In Exercises 1-6, the given set is a basis for a subspace W. Use the Gram-Schmidt process to produce an orthogonal basis for W.

2. \(\left( {\begin{aligned}{{}{}}0\\4\\2\end{aligned}} \right),\left( {\begin{aligned}{{}{}}5\\6\\{ - 7}\end{aligned}} \right)\)

Short Answer

Expert verified

\(\left\{ {\left( {\begin{aligned}{{}{}}0\\4\\2\end{aligned}} \right),\left( {\begin{aligned}{{}{}}5\\4\\{ - 8}\end{aligned}} \right)} \right\}\) is an orthogonal basis for \(W\).

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

Use a Gram-Schmidt process to produce an orthogonal basis for W

Let \({{\bf{x}}_1} = \left( {\begin{aligned}{{}{}}0\\4\\2\end{aligned}} \right),{{\bf{x}}_2} = \left( {\begin{aligned}{{}{}}5\\6\\{ - 7}\end{aligned}} \right)\).

Use a Gram-Schmidt process and let \({{\bf{x}}_1} = {{\bf{v}}_1}\) to calculate \({{\bf{v}}_2}\) as shown below:

\(\begin{aligned}{}{{\bf{v}}_2} &= {{\bf{x}}_2} - \frac{{{{\bf{x}}_2} \cdot {{\bf{v}}_1}}}{{{{\bf{v}}_1} \cdot {{\bf{v}}_1}}}{{\bf{v}}_2}\\ & = {{\bf{x}}_2} - \frac{{10}}{{20}}{{\bf{v}}_1}\\ & = {{\bf{x}}_2} - \frac{1}{2}{{\bf{v}}_1}\\ & = \left( {\begin{aligned}{{}{}}5\\6\\{ - 7}\end{aligned}} \right) - \frac{1}{2}\left( {\begin{aligned}{{}{}}0\\4\\2\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}{}}{5 - 0}\\{6 - 2}\\{ - 7 - 1}\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}{}}5\\4\\{ - 8}\end{aligned}} \right)\end{aligned}\)

Thus, \(\left\{ {\left( {\begin{aligned}{{}{}}0\\4\\2\end{aligned}} \right),\left( {\begin{aligned}{{}{}}5\\4\\{ - 8}\end{aligned}} \right)} \right\}\) is an orthogonal basis for \(W\).

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Most popular questions from this chapter

In Exercises 1-6, the given set is a basis for a subspace W. Use the Gram-Schmidt process to produce an orthogonal basis for W.

6. \(\left( {\begin{aligned}{{}}3\\{ - 1}\\2\\{ - 1}\end{aligned}} \right),\left( {\begin{aligned}{{}}{ - 5}\\9\\{ - 9}\\3\end{aligned}} \right)\)

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\(\begin{aligned}{l}\sum x = \sum\nolimits_{i = 1}^n {{x_i}} ,{\rm{ }}\sum {{x^2}} = \sum\nolimits_{i = 1}^n {x_i^2} ,\\\sum y = \sum\nolimits_{i = 1}^n {{y_i}} ,{\rm{ }}\sum {xy} = \sum\nolimits_{i = 1}^n {{x_i}{y_i}} \end{aligned}\)

The normal equations for a least-squares line \(y = {\hat \beta _0} + {\hat \beta _1}x\)may be written in the form

\(\begin{aligned}{{\hat \beta }_0} + {{\hat \beta }_1}\sum x = \sum y \\{{\hat \beta }_0}\sum x + {{\hat \beta }_1}\sum {{x^2}} = \sum {xy} {\rm{ (7)}}\end{aligned}\)

16. Use a matrix inverse to solve the system of equations in (7) and thereby obtain formulas for \({\hat \beta _0}\) , and that appear in many statistics texts.

To measure the take-off performance of an airplane, the horizontal position of the plane was measured every second, from \(t = 0\) to \(t = 12\). The positions (in feet) were: 0, 8.8, 29.9, 62.0, 104.7, 159.1, 222.0, 294.5, 380.4, 471.1, 571.7, 686.8, 809.2.

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