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Chapter 6: Q16E (page 331)

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

16. If \(\left\{ {{\rm{u,}}\,{\rm{v}}} \right\}\) is an orthonormal set in \(V\), then \(\left\| {{\rm{u}} - {\rm{v}}} \right\| = \sqrt 2 \).

Short Answer

Expert verified

The statement \(\left\| {{\rm{u}} - {\rm{v}}} \right\| = \sqrt 2 \) is verified.

Step by step solution

01

Apply axiom 2

According to axiom 2, if \({\bf{u}}\) and \({\rm{v}}\) be pair of vectors in a vector space\(V\), then, the inner product on \(V\), relates a real number \(\left\langle {{\bf{u}},{\rm{v}}} \right\rangle \)and satisfies the axiom, \(\left\langle {{\bf{u}} + {\rm{v,}}\,{\rm{w}}} \right\rangle = \left\langle {{\rm{u,w}}} \right\rangle + \left\langle {{\rm{v,w}}} \right\rangle \) for all \({\bf{u}}\), \({\rm{v}}\),\({\rm{w}}\)and scalars \(c\).

Apply axiom 2 to the left side of the given equation, as follows:

\(\begin{aligned}{}{\left\| {{\rm{u}} - {\rm{v}}} \right\|^2} &= \left\langle {{\rm{u}} - {\rm{v,}}\,{\rm{u}} - {\rm{v}}} \right\rangle \\ &= \left\langle {{\rm{u,}}\,{\rm{u}} - {\rm{v}}} \right\rangle - \left\langle {{\rm{v,}}\,{\rm{u}} - {\rm{v}}} \right\rangle \\ &= \left\langle {{\rm{u,}}\,{\rm{u}}} \right\rangle - \left\langle {{\rm{u,v}}} \right\rangle - \left\langle {{\rm{v,}}\,{\rm{u}}} \right\rangle ,\left\langle {{\rm{v,}}\,{\rm{v}}} \right\rangle \end{aligned}\)

02

Apply axiom 1

According to axiom 1, if \({\bf{u}}\) and \({\rm{v}}\) be pair of vectors in a vector space \(V\), then, the inner product on \(V\), relates a real number \(\left\langle {{\bf{u}},{\rm{v}}} \right\rangle \)and satisfies the axiom, \(\left\langle {{\bf{u}},{\rm{v}}} \right\rangle = \left\langle {{\rm{v}},{\rm{u}}} \right\rangle \)for all \({\bf{u}}\), \({\rm{v}}\), \({\rm{w}}\)and scalars \(c\).

Apply axiom 1 to the left side of the given equation, as follows:

\(\begin{aligned}{}{\left\| {{\rm{u}} - {\rm{v}}} \right\|^2} &= \left\langle {{\rm{u,}}\,{\rm{u}}} \right\rangle - \left\langle {{\rm{u,v}}} \right\rangle - \left\langle {{\rm{u,v}}} \right\rangle + \left\langle {{\rm{v,}}\,{\rm{v}}} \right\rangle \\ &= \left\langle {{\rm{u,}}\,{\rm{u}}} \right\rangle - {\rm{2}}\left\langle {{\rm{u,v}}} \right\rangle + \left\langle {{\rm{v,}}\,{\rm{v}}} \right\rangle \end{aligned}\)

03

Use the given information

It is given that \(\left\{ {{\bf{u}},{\rm{v}}} \right\}\) is orthonormal. So, \(\left\langle {{\bf{u}},{\rm{v}}} \right\rangle = 0\)and \({\left\| {\bf{u}} \right\|^2} = {\left\| {\rm{v}} \right\|^2} = 1\).

Apply the obtained results into the resulting equation, as follows:

\(\begin{aligned}{}{\left\| {{\rm{u}} - {\rm{v}}} \right\|^2} &= {\left\| {\rm{u}} \right\|^2} - {\rm{2}}\left\langle {{\rm{u,v}}} \right\rangle + {\left\| {\rm{v}} \right\|^2}\\ &= 1 - 2\left( 0 \right) + 1\\ &= 2\end{aligned}\)

So,\(\left\| {{\rm{u}} - {\rm{v}}} \right\| = \sqrt 2 \)

Thus, the statement\(\left\| {{\rm{u}} - {\rm{v}}} \right\| = \sqrt 2 \) is verified.

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Most popular questions from this chapter

In Exercises 3鈥�6, verify that\[\left\{ {{{\bf{u}}_1},{{\bf{u}}_2}} \right\}\]is an orthogonal set, and then find the orthogonal projection of\[{\bf{y}}\]onto Span\[\left\{ {{{\bf{u}}_1},{{\bf{u}}_2}} \right\}\].

5.\[y = \left[ {\begin{aligned}{ - 1}\\2\\6\end{aligned}} \right]\],\[{{\bf{u}}_1} = \left[ {\begin{aligned}3\\{ - 1}\\2\end{aligned}} \right]\],\[{{\bf{u}}_2} = \left[ {\begin{aligned}1\\{ - 1}\\{ - 2}\end{aligned}} \right]\]

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\[W\].

10.\[y = \left[ {\begin{aligned}3\\4\\5\\6\end{aligned}} \right]\],\[{{\bf{u}}_1} = \left[ {\begin{aligned}1\\1\\0\\{ - 1}\end{aligned}} \right]\],\[{{\bf{u}}_2} = \left[ {\begin{aligned}1\\0\\1\\1\end{aligned}} \right]\],\[{{\bf{u}}_3} = \left[ {\begin{aligned}0\\{ - 1}\\1\\{ - 1}\end{aligned}} \right]\]

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

\(\left[ {\begin{array}{*{20}{c}}5\\{ - 4}\\0\\3\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}{ - 4}\\1\\{ - 3}\\8\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}3\\3\\5\\{ - 1}\end{array}} \right]\)

Find an orthonormal basis of the subspace spanned by the vectors in Exercise 3.

Given data for a least-squares problem, \(\left( {{x_1},{y_1}} \right), \ldots ,\left( {{x_n},{y_n}} \right)\), the following abbreviations are helpful:

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

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

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

Derive the normal equations (7) from the matrix form given in this section.
See all solutions

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