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

Suppose the x-coordinates of the data \(\left( {{x_1},{y_1}} \right), \ldots ,\left( {{x_n},{y_n}} \right)\) are in mean deviation form, so that \(\sum {{x_i}} = 0\). Show that if \(X\) is the design matrix for the least-squares line in this case, then \({X^T}X\) is a diagonal matrix.

Determine which pairs of vectors in Exercises 15-18 are orthogonal.

15. \({\mathop{\rm a}\nolimits} = \left( {\begin{aligned}{*{20}{c}}8\\{ - 5}\end{aligned}} \right),{\rm{ }}{\mathop{\rm b}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 2}\\{ - 3}\end{aligned}} \right)\)

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\[W\].

7.\[y = \left[ {\begin{aligned}1\\3\\5\end{aligned}} \right]\],\[{{\bf{u}}_1} = \left[ {\begin{aligned}1\\3\\{ - 2}\end{aligned}} \right]\],\[{{\bf{u}}_2} = \left[ {\begin{aligned}5\\1\\4\end{aligned}} \right]\]

Compute the quantities in Exercises 1-8 using the vectors

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

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