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 9E from Chapter 6.4 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 9E
Chapter:
Problem:
Find an orthogonal basis for the column space of each matrix in Exercises 9–12.
Step-by-Step Solution
Given Information
We are given with a matrix A. \[A = \left[ {\begin{array}{*{20}{r}}3&{ - 5}&1\\1&1&1\\{ - 1}&5&{ - 2}\\3&{ - 7}&8\end{array}} \right]\]We have to find the orthogonal basis for the column space of A. Let {$v_1,v_2,v_3$} be an orthogonal basis for the column space W of A.
Step-1: First vector
The first vector is given by the first column of matrix A\[{v_1} = \left[ {\begin{array}{*{20}{c}}3\\1\\{ - 1}\\3\end{array}} \right]\]
Step-2: Second vector
The second vector an be obtained as: \[\begin{array}{l}{{\bf{v}}_{\bf{2}}} = {{\bf{x}}_{\bf{2}}} - {\rm{pro}}{{\rm{j}}_{{w_2}}}{{\bf{x}}_{\bf{2}}}\\ = {{\bf{x}}_{\bf{2}}} - \dfrac{{{{\bf{x}}_{\bf{2}}}{\bf{.}}{{\bf{v}}_{\bf{1}}}}}{{{{\bf{v}}_{\bf{1}}}{\bf{.}}{{\bf{v}}_{\bf{1}}}}}{{\bf{v}}_{\bf{1}}}\\ = {{\bf{x}}_{\bf{2}}} - \dfrac{{\left( {{{\bf{x}}_{\bf{2}}}{\bf{.}}{{\bf{x}}_{\bf{1}}}} \right)}}{{{{\bf{v}}_{\bf{1}}}{\bf{.}}{{\bf{v}}_{\bf{1}}}}}{{\bf{v}}_{\bf{1}}}\\ = {{\bf{x}}_{\bf{2}}} - \dfrac{{\left( { - 5 \times 3 + 1 \times 1 - 1 \times 5 + 3 \times \left( { - 7} \right)} \right)}}{{{3^2} + {1^2} + {1^2} + {3^2}}}{{\bf{v}}_{\bf{1}}}\\ = {{\bf{x}}_{\bf{2}}} - \dfrac{{\left( { - 40} \right)}}{{20}}{{\bf{v}}_{\bf{1}}}\\ = {{\bf{x}}_{\bf{2}}} + {2_{\bf{1}}}{{\bf{v}}_{\bf{1}}}\\ = \left[ {\begin{array}{*{20}{c}}{ - 5}\\1\\5\\{ - 7}\end{array}} \right] + {2_{\bf{1}}}\left[ {\begin{array}{*{20}{c}}3\\1\\{ - 1}\\3\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}1\\3\\3\\{ - 1}\end{array}} \right]\end{array}\]
Step-3: Third vector
\[\begin{array}{l}{{\bf{v}}_3} = {{\bf{x}}_3} - {\rm{pro}}{{\rm{j}}_{{w_1}}}{{\bf{x}}_3} - {\rm{pro}}{{\rm{j}}_{{w_2}}}{{\bf{x}}_3}\\ = {{\bf{x}}_3} - \dfrac{{{{\bf{x}}_3}{\bf{.}}{{\bf{v}}_{\bf{1}}}}}{{{{\bf{v}}_{\bf{1}}}{\bf{.}}{{\bf{v}}_{\bf{1}}}}}{{\bf{v}}_{\bf{1}}} - \dfrac{{{{\bf{x}}_3}{\bf{.}}{{\bf{v}}_2}}}{{{{\bf{v}}_2}{\bf{.}}{{\bf{v}}_2}}}{{\bf{v}}_2}\\ = {{\bf{x}}_3} - \dfrac{{\left( {3 + 1 + 2 + 24} \right)}}{{{3^2} + {1^2} + {1^2} + {3^2}}}{{\bf{v}}_{\bf{1}}} - \dfrac{{\left( {1 + 3 - 6 - 8} \right)}}{{1 + 9 + 9 + 1}}{{\bf{v}}_2}\\ = {{\bf{x}}_3} - \dfrac{3}{2}{{\bf{v}}_{\bf{1}}} - \left( { - \dfrac{1}{2}} \right){{\bf{v}}_2}\\ = {{\bf{x}}_3} - \dfrac{3}{2}{{\bf{v}}_{\bf{1}}} + \dfrac{1}{2}{{\bf{v}}_2}\\ = \left[ {\begin{array}{*{20}{c}}1\\1\\{ - 2}\\8\end{array}} \right] - \dfrac{3}{2}\left[ {\begin{array}{*{20}{c}}3\\1\\{ - 1}\\3\end{array}} \right] + \dfrac{1}{2}\left[ {\begin{array}{*{20}{c}}1\\3\\3\\{ - 1}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{ - 3}\\1\\1\\3\end{array}} \right]\end{array}\]Therefore the orthogonal basis is:
\[\beta = \left\{ {\left[ {\begin{array}{*{20}{c}}3\\1\\{ - 1}\\3\end{array}} \right],\left[ {\begin{array}{*{20}{c}}1\\3\\3\\{ - 1}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{ - 3}\\1\\1\\3\end{array}} \right]} \right\}\]
We are given with a matrix A. \[A = \left[ {\begin{array}{*{20}{r}}3&{ - 5}&1\\1&1&1\\{ - 1}&5&{ - 2}\\3&{ - 7}&8\end{array}} \right]\]We have to find the orthogonal basis for the column space of A. Let {$v_1,v_2,v_3$} be an orthogonal basis for the column space W of A.
Step-1: First vector
The first vector is given by the first column of matrix A\[{v_1} = \left[ {\begin{array}{*{20}{c}}3\\1\\{ - 1}\\3\end{array}} \right]\]
Step-2: Second vector
The second vector an be obtained as: \[\begin{array}{l}{{\bf{v}}_{\bf{2}}} = {{\bf{x}}_{\bf{2}}} - {\rm{pro}}{{\rm{j}}_{{w_2}}}{{\bf{x}}_{\bf{2}}}\\ = {{\bf{x}}_{\bf{2}}} - \dfrac{{{{\bf{x}}_{\bf{2}}}{\bf{.}}{{\bf{v}}_{\bf{1}}}}}{{{{\bf{v}}_{\bf{1}}}{\bf{.}}{{\bf{v}}_{\bf{1}}}}}{{\bf{v}}_{\bf{1}}}\\ = {{\bf{x}}_{\bf{2}}} - \dfrac{{\left( {{{\bf{x}}_{\bf{2}}}{\bf{.}}{{\bf{x}}_{\bf{1}}}} \right)}}{{{{\bf{v}}_{\bf{1}}}{\bf{.}}{{\bf{v}}_{\bf{1}}}}}{{\bf{v}}_{\bf{1}}}\\ = {{\bf{x}}_{\bf{2}}} - \dfrac{{\left( { - 5 \times 3 + 1 \times 1 - 1 \times 5 + 3 \times \left( { - 7} \right)} \right)}}{{{3^2} + {1^2} + {1^2} + {3^2}}}{{\bf{v}}_{\bf{1}}}\\ = {{\bf{x}}_{\bf{2}}} - \dfrac{{\left( { - 40} \right)}}{{20}}{{\bf{v}}_{\bf{1}}}\\ = {{\bf{x}}_{\bf{2}}} + {2_{\bf{1}}}{{\bf{v}}_{\bf{1}}}\\ = \left[ {\begin{array}{*{20}{c}}{ - 5}\\1\\5\\{ - 7}\end{array}} \right] + {2_{\bf{1}}}\left[ {\begin{array}{*{20}{c}}3\\1\\{ - 1}\\3\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}1\\3\\3\\{ - 1}\end{array}} \right]\end{array}\]
Step-3: Third vector
\[\begin{array}{l}{{\bf{v}}_3} = {{\bf{x}}_3} - {\rm{pro}}{{\rm{j}}_{{w_1}}}{{\bf{x}}_3} - {\rm{pro}}{{\rm{j}}_{{w_2}}}{{\bf{x}}_3}\\ = {{\bf{x}}_3} - \dfrac{{{{\bf{x}}_3}{\bf{.}}{{\bf{v}}_{\bf{1}}}}}{{{{\bf{v}}_{\bf{1}}}{\bf{.}}{{\bf{v}}_{\bf{1}}}}}{{\bf{v}}_{\bf{1}}} - \dfrac{{{{\bf{x}}_3}{\bf{.}}{{\bf{v}}_2}}}{{{{\bf{v}}_2}{\bf{.}}{{\bf{v}}_2}}}{{\bf{v}}_2}\\ = {{\bf{x}}_3} - \dfrac{{\left( {3 + 1 + 2 + 24} \right)}}{{{3^2} + {1^2} + {1^2} + {3^2}}}{{\bf{v}}_{\bf{1}}} - \dfrac{{\left( {1 + 3 - 6 - 8} \right)}}{{1 + 9 + 9 + 1}}{{\bf{v}}_2}\\ = {{\bf{x}}_3} - \dfrac{3}{2}{{\bf{v}}_{\bf{1}}} - \left( { - \dfrac{1}{2}} \right){{\bf{v}}_2}\\ = {{\bf{x}}_3} - \dfrac{3}{2}{{\bf{v}}_{\bf{1}}} + \dfrac{1}{2}{{\bf{v}}_2}\\ = \left[ {\begin{array}{*{20}{c}}1\\1\\{ - 2}\\8\end{array}} \right] - \dfrac{3}{2}\left[ {\begin{array}{*{20}{c}}3\\1\\{ - 1}\\3\end{array}} \right] + \dfrac{1}{2}\left[ {\begin{array}{*{20}{c}}1\\3\\3\\{ - 1}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{ - 3}\\1\\1\\3\end{array}} \right]\end{array}\]Therefore the orthogonal basis is:
\[\beta = \left\{ {\left[ {\begin{array}{*{20}{c}}3\\1\\{ - 1}\\3\end{array}} \right],\left[ {\begin{array}{*{20}{c}}1\\3\\3\\{ - 1}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{ - 3}\\1\\1\\3\end{array}} \right]} \right\}\]