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Chapter 6: Q6E (page 331)

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

Short Answer

Expert verified

\(\left\{ {\left( {\begin{aligned}{{}}3\\{ - 1}\\2\\{ - 1}\end{aligned}} \right),\left( {\begin{aligned}{{}}4\\6\\{ - 3}\\0\end{aligned}} \right)} \right\}\) 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 \({{\bf{x}}_1} = \left( {\begin{aligned}{{}}3\\{ - 1}\\2\\{ - 1}\end{aligned}} \right),{{\bf{x}}_2} = \left( {\begin{aligned}{{}}{ - 5}\\9\\{ - 9}\\3\end{aligned}} \right)\).

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

\(\begin{aligned}{}{{\bf{v}}_2} & = {{\bf{x}}_2} - \frac{{{{\bf{x}}_2} \cdot {{\bf{v}}_1}}}{{{{\bf{v}}_1} \cdot {{\bf{v}}_1}}}{{\bf{v}}_2}\\ & = {{\bf{x}}_2} - \frac{{ - 45}}{{15}}{{\bf{v}}_1}\\ &= {{\bf{x}}_2} - \left( { - 3} \right){{\bf{v}}_1}\\ & = \left( {\begin{aligned}{{}}{ - 5}\\9\\{ - 9}\\3\end{aligned}} \right) + 3\left( {\begin{aligned}{{}}3\\{ - 1}\\2\\{ - 1}\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}}{ - 5 + 9}\\{9 - 3}\\{ - 9 + 6}\\{3 + 3}\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}}4\\6\\{ - 3}\\0\end{aligned}} \right)\end{aligned}\)

Hence, an orthogonal basis for \(W\) is \(\left\{ {\left( {\begin{aligned}{{}}3\\{ - 1}\\2\\{ - 1}\end{aligned}} \right),\left( {\begin{aligned}{{}}4\\6\\{ - 3}\\0\end{aligned}} \right)} \right\}\).

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Most popular questions from this chapter

Exercises 19 and 20 involve a design matrix \(X\) with two or more columns and a least-squares solution \(\hat \beta \) of \({\bf{y}} = X\beta \). Consider the following numbers.

(i) \({\left\| {X\hat \beta } \right\|^2}\)鈥攖he sum of the squares of the 鈥渞egression term.鈥� Denote this number by .

(ii) \({\left\| {{\bf{y}} - X\hat \beta } \right\|^2}\)鈥攖he sum of the squares for error term. Denote this number by \(SS\left( E \right)\).

(iii) \({\left\| {\bf{y}} \right\|^2}\)鈥攖he 鈥渢otal鈥� sum of the squares of the \(y\)-values. Denote this number by \(SS\left( T \right)\).

Every statistics text that discusses regression and the linear model \(y = X\beta + \in \) introduces these numbers, though terminology and notation vary somewhat. To simplify matters, assume that the mean of the -values is zero. In this case, \(SS\left( T \right)\) is proportional to what is called the variance of the set of -values.

19. Justify the equation \(SS\left( T \right) = SS\left( R \right) + SS\left( E \right)\). (Hint: Use a theorem, and explain why the hypotheses of the theorem are satisfied.) This equation is extremely important in statistics, both in regression theory and in the analysis of variance.

In Exercises 1-4, find the equation \(y = {\beta _0} + {\beta _1}x\) of the least-square line that best fits the given data points.

  1. \(\left( { - 1,0} \right),\left( {0,1} \right),\left( {1,2} \right),\left( {2,4} \right)\)

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

11. \(\left( {\begin{aligned}{{}{}}1&2&5\\{ - 1}&1&{ - 4}\\{ - 1}&4&{ - 3}\\1&{ - 4}&7\\1&2&1\end{aligned}} \right)\)

Compute the quantities in Exercises 1-8 using the vectors

\({\mathop{\rm u}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 1}\\2\end{aligned}} \right),{\rm{ }}{\mathop{\rm v}\nolimits} = \left( {\begin{aligned}{*{20}{c}}4\\6\end{aligned}} \right),{\rm{ }}{\mathop{\rm w}\nolimits} = \left( {\begin{aligned}{*{20}{c}}3\\{ - 1}\\{ - 5}\end{aligned}} \right),{\rm{ }}{\mathop{\rm x}\nolimits} = \left( {\begin{aligned}{*{20}{c}}6\\{ - 2}\\3\end{aligned}} \right)\)

4. \(\frac{1}{{{\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} }}{\mathop{\rm u}\nolimits} \)

In Exercises 9-12, find a unit vector in the direction of the given vector.

9. \(\left( {\begin{aligned}{*{20}{c}}{ - 30}\\{40}\end{aligned}} \right)\)
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