Linear Algebra and Its Applications, 5th Edition
Authors: David C. Lay, Steven R. Lay, Judi J. McDonald
ISBN-13: 978-0321982384
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See our solution for Question 6E from Chapter 6.3 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 6E
Chapter:
Problem:
In Exercises 3–6, verify that {u1, u2} is an orthogonal set, and then find the orthogonal projection of y onto Span {u1, u2}.
Step-by-Step Solution
Given Information
We are given with three vectors: \[ \mathbf { u } _ { 1 } = \left[ \begin{array} { c } { - 4 } \\ { - 1 } \\ { 1 } \end{array} \right] , \mathbf { u } _ { 2 } = \left[ \begin{array} { l } { 0 } \\ { 1 } \\ { 1 } \end{array} \right] \text { and } \mathbf { y } = \left[ \begin{array} { l } { 6 } \\ { 4 } \\ { 1 } \end{array} \right] \] We have to verify that vectors $u_1$ and $u_2$ are orthogonal and then have to find the projection of y on Span{$u_1$,$u_2$}
Step-1: Orthogonality of $u_1$ and $u_2$
Find the product of these vectors. \[ \begin{aligned} \mathbf { u } _ { 1 } \cdot \mathbf { u } _ { 2 } & = \mathbf { u } _ { 1 } ^ { T } \mathbf { u } _ { 2 } \\ & = \left[ \begin{array} { c c c } { - 4 } & { - 1 } & { 1 } \end{array} \right] \left[ \begin{array} { c } { 0 } \\ { 1 } \\ { 1 } \end{array} \right] \\ & = ( - 4 ) 0 + ( - 1 ) 1 + 1 ( 1 ) \\ & = 0 \end{aligned} \] As, the product of $u_1$ and $u_2$ is zero, they are orthogonal to each other.
Step-2: Orthogonal projection of y on Span{$u_1$,$u_2$}
\[ \hat { \mathbf { y } } = \dfrac { \mathbf { y } \cdot \mathbf { u } _ { 1 } } { \mathbf { u } _ { 1 } \cdot \mathbf { u } _ { 1 } } \mathbf { u } _ { 1 } + \dfrac { \mathbf { y } \cdot \mathbf { u } _ { 2 } } { \mathbf { u } _ { 2 } \cdot \mathbf { u } _ { 2 } } \mathbf { u } _ { 2 } \] Here, \[{\bf{y}}\cdot{{\bf{u}}_1} = 6( - 4) + 4( - 1) + 1(1) = - 27\] \[ \mathbf { y } \cdot \mathbf { u } _ { 2 } = 6 ( 0 ) + 4 ( 1 ) + 1 ( 1 ) = 5 \] \[ \mathbf { u } _ { 1 } \cdot \mathbf { u } _ { 1 } = ( - 4 ) ^ { 2 } + ( - 1 ) ^ { 2 } + 1 ^ { 2 } = 18 \] \[ \mathbf { u } _ { 2 } \cdot \mathbf { u } _ { 2 } = 0 ^ { 2 } + 1 ^ { 2 } + 1 ^ { 2 } = 2 \]
Step-3: The projection of y on Span{$u_1$,$u_2$}
\[\begin{array}{l} \widehat {\bf{y}} = - \dfrac{{27}}{{18}}{{\bf{u}}_1} + \dfrac{5}{2}{{\bf{u}}_2}\\ = \dfrac{{ - 3}}{2}\left[ {\begin{array}{*{20}{c}} { - 4}\\ { - 1}\\ 1 \end{array}} \right] + \dfrac{5}{2}\left[ {\begin{array}{*{20}{l}} 0\\ 1\\ 1 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{l}} 6\\ 4\\ 1 \end{array}} \right] \end{array}\]
\[\widehat y = \left[ {\begin{array}{*{20}{l}} 6\\ 4\\ 1 \end{array}} \right]\]
We are given with three vectors: \[ \mathbf { u } _ { 1 } = \left[ \begin{array} { c } { - 4 } \\ { - 1 } \\ { 1 } \end{array} \right] , \mathbf { u } _ { 2 } = \left[ \begin{array} { l } { 0 } \\ { 1 } \\ { 1 } \end{array} \right] \text { and } \mathbf { y } = \left[ \begin{array} { l } { 6 } \\ { 4 } \\ { 1 } \end{array} \right] \] We have to verify that vectors $u_1$ and $u_2$ are orthogonal and then have to find the projection of y on Span{$u_1$,$u_2$}
Step-1: Orthogonality of $u_1$ and $u_2$
Find the product of these vectors. \[ \begin{aligned} \mathbf { u } _ { 1 } \cdot \mathbf { u } _ { 2 } & = \mathbf { u } _ { 1 } ^ { T } \mathbf { u } _ { 2 } \\ & = \left[ \begin{array} { c c c } { - 4 } & { - 1 } & { 1 } \end{array} \right] \left[ \begin{array} { c } { 0 } \\ { 1 } \\ { 1 } \end{array} \right] \\ & = ( - 4 ) 0 + ( - 1 ) 1 + 1 ( 1 ) \\ & = 0 \end{aligned} \] As, the product of $u_1$ and $u_2$ is zero, they are orthogonal to each other.
Step-2: Orthogonal projection of y on Span{$u_1$,$u_2$}
\[ \hat { \mathbf { y } } = \dfrac { \mathbf { y } \cdot \mathbf { u } _ { 1 } } { \mathbf { u } _ { 1 } \cdot \mathbf { u } _ { 1 } } \mathbf { u } _ { 1 } + \dfrac { \mathbf { y } \cdot \mathbf { u } _ { 2 } } { \mathbf { u } _ { 2 } \cdot \mathbf { u } _ { 2 } } \mathbf { u } _ { 2 } \] Here, \[{\bf{y}}\cdot{{\bf{u}}_1} = 6( - 4) + 4( - 1) + 1(1) = - 27\] \[ \mathbf { y } \cdot \mathbf { u } _ { 2 } = 6 ( 0 ) + 4 ( 1 ) + 1 ( 1 ) = 5 \] \[ \mathbf { u } _ { 1 } \cdot \mathbf { u } _ { 1 } = ( - 4 ) ^ { 2 } + ( - 1 ) ^ { 2 } + 1 ^ { 2 } = 18 \] \[ \mathbf { u } _ { 2 } \cdot \mathbf { u } _ { 2 } = 0 ^ { 2 } + 1 ^ { 2 } + 1 ^ { 2 } = 2 \]
Step-3: The projection of y on Span{$u_1$,$u_2$}
\[\begin{array}{l} \widehat {\bf{y}} = - \dfrac{{27}}{{18}}{{\bf{u}}_1} + \dfrac{5}{2}{{\bf{u}}_2}\\ = \dfrac{{ - 3}}{2}\left[ {\begin{array}{*{20}{c}} { - 4}\\ { - 1}\\ 1 \end{array}} \right] + \dfrac{5}{2}\left[ {\begin{array}{*{20}{l}} 0\\ 1\\ 1 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{l}} 6\\ 4\\ 1 \end{array}} \right] \end{array}\]
\[\widehat y = \left[ {\begin{array}{*{20}{l}} 6\\ 4\\ 1 \end{array}} \right]\]