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Chapter 6: Q6.2-12E (page 331)

Question 12: Compute the orthogonal projection of \(\left( {\begin{array}{*{20}{c}}1\\{ - 1}\end{array}} \right)\) onto the line through \(\left( {\begin{array}{*{20}{c}}{ - 1}\\3\end{array}} \right)\) and the origin.

Short Answer

Expert verified

The orthogonal projection of \({\mathop{\rm y}\nolimits} \) onto \({\mathop{\rm u}\nolimits} \) is \(\widehat {\mathop{\rm y}\nolimits} = \left( {\begin{array}{*{20}{c}}{\frac{2}{5}}\\{\frac{{ - 6}}{5}}\end{array}} \right)\).

Step by step solution

01

Definition of orthogonal projection

The equation \({\mathop{\rm y}\nolimits} = \widehat {\mathop{\rm y}\nolimits} + {\mathop{\rm z}\nolimits} \) is satisfied with z orthogonal to u such that if \(\alpha = \frac{{{\mathop{\rm y}\nolimits} \cdot {\mathop{\rm u}\nolimits} }}{{{\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} }}\), and \(\widehat {\mathop{\rm y}\nolimits} = \frac{{{\mathop{\rm y}\nolimits} \cdot {\mathop{\rm u}\nolimits} }}{{{\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} }}{\mathop{\rm u}\nolimits} \).

The vector \(\widehat {\mathop{\rm y}\nolimits} \) is known as theorthogonal projection of y onto uand, the vector z is known as thecomponentof y orthogonal to u. Typically, \(\widehat {\mathop{\rm y}\nolimits} \) is represented by \({{\mathop{\rm proj}\nolimits} _L}{\mathop{\rm y}\nolimits} \) and is called the orthogonal projection of y onto L.

\(\widehat {\mathop{\rm y}\nolimits} = {{\mathop{\rm proj}\nolimits} _L}{\mathop{\rm y}\nolimits} = \frac{{{\mathop{\rm y}\nolimits} \cdot {\mathop{\rm u}\nolimits} }}{{{\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} }}{\mathop{\rm u}\nolimits} \)

02

Compute the orthogonal projection of y onto u

Consider \({\mathop{\rm y}\nolimits} = \left( {\begin{array}{*{20}{c}}1\\{ - 1}\end{array}} \right)\) and \({\mathop{\rm u}\nolimits} = \left( {\begin{array}{*{20}{c}}{ - 1}\\3\end{array}} \right)\).

Compute \({\mathop{\rm y}\nolimits} \cdot {\mathop{\rm u}\nolimits} \) and \({\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} \) as shown below:

\(\begin{array}{c}{\mathop{\rm u}\nolimits} \cdot {\mathop{\rm y}\nolimits} = \left( {\begin{array}{*{20}{c}}{ - 1}\\3\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}1\\{ - 1}\end{array}} \right)\\ = - 1 - 3\\ = - 4\\{\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} = \left( {\begin{array}{*{20}{c}}{ - 1}\\3\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}{ - 1}\\3\end{array}} \right)\\ = 1 + 9\\ = 10\end{array}\)

It is observed that the orthogonal projection of \({\mathop{\rm y}\nolimits} = \left( {\begin{array}{*{20}{c}}1\\{ - 1}\end{array}} \right)\) onto the line through \({\mathop{\rm u}\nolimits} = \left( {\begin{array}{*{20}{c}}{ - 1}\\3\end{array}} \right)\) and the origin is the orthogonal projection of \({\mathop{\rm y}\nolimits} \) onto \({\mathop{\rm u}\nolimits} \).

Obtain the orthogonal projection of \({\mathop{\rm y}\nolimits} \) onto \({\mathop{\rm u}\nolimits} \) as shown below:

\(\begin{array}{c}\widehat {\mathop{\rm y}\nolimits} = \frac{{{\mathop{\rm y}\nolimits} \cdot {\mathop{\rm u}\nolimits} }}{{{\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} }}{\mathop{\rm u}\nolimits} \\ = \frac{{ - 4}}{{10}}\left( {\begin{array}{*{20}{c}}{ - 1}\\3\end{array}} \right)\\ = \frac{{ - 2}}{5}\left( {\begin{array}{*{20}{c}}{ - 1}\\3\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{\frac{2}{5}}\\{\frac{{ - 6}}{5}}\end{array}} \right)\end{array}\)

Thus, the orthogonal projection of \({\mathop{\rm y}\nolimits} \) onto \({\mathop{\rm u}\nolimits} \) is \(\widehat {\mathop{\rm y}\nolimits} = \left( {\begin{array}{*{20}{c}}{\frac{2}{5}}\\{\frac{{ - 6}}{5}}\end{array}} \right)\).

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Most popular questions from this chapter

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.)

Compute the quantities in Exercises 1-8 using the vectors

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

6. \(\left( {\frac{{{\mathop{\rm x}\nolimits} \cdot {\mathop{\rm w}\nolimits} }}{{{\mathop{\rm x}\nolimits} \cdot {\mathop{\rm x}\nolimits} }}} \right){\mathop{\rm x}\nolimits} \)

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 \({\bf{\hat x}}\).

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

Question: In Exercises 1 and 2, you may assume that\(\left\{ {{{\bf{u}}_{\bf{1}}},...,{{\bf{u}}_{\bf{4}}}} \right\}\)is an orthogonal basis for\({\mathbb{R}^{\bf{4}}}\).

2.\({{\bf{u}}_{\bf{1}}} = \left[ {\begin{aligned}{\bf{1}}\\{\bf{2}}\\{\bf{1}}\\{\bf{1}}\end{aligned}} \right]\),\({{\bf{u}}_{\bf{2}}} = \left[ {\begin{aligned}{ - {\bf{2}}}\\{\bf{1}}\\{ - {\bf{1}}}\\{\bf{1}}\end{aligned}} \right]\),\({{\bf{u}}_{\bf{3}}} = \left[ {\begin{aligned}{\bf{1}}\\{\bf{1}}\\{ - {\bf{2}}}\\{ - {\bf{1}}}\end{aligned}} \right]\),\({{\bf{u}}_{\bf{4}}} = \left[ {\begin{aligned}{ - {\bf{1}}}\\{\bf{1}}\\{\bf{1}}\\{ - {\bf{2}}}\end{aligned}} \right]\),\({\bf{x}} = \left[ {\begin{aligned}{\bf{4}}\\{\bf{5}}\\{ - {\bf{3}}}\\{\bf{3}}\end{aligned}} \right]\)

Write v as the sum of two vectors, one in\({\bf{Span}}\left\{ {{{\bf{u}}_1}} \right\}\)and the other in\({\bf{Span}}\left\{ {{{\bf{u}}_2},{{\bf{u}}_3},{{\bf{u}}_{\bf{4}}}} \right\}\).

Let \(T:{\mathbb{R}^n} \to {\mathbb{R}^n}\) be a linear transformation that preserves lengths; that is, \(\left\| {T\left( {\bf{x}} \right)} \right\| = \left\| {\bf{x}} \right\|\) for all x in \({\mathbb{R}^n}\).

  1. Show that T also preserves orthogonality; that is, \(T\left( {\bf{x}} \right) \cdot T\left( {\bf{y}} \right) = 0\) whenever \({\bf{x}} \cdot {\bf{y}} = 0\).
  2. Show that the standard matrix of T is an orthogonal matrix.
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