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 2E from Chapter 6.4 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 2E
Chapter:
Problem:
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.
Step-by-Step Solution
Given Information
We are given following basis for a subspace W. \[ \mathbf { x } _ { 1 } = \left[ \begin{array} { l } { 0 } \\ { 4 } \\ { 2 } \end{array} \right] \text { and } \mathbf { x } _ { 2 } = \left[ \begin{array} { c } { 5 } \\ { 6 } \\ { - 7 } \end{array} \right] \] We have to find the orthogonal basis for W.
Step-1:
Let, \[ \mathbf { v } _ { 1 } = \mathbf { x } _ { 1 } \text { and } W _ { 1 } = \operatorname { Span } \left\{ \mathbf { x } _ { 1 } \right\} = \operatorname { Span } \left\{ \mathbf { v } _ { 1 } \right\} \] Also, let $v_2$ is the vector produced by subtracting from $x_2$ its projection onto the subspace $W_1$. Then $v_2$ is given by: \[\begin{array}{l} {{\bf{v}}_2} = {{\bf{x}}_2} - proj{ _{{W_1}}}{{\bf{x}}_2}\\ = {{\bf{x}}_2} - \dfrac{{{{\bf{x}}_2}\cdot{{\bf{v}}_1}}}{{{{\bf{v}}_1}\cdot{{\bf{v}}_1}}}{{\bf{v}}_1} \end{array}\]
Step-2: The required products
Product of $x_2$ and $v_1$ \[ \begin{aligned} \mathbf { x } _ { 2 } \cdot \mathbf { v } _ { 1 } & = \mathbf { x } _ { 2 } ^ { T } \mathbf { v } _ { 1 } \\ & = \left[ \begin{array} { c c c } { 5 } & { 6 } & { - 7 } \end{array} \right] \left[ \begin{array} { c } { 0 } \\ { 4 } \\ { 2 } \end{array} \right] \\ & = ( 5 ) ( 0 ) + ( 6 ) ( 4 ) + ( - 7 ) ( 2 ) \\ & = 10 \end{aligned} \] Product of $v_1$ and $v_1$ \[ \begin{aligned} \mathbf { v } _ { 1 } \cdot \mathbf { v } _ { 1 } & = \mathbf { v } _ { 1 } ^ { T } \mathbf { v } _ { 1 } \\ & = \left[ \begin{array} { c c c } { 0 } & { 4 } & { 2 } \end{array} \right] \left[ \begin{array} { l } { 0 } \\ { 4 } \\ { 2 } \end{array} \right] \\ & = ( 0 ) ( 0 ) + ( 4 ) ( 4 ) + ( 2 ) ( 2 ) \\ & = 20 \end{aligned} \] Thus, \[ \dfrac { \mathbf { x } _ { 2 } \cdot \mathbf { v } _ { 1 } } { \mathbf { v } _ { 1 } \cdot \mathbf { v } _ { 1 } } = \dfrac { 10 } { 20 } = \dfrac { 1 } { 2 } \]
Step-3: The orthogonal basis
Find the vector $v_2$: \[\begin{array}{l} {{\bf{v}}_2} = {{\bf{x}}_2} - \dfrac{{{{\bf{x}}_2}\cdot{{\bf{v}}_1}}}{{{{\bf{v}}_1}\cdot{{\bf{v}}_1}}}{{\bf{v}}_1}\\ = \left[ {\begin{array}{*{20}{c}} 5\\ 6\\ { - 7} \end{array}} \right] - \dfrac{1}{2}\left[ {\begin{array}{*{20}{l}} 0\\ 4\\ 2 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 5\\ 6\\ { - 7} \end{array}} \right] - \left[ {\begin{array}{*{20}{l}} 0\\ 2\\ 1 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 5\\ 4\\ { - 8} \end{array}} \right] \end{array}\] Therefore, the orthogonal basis is:
\[ \left\{ \mathbf { v } _ { 1 } , \mathbf { v } _ { 2 } \right\} = \left\{ \left[ \begin{array} { l } { 0 } \\ { 4 } \\ { 2 } \end{array} \right] , \left[ \begin{array} { c } { 5 } \\ { 4 } \\ { - 8 } \end{array} \right] \right. \]
We are given following basis for a subspace W. \[ \mathbf { x } _ { 1 } = \left[ \begin{array} { l } { 0 } \\ { 4 } \\ { 2 } \end{array} \right] \text { and } \mathbf { x } _ { 2 } = \left[ \begin{array} { c } { 5 } \\ { 6 } \\ { - 7 } \end{array} \right] \] We have to find the orthogonal basis for W.
Step-1:
Let, \[ \mathbf { v } _ { 1 } = \mathbf { x } _ { 1 } \text { and } W _ { 1 } = \operatorname { Span } \left\{ \mathbf { x } _ { 1 } \right\} = \operatorname { Span } \left\{ \mathbf { v } _ { 1 } \right\} \] Also, let $v_2$ is the vector produced by subtracting from $x_2$ its projection onto the subspace $W_1$. Then $v_2$ is given by: \[\begin{array}{l} {{\bf{v}}_2} = {{\bf{x}}_2} - proj{ _{{W_1}}}{{\bf{x}}_2}\\ = {{\bf{x}}_2} - \dfrac{{{{\bf{x}}_2}\cdot{{\bf{v}}_1}}}{{{{\bf{v}}_1}\cdot{{\bf{v}}_1}}}{{\bf{v}}_1} \end{array}\]
Step-2: The required products
Product of $x_2$ and $v_1$ \[ \begin{aligned} \mathbf { x } _ { 2 } \cdot \mathbf { v } _ { 1 } & = \mathbf { x } _ { 2 } ^ { T } \mathbf { v } _ { 1 } \\ & = \left[ \begin{array} { c c c } { 5 } & { 6 } & { - 7 } \end{array} \right] \left[ \begin{array} { c } { 0 } \\ { 4 } \\ { 2 } \end{array} \right] \\ & = ( 5 ) ( 0 ) + ( 6 ) ( 4 ) + ( - 7 ) ( 2 ) \\ & = 10 \end{aligned} \] Product of $v_1$ and $v_1$ \[ \begin{aligned} \mathbf { v } _ { 1 } \cdot \mathbf { v } _ { 1 } & = \mathbf { v } _ { 1 } ^ { T } \mathbf { v } _ { 1 } \\ & = \left[ \begin{array} { c c c } { 0 } & { 4 } & { 2 } \end{array} \right] \left[ \begin{array} { l } { 0 } \\ { 4 } \\ { 2 } \end{array} \right] \\ & = ( 0 ) ( 0 ) + ( 4 ) ( 4 ) + ( 2 ) ( 2 ) \\ & = 20 \end{aligned} \] Thus, \[ \dfrac { \mathbf { x } _ { 2 } \cdot \mathbf { v } _ { 1 } } { \mathbf { v } _ { 1 } \cdot \mathbf { v } _ { 1 } } = \dfrac { 10 } { 20 } = \dfrac { 1 } { 2 } \]
Step-3: The orthogonal basis
Find the vector $v_2$: \[\begin{array}{l} {{\bf{v}}_2} = {{\bf{x}}_2} - \dfrac{{{{\bf{x}}_2}\cdot{{\bf{v}}_1}}}{{{{\bf{v}}_1}\cdot{{\bf{v}}_1}}}{{\bf{v}}_1}\\ = \left[ {\begin{array}{*{20}{c}} 5\\ 6\\ { - 7} \end{array}} \right] - \dfrac{1}{2}\left[ {\begin{array}{*{20}{l}} 0\\ 4\\ 2 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 5\\ 6\\ { - 7} \end{array}} \right] - \left[ {\begin{array}{*{20}{l}} 0\\ 2\\ 1 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 5\\ 4\\ { - 8} \end{array}} \right] \end{array}\] Therefore, the orthogonal basis is:
\[ \left\{ \mathbf { v } _ { 1 } , \mathbf { v } _ { 2 } \right\} = \left\{ \left[ \begin{array} { l } { 0 } \\ { 4 } \\ { 2 } \end{array} \right] , \left[ \begin{array} { c } { 5 } \\ { 4 } \\ { - 8 } \end{array} \right] \right. \]