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Chapter 6: Q20E (page 331)

Let \({\rm{u}} = \left( \begin{array}{l}a\\b\end{array} \right)\)and\({\rm{v}} = \left( \begin{array}{l}1\\1\end{array} \right)\). Use the Cauchy–Schwarz inequality to show that \({\left( {\frac{{a + b}}{2}} \right)^2} \le \frac{{{a^2} + {b^2}}}{2}\).

Short Answer

Expert verified

The inequality \({\left( {\frac{{a + b}}{2}} \right)^2} \le \frac{{{a^2} + {b^2}}}{2}\) is proved to be true.

Step by step solution

01

Use the given information

It is given that \({\rm{u}} = \left( \begin{array}{l}a\\b\end{array} \right)\) and \({\rm{v}} = \left( \begin{array}{l}1\\1\end{array} \right)\). So, according to inner product axioms, \({\left\| {\rm{u}} \right\|^2} = {a^2} + {b^2}\) and \({\left\| {\rm{v}} \right\|^2} = 2\)and\(\left\langle {{\rm{u,v}}} \right\rangle = a + b\).

02

Modify the Cauchy Schwarz inequality

According to Cauchy Schwarz inequality, an inner product on a vector space \(V\) is a function that, to each pair of vectors \({\bf{u}}\) and \({\rm{v}}\) in \(V\), associates a real number \(\left\langle {{\bf{u}},{\rm{v}}} \right\rangle \)and satisfies the axiom, \(\left| {\left\langle {{\bf{u}},{\rm{v}}} \right\rangle } \right| \le \left\| {\rm{u}} \right\|\left\| {\rm{v}} \right\|\) for all \({\bf{u}}\) and \({\rm{v}}\)in \(V\).

Divide both sides of Cauchy Schwarz inequality by 2 and thence squaring both sides yieldthe following result:

\(\begin{array}{c}{\left\langle {{\rm{u,v}}} \right\rangle ^2} \le {\left\| {\rm{u}} \right\|^2}{\left\| {\rm{v}} \right\|^2}\\\frac{{{{\left\langle {{\rm{u,v}}} \right\rangle }^2}}}{4} \le \frac{{{{\left\| {\rm{u}} \right\|}^2}{{\left\| {\rm{v}} \right\|}^2}}}{4}\\{\left( {\frac{{\left\langle {{\rm{u,v}}} \right\rangle }}{2}} \right)^2} \le \frac{{{{\left\| {\rm{u}} \right\|}^2}{{\left\| {\rm{v}} \right\|}^2}}}{4}\end{array}\)

03

Use Cauchy Schwarz inequality

Plug the above obtained values into theinequality \({\left( {\frac{{\left\langle {{\rm{u,v}}} \right\rangle }}{2}} \right)^2} \le \frac{{{{\left\| {\rm{u}} \right\|}^2}{{\left\| {\rm{v}} \right\|}^2}}}{4}\), as follows:

\(\begin{array}{c}{\left( {\frac{{\left\langle {{\rm{u,v}}} \right\rangle }}{2}} \right)^2} \le \frac{{{{\left\| {\rm{u}} \right\|}^2}{{\left\| {\rm{v}} \right\|}^2}}}{4}\\{\left( {\frac{{a + b}}{2}} \right)^2} \le \frac{{\left( {{a^2} + {b^2}} \right)\left( 2 \right)}}{4}\\{\left( {\frac{{a + b}}{2}} \right)^2} \le \frac{{{a^2} + {b^2}}}{2}\end{array}\)

Thus, the inequality\({\left( {\frac{{a + b}}{2}} \right)^2} \le \frac{{{a^2} + {b^2}}}{2}\)is proved to be true.

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Most popular questions from this chapter

In Exercises 5 and 6, describe all least squares solutions of the equation \(A{\bf{x}} = {\bf{b}}\).

6.\(A = \left( {\begin{aligned}{{}{}}{\bf{1}}&{\bf{1}}&{\bf{0}}\\{\bf{1}}&{\bf{1}}&{\bf{0}}\\{\bf{1}}&{\bf{1}}&{\bf{0}}\\{\bf{1}}&{\bf{0}}&{\bf{1}}\\{\bf{1}}&{\bf{0}}&{\bf{1}}\\{\bf{1}}&{\bf{0}}&{\bf{1}}\end{aligned}} \right)\),\({\bf{b}} = \left( {\begin{aligned}{{}{}}{\bf{7}}\\{\bf{2}}\\{\bf{3}}\\{\bf{6}}\\{\bf{5}}\\{\bf{4}}\end{aligned}} \right)\)

Question: In Exercises 3-6, verify that\(\left\{ {{{\bf{u}}_{\bf{1}}},{{\bf{u}}_{\bf{2}}}} \right\}\)is an orthogonal set, and then find the orthogonal projection of y onto\({\bf{Span}}\left\{ {{{\bf{u}}_{\bf{1}}},{{\bf{u}}_{\bf{2}}}} \right\}\).

4.\(y = \left[ {\begin{aligned}{\bf{6}}\\{\bf{3}}\\{ - {\bf{2}}}\end{aligned}} \right]\),\({{\bf{u}}_{\bf{1}}} = \left[ {\begin{aligned}{\bf{3}}\\{\bf{4}}\\{\bf{0}}\end{aligned}} \right]\),\({{\bf{u}}_{\bf{2}}} = \left[ {\begin{aligned}{ - {\bf{4}}}\\{\bf{3}}\\{\bf{0}}\end{aligned}} \right]\)

For a matrix program, the Gram–Schmidt process worksbetter with orthonormal vectors. Starting with \({x_1},......,{x_p}\) asin Theorem 11, let \(A = \left\{ {{x_1},......,{x_p}} \right\}\) . Suppose \(Q\) is an\(n \times k\)matrix whose columns form an orthonormal basis for

the subspace \({W_k}\) spanned by the first \(k\) columns of A. Thenfor \(x\) in \({\mathbb{R}^n}\), \(Q{Q^T}x\) is the orthogonal projection of x onto \({W_k}\) (Theorem 10 in Section 6.3). If \({x_{k + 1}}\) is the next column of \(A\),then equation (2) in the proof of Theorem 11 becomes

\({v_{k + 1}} = {x_{k + 1}} - Q\left( {{Q^T}T {x_{k + 1}}} \right)\)

(The parentheses above reduce the number of arithmeticoperations.) Let \({u_{k + 1}} = \frac{{{v_{k + 1}}}}{{\left\| {{v_{k + 1}}} \right\|}}\). The new \(Q\) for thenext step is \(\left( {\begin{aligned}{{}{}}Q&{{u_{k + 1}}}\end{aligned}} \right)\). Use this procedure to compute the\(QR\)factorization of the matrix in Exercise 24. Write thekeystrokes or commands you use.

Compute the quantities in Exercises 1-8 using the vectors

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

5. \(\left( {\frac{{{\mathop{\rm u}\nolimits} \cdot {\mathop{\rm v}\nolimits} }}{{{\mathop{\rm v}\nolimits} \cdot {\mathop{\rm v}\nolimits} }}} \right){\mathop{\rm v}\nolimits} \)

Given \(A = QR\) as in Theorem 12, describe how to find an orthogonal\(m \times m\)(square) matrix \({Q_1}\) and an invertible \(n \times n\) upper triangular matrix \(R\) such that

\(A = {Q_1}\left[ {\begin{aligned}{{}{}}R\\0\end{aligned}} \right]\)

The MATLAB qr command supplies this “full” QR factorization

when rank \(A = n\).
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