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6E

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Linear Algebra and its Applications
Found in: Page 331
Linear Algebra and its Applications

Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

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Short Answer

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

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

The projection of \[{\bf{y}}\] onto Span \[\left\{ {{{\bf{u}}_1},{{\bf{u}}_2}} \right\}\] is \[\left[ {\begin{aligned}6\\4\\1\end{aligned}} \right]\].

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Write the definition

Orthogonal set: If \[{{\bf{u}}_i} \cdot {{\bf{u}}_j} = 0\] for \[i \ne j\], then the set of vectors \[\left\{ {{{\bf{u}}_1}, \ldots ,{{\bf{u}}_p}} \right\} \in {\mathbb{R}^n}\] is said to be orthogonal.

Step 2: Check the set of given vectors is orthogonal or not

Given vectors are, \[{{\bf{u}}_1} = \left[ {\begin{aligned}{ - 4}\\{ - 1}\\1\end{aligned}} \right]\] and \[{{\bf{u}}_2} = \left[ {\begin{aligned}0\\1\\1\end{aligned}} \right]\].

Find the product \[{{\bf{u}}_1} \cdot {{\bf{u}}_2}\] as:

\[\begin{aligned}{c}{{\bf{u}}_1} \cdot {{\bf{u}}_2} &= \left[ {\begin{aligned}{ - 4}\\{ - 1}\\1\end{aligned}} \right] \cdot \left[ {\begin{aligned}0\\1\\1\end{aligned}} \right]\\ &= - 4\left( 0 \right) + \left( { - 1} \right)\left( 1 \right) + 1\left( 1 \right)\\ &= 0\end{aligned}\]

As \[{{\bf{u}}_1} \cdot {{\bf{u}}_2} = 0\], so the set of vectors \[\left\{ {{{\bf{u}}_1},{{\bf{u}}_2}} \right\}\] is orthogonal.

Step 3: The Orthogonal Decomposition Theorem

If \[\left\{ {{{\bf{u}}_1}, \ldots ,{{\bf{u}}_p}} \right\}\] is any orthogonal basis of \[w\], then \[{\bf{\hat y}} = \frac{{{\bf{y}} \cdot {{\bf{u}}_1}}}{{{{\bf{u}}_1} \cdot {{\bf{u}}_1}}}{{\bf{u}}_1} + \cdots + \frac{{{\bf{y}} \cdot {{\bf{u}}_p}}}{{{{\bf{u}}_p} \cdot {{\bf{u}}_p}}}{{\bf{u}}_p}\], where \[w\] is a subspace of \[{\mathbb{R}^n}\] and \[{\bf{\hat y}} \in w\], for which each \[{\bf{y}}\] can be written as \[{\bf{y}} = {\bf{\hat y}} + {\bf{z}}\] when \[{\bf{z}} \in {w^ \bot }\].

Step 4: Find the Orthogonal projection of \[{\bf{y}}\] onto Span \[\left\{ {{{\bf{u}}_1},{{\bf{u}}_2}} \right\}\]

Find \[{\bf{y}} \cdot {{\bf{u}}_1}\], \[{\bf{y}} \cdot {{\bf{u}}_2}\], \[{{\bf{u}}_1} \cdot {{\bf{u}}_1}\] and \[{{\bf{u}}_2} \cdot {{\bf{u}}_2}\] first, where \[{\bf{y}} = \left[ {\begin{aligned}6\\4\\1\end{aligned}} \right]\].

\[\begin{aligned}{\bf{y}} \cdot {{\bf{u}}_1} &= \left[ {\begin{aligned}6\\4\\1\end{aligned}} \right] \cdot \left[ {\begin{aligned}{ - 4}\\{ - 1}\\1\end{aligned}} \right]\\ &= \left( 6 \right)\left( { - 4} \right) + 4\left( { - 1} \right) + 1\left( 1 \right)\\ &= - 27\end{aligned}\]

\[\begin{aligned}{\bf{y}} \cdot {{\bf{u}}_2} &= \left[ {\begin{aligned}6\\4\\1\end{aligned}} \right] \cdot \left[ {\begin{aligned}0\\1\\1\end{aligned}} \right]\\ &= \left( 6 \right)0 + 4\left( 1 \right) + 1\left( 1 \right)\\ &= 5\end{aligned}\]

\[\begin{aligned}{{\bf{u}}_1} \cdot {{\bf{u}}_1} &= \left[ {\begin{aligned}{ - 4}\\{ - 1}\\1\end{aligned}} \right] \cdot \left[ {\begin{aligned}{ - 4}\\{ - 1}\\1\end{aligned}} \right]\\ &= - 4\left( { - 4} \right) + \left( { - 1} \right)\left( { - 1} \right) + 1\left( 1 \right)\\ &= 18\end{aligned}\]

\[\begin{aligned}{{\bf{u}}_2} \cdot {{\bf{u}}_2} &= \left[ {\begin{aligned}0\\1\\1\end{aligned}} \right] \cdot \left[ {\begin{aligned}0\\1\\1\end{aligned}} \right]\\ &= 0\left( 0 \right) + \left( 1 \right)\left( 1 \right) + \left( 1 \right)\left( 1 \right)\\ &= 2\end{aligned}\]

Now, find the projection of \[{\bf{y}}\] onto Span \[\left\{ {{{\bf{u}}_1},{{\bf{u}}_2}} \right\}\] by substituting the obtained values into \[{\bf{\hat y}} = \frac{{{\bf{y}} \cdot {{\bf{u}}_1}}}{{{{\bf{u}}_1} \cdot {{\bf{u}}_1}}}{{\bf{u}}_1} + \cdots + \frac{{{\bf{y}} \cdot {{\bf{u}}_p}}}{{{{\bf{u}}_p} \cdot {{\bf{u}}_p}}}{{\bf{u}}_p}\].

\[\begin{aligned}{\bf{\hat y}} &= - \frac{{27}}{{18}}{{\bf{u}}_1} + \frac{5}{2}{{\bf{u}}_2}\\ &= - \frac{3}{2}\left[ {\begin{aligned}{ - 4}\\{ - 1}\\1\end{aligned}} \right] + \frac{5}{2}\left[ {\begin{aligned}0\\1\\1\end{aligned}} \right]\\ &= \left[ {\begin{aligned}6\\4\\1\end{aligned}} \right]\end{aligned}\]

Hence, the projection of \[{\bf{y}}\] onto Span \[\left\{ {{{\bf{u}}_1},{{\bf{u}}_2}} \right\}\] is \[\left[ {\begin{aligned}6\\4\\1\end{aligned}} \right]\].

Most popular questions for Math Textbooks

In Exercises 1-4, find the equation \(y = {\beta _0} + {\beta _1}x\) of the least-square line that best fits the given data points.

  1. \(\left( {1,0} \right),\left( {2,1} \right),\left( {4,2} \right),\left( {5,3} \right)\)

A Householder matrix, or an elementary reflector, has the form \(Q = I - 2{\bf{u}}{{\bf{u}}^T}\) where u is a unit vector. (See Exercise 13 in the Supplementary Exercise for Chapter 2.) Show that Q is an orthogonal matrix. (Elementary reflectors are often used in computer programs to produce a QR factorization of a matrix A. If A has linearly independent columns, then left-multiplication by a sequence of elementary reflectors can produce an upper triangular matrix.)

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.

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

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