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Chapter 6: Q22E (page 331)

Let \({\bf{u}} = \left( {{u_1},{u_2},{u_3}} \right)\). Explain why \({\bf{u}} \cdot {\bf{u}} \ge 0\). When is \({\bf{u}} \cdot {\bf{u}} = 0\)?

Short Answer

Expert verified

The expression \({\bf{u}} \cdot {\bf{u}}\) means the sum of squares of element which will always be greater than zero. Their product will be zero if and only if the elements themselves are zero.

Step by step solution

01

Find the product \({\bf{u}} \cdot {\bf{u}}\) 

The given vector can be written as,\({\bf{u}} = \left( {\begin{aligned}{*{20}{c}}{{u_1}}\\{{u_2}}\\{{u_3}}\end{aligned}} \right)\).

Find \({\bf{u}} \cdot {\bf{u}}\).

\(\begin{aligned}{c}{\bf{u}} \cdot {\bf{u}} &= {\left( {\begin{aligned}{*{20}{c}}{{u_1}}\\{{u_2}}\\{{u_3}}\end{aligned}} \right)^T} \cdot \left( {\begin{aligned}{*{20}{c}}{{u_1}}\\{{u_2}}\\{{u_3}}\end{aligned}} \right)\\ &= u_1^2 + u_2^2 + u_3^2\end{aligned}\)

02

 Verification of statement

Since, the product \({\bf{u}} \cdot {\bf{u}}\) is the sum of squares of element which will always be greater than zero.

Therefore,

\({\bf{u}} \cdot {\bf{u}} \ge 0\)

The equation \({\bf{u}} \cdot {\bf{u}} = 0\) will be valid only when all the elements of the vector will be equal to zero.

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Most popular questions from this chapter

In Exercises 1-4, find a least-sqaures solution of \(A{\bf{x}} = {\bf{b}}\) by (a) constructing a normal equations for \({\bf{\hat x}}\) and (b) solving for \({\bf{\hat x}}\).

1. \(A = \left[ {\begin{aligned}{{}{}}{ - {\bf{1}}}&{\bf{2}}\\{\bf{2}}&{ - {\bf{3}}}\\{ - {\bf{1}}}&{\bf{3}}\end{aligned}} \right]\), \({\bf{b}} = \left[ {\begin{aligned}{{}{}}{\bf{4}}\\{\bf{1}}\\{\bf{2}}\end{aligned}} \right]\)

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

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

[M] Let \({f_{\bf{4}}}\) and \({f_{\bf{5}}}\) be the fourth-order and fifth order Fourier approximations in \(C\left[ {{\bf{0}},{\bf{2}}\pi } \right]\) to the square wave function in Exercise 10. Produce separate graphs of \({f_{\bf{4}}}\) and \({f_{\bf{5}}}\) on the interval \(\left[ {{\bf{0}},{\bf{2}}\pi } \right]\), and produce graph of \({f_{\bf{5}}}\) on \(\left[ { - {\bf{2}}\pi ,{\bf{2}}\pi } \right]\).

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.

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

Let \(X\) be the design matrix in Example 2 corresponding to a least-square fit of parabola to data \(\left( {{x_1},{y_1}} \right), \ldots ,\left( {{x_n},{y_n}} \right)\). Suppose \({x_1}\), \({x_2}\) and \({x_3}\) are distinct. Explain why there is only one parabola that best, in a least-square sense. (See Exercise 5.)
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