}}_1} & = {{\bf{x}}_1}\\{{\bf{v}}_2} & = {{\bf{x}}_2} - \frac{{{{\bf{x}}_2} \cdot {{\bf{v}}_1}}}{{{{\bf{v}}_1} \cdot {{\bf{v}}_1}}}{{\bf{v}}_1}\\ & = {{\bf{x}}_2} - \frac{{ - 36}}{{12}}{{\bf{v}}_1}\\ & = {{\bf{x}}_2} - \left( { - 3} \right){{\bf{v}}_1}\\ & = \left( {\begin{aligned}{{}{}}6\\{ - 8}\\{ - 2}\\{ - 4}\end{aligned}} \right) + 3\left( {\begin{aligned}{{}{}}{ - 1}\\3\\1\\1\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}{}}{6 - 3}\\{ - 8 + 9}\\{ - 2 + 3}\\{ - 4 + 3}\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}{}}3\\1\\1\\{ - 1}\end{aligned}} \right)\end{aligned}\)

And,

\(\begin{aligned}{}{{\bf{v}}_3} & = {{\bf{x}}_3} - \frac{{{{\bf{x}}_3} \cdot {{\bf{v}}_1}}}{{{{\bf{v}}_1} \cdot {{\bf{v}}_1}}}{{\bf{v}}_1} - \frac{{{{\bf{x}}_3} \cdot {{\bf{v}}_2}}}{{{{\bf{v}}_2} \cdot {{\bf{v}}_2}}}{{\bf{v}}_2}\\ & = {{\bf{x}}_3} - \frac{6}{{12}}{{\bf{v}}_1} - \left( {\frac{{30}}{{12}}} \right){{\bf{v}}_2}\\ & = {{\bf{x}}_3} - \frac{1}{2}{{\bf{v}}_1} - \left( {\frac{5}{2}} \right){{\bf{v}}_2}\\ & = \left( {\begin{aligned}{{}{}}6\\3\\6\\{ - 3}\end{aligned}} \right) - \frac{1}{2}\left( {\begin{aligned}{{}{}}{ - 1}\\3\\1\\1\end{aligned}} \right) - \frac{5}{2}\left( {\begin{aligned}{{}{}}3\\1\\1\\{ - 1}\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}{}}{6 + \frac{1}{2} - \frac{{15}}{2}}\\{3 - \frac{3}{2} - \frac{5}{2}}\\{6 - \frac{1}{2} - \frac{5}{2}}\\{ - 3 - \frac{1}{2} + \frac{5}{2}}\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}{}}{ - 1}\\{ - 1}\\3\\{ - 1}\end{aligned}} \right)\end{aligned}\)

Hence, an orthogonal basis for the column space is \(\left\{ {\left( {\begin{aligned}{{}{}}{ - 1}\\3\\1\\1\end{aligned}} \right),\left( {\begin{aligned}{{}{}}3\\1\\1\\{ - 1}\end{aligned}} \right),\left( {\begin{aligned}{{}{}}{ - 1}\\{ - 1}\\3\\{ - 1}\end{aligned}} \right)} \right\}\).

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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.

According to Kepler鈥檚 first law, a comet should have an elliptic, parabolic, or hyperbolic orbit (with gravitational attractions from the planets ignored). In suitable polar coordinates, the position \(\left( {r,\vartheta } \right)\) of a comet satisfies an equation of the form

\(r = \beta + e\left( {r \cdot \cos \vartheta } \right)\)

Where \(\beta \) is a constant and \(e\) is the eccentricity of the orbit, with \(0 \le e < 1\) for an ellipse, \(e = 1\) for a parabola, and \(e > 1\) for a hyperbola. Suppose observations of a newly discovered comet provide the data below. Determine the type of orbit, and predict where the comet will be when \(\vartheta = 4.6\left( {{\rm{radians}}} \right)\).

\(\begin{array}{{}{}} \vartheta & & {.88}&{1.10}&{1.42}&{1.77}&{2.14} \\ \hline r& {3.00}&{2.30}&{1.65}&{1.25}&{1.01} \end{array}\)

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

In Exercises 9-12 find (a) the orthogonal projection of b onto \({\bf{Col}}A\) and (b) a least-squares solution of \(A{\bf{x}} = {\bf{b}}\).

12. \(A = \left[ {\begin{array}{{}{}}{\bf{1}}&{\bf{1}}&{\bf{0}}\\{\bf{1}}&{\bf{0}}&{ - {\bf{1}}}\\{\bf{0}}&{\bf{1}}&{\bf{1}}\\{ - {\bf{1}}}&{\bf{1}}&{ - {\bf{1}}}\end{array}} \right]\), \({\bf{b}} = \left( {\begin{array}{{}{}}{\bf{2}}\\{\bf{5}}\\{\bf{6}}\\{\bf{6}}\end{array}} \right)\)

In Exercises 11 and 12, find the closest point to\[{\bf{y}}\]in the subspace\[W\]spanned by\[{{\bf{v}}_1}\], and\[{{\bf{v}}_2}\].

11.\[y = \left[ {\begin{aligned}3\\1\\5\\1\end{aligned}} \right]\],\[{{\bf{v}}_1} = \left[ {\begin{aligned}3\\1\\{ - 1}\\1\end{aligned}} \right]\],\[{{\bf{v}}_2} = \left[ {\begin{aligned}1\\{ - 1}\\1\\{ - 1}\end{aligned}} \right]\]
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