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 19E from Chapter 4.7 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 19E
Chapter:
Problem:
a. Find a basis for such that P is the change-of-coordinates matrix from to the basis . [Hint: What do the columns of represent?] b. Find a basis such that P is the change-of- coordinates matrix from
Step-by-Step Solution
Given Information
We are given with following matrix and vectors \[\begin{array}{l} P = \left[ {\begin{array}{*{20}{c}} 1&2&{ - 1}\\ { - 3}&{ - 5}&0\\ 4&6&1 \end{array}} \right]\\ {{\bf{v}}_1} = \left[ {\begin{array}{*{20}{c}} { - 2}\\ 2\\ 3 \end{array}} \right],\,\,{{\bf{v}}_2} = \left[ {\begin{array}{*{20}{c}} { - 8}\\ 5\\ 2 \end{array}} \right],\,\,{{\bf{v}}_3} = \left[ {\begin{array}{*{20}{c}} { - 7}\\ 2\\ 6 \end{array}} \right] \end{array}\] (a) We have to find the basis $\left\{ \mathbf { u } _ { 1 } , \mathbf { u } _ { 2 } , \mathbf { u } _ { 3 } \right\}$ for $R^3$ such that $P$ is the change of coordinate matrix from $\left\{ \mathbf { u } _ { 1 } , \mathbf { u } _ { 2 } , \mathbf { u } _ { 3 } \right\}$ to the basis $\left\{ \mathbf { v } _ { 1 } , \mathbf { v } _ { 2 } , \mathbf { v } _ { 3 } \right\}$
(b) We also have to find the basis $\left\{ \mathbf { w } _ { 1 } , \mathbf { w } _ { 2 } , \mathbf { w } _ { 3 } \right\}$ for $R^3$such that $P$ is the change of coordinate matrix from $\left\{ \mathbf { v } _ { 1 } , \mathbf { v } _ { 2 } , \mathbf { v } _ { 3 } \right\}$ to the basis $\left\{ \mathbf { w } _ { 1 } , \mathbf { w } _ { 2 } , \mathbf { w } _ { 3 } \right\} .$
Step-1: (a)
Using the definition of C-coordinate vectors: \[\begin{array}{l} \left[ {\begin{array}{*{20}{l}} {{{\bf{u}}_1}}&{{{\bf{u}}_2}}&{{{\bf{u}}_3}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {{{\bf{v}}_1}}&{{{\bf{v}}_2}}&{{{\bf{v}}_3}} \end{array}} \right]\left[ {{{\left[ {{{\bf{u}}_1}} \right]}_c}\quad {{\left[ {{{\bf{u}}_2}} \right]}_c}\quad {{\left[ {{{\bf{u}}_2}} \right]}_c}} \right]\\ = \left[ {\begin{array}{*{20}{l}} {{{\bf{v}}_1}}&{{{\bf{v}}_2}}&{{{\bf{v}}_3}} \end{array}} \right]P\\ = \left[ {\begin{array}{*{20}{c}} { - 2}&{ - 8}&{ - 7}\\ 2&5&2\\ 3&2&6 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&2&{ - 1}\\ { - 3}&{ - 5}&0\\ 4&6&1 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} { - 6}&{ - 6}&{ - 5}\\ { - 5}&{ - 9}&0\\ {21}&{32}&3 \end{array}} \right] \end{array}\] The basis is given by the columns:
\[ \left\{ \mathbf { u } _ { 1 } = \left[ \begin{array} { c } { - 6 } \\ { - 5 } \\ { 21 } \end{array} \right] , \mathbf { u } _ { 2 } = \left[ \begin{array} { c } { - 6 } \\ { - 9 } \\ { 32 } \end{array} \right] , \mathbf { u } _ { 3 } = \left[ \begin{array} { c } { - 5 } \\ { 0 } \\ { 3 } \end{array} \right] \right\} \]
Step-2: (b)
Using the definition of C-coordinate vectors: \[\begin{array}{l} \left[ {\begin{array}{*{20}{l}} {{{\bf{w}}_1}}&{{{\bf{w}}_2}}&{{{\bf{w}}_3}} \end{array}} \right]\left[ {{{\left[ {{{\bf{v}}_1}} \right]}_c}\,\,{{\left[ {{{\bf{v}}_2}} \right]}_c}\,\,{{\left[ {{{\bf{v}}_3}} \right]}_c}} \right] = \left[ {\begin{array}{*{20}{l}} {{{\bf{v}}_1}}&{{{\bf{v}}_2}}&{{{\bf{v}}_3}} \end{array}} \right]\\ \left[ {\begin{array}{*{20}{l}} {{{\bf{w}}_1}}&{{{\bf{w}}_2}}&{{{\bf{w}}_3}} \end{array}} \right]P = \left[ {\begin{array}{*{20}{l}} {{{\bf{v}}_1}}&{{{\bf{v}}_2}}&{{{\bf{v}}_3}} \end{array}} \right]\\ \left[ {\begin{array}{*{20}{l}} {{{\bf{w}}_1}}&{{{\bf{w}}_2}}&{{{\bf{w}}_3}} \end{array}} \right]P{P^{ - 1}} = \left[ {\begin{array}{*{20}{l}} {{{\bf{v}}_1}}&{{{\bf{v}}_2}}&{{{\bf{v}}_3}} \end{array}} \right]{P^{ - 1}}\\ \left[ {\begin{array}{*{20}{l}} {{{\bf{w}}_1}}&{{{\bf{w}}_2}}&{{{\bf{w}}_3}} \end{array}} \right] = \left[ {\begin{array}{*{20}{l}} {{{\bf{v}}_1}}&{{{\bf{v}}_2}}&{{{\bf{v}}_3}} \end{array}} \right]{P^{ - 1}}\\ = \left[ {\begin{array}{*{20}{c}} { - 2}&{ - 8}&{ - 7}\\ 2&5&2\\ 3&2&6 \end{array}} \right]{\left[ {\begin{array}{*{20}{c}} 1&2&{ - 1}\\ { - 3}&{ - 5}&0\\ 4&6&1 \end{array}} \right]^{ - 1}}\\ = \left[ {\begin{array}{*{20}{c}} { - 2}&{ - 8}&{ - 7}\\ 2&5&2\\ 3&2&6 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 5&8&5\\ { - 3}&{ - 5}&{ - 3}\\ { - 2}&{ - 2}&{ - 1} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {28}&{38}&{21}\\ { - 9}&{ - 13}&{ - 7}\\ { - 3}&2&3 \end{array}} \right] \end{array}\] The basis is given by the columns:
\[\left\{ {{{\bf{w}}_1} = \left[ {\begin{array}{*{20}{c}} {28}\\ { - 9}\\ { - 3} \end{array}} \right],{{\bf{w}}_2} = \left[ {\begin{array}{*{20}{c}} {38}\\ { - 13}\\ 2 \end{array}} \right],{{\bf{w}}_3} = \left[ {\begin{array}{*{20}{c}} {21}\\ { - 7}\\ 3 \end{array}} \right]} \right\}\]
We are given with following matrix and vectors \[\begin{array}{l} P = \left[ {\begin{array}{*{20}{c}} 1&2&{ - 1}\\ { - 3}&{ - 5}&0\\ 4&6&1 \end{array}} \right]\\ {{\bf{v}}_1} = \left[ {\begin{array}{*{20}{c}} { - 2}\\ 2\\ 3 \end{array}} \right],\,\,{{\bf{v}}_2} = \left[ {\begin{array}{*{20}{c}} { - 8}\\ 5\\ 2 \end{array}} \right],\,\,{{\bf{v}}_3} = \left[ {\begin{array}{*{20}{c}} { - 7}\\ 2\\ 6 \end{array}} \right] \end{array}\] (a) We have to find the basis $\left\{ \mathbf { u } _ { 1 } , \mathbf { u } _ { 2 } , \mathbf { u } _ { 3 } \right\}$ for $R^3$ such that $P$ is the change of coordinate matrix from $\left\{ \mathbf { u } _ { 1 } , \mathbf { u } _ { 2 } , \mathbf { u } _ { 3 } \right\}$ to the basis $\left\{ \mathbf { v } _ { 1 } , \mathbf { v } _ { 2 } , \mathbf { v } _ { 3 } \right\}$
(b) We also have to find the basis $\left\{ \mathbf { w } _ { 1 } , \mathbf { w } _ { 2 } , \mathbf { w } _ { 3 } \right\}$ for $R^3$such that $P$ is the change of coordinate matrix from $\left\{ \mathbf { v } _ { 1 } , \mathbf { v } _ { 2 } , \mathbf { v } _ { 3 } \right\}$ to the basis $\left\{ \mathbf { w } _ { 1 } , \mathbf { w } _ { 2 } , \mathbf { w } _ { 3 } \right\} .$
Step-1: (a)
Using the definition of C-coordinate vectors: \[\begin{array}{l} \left[ {\begin{array}{*{20}{l}} {{{\bf{u}}_1}}&{{{\bf{u}}_2}}&{{{\bf{u}}_3}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {{{\bf{v}}_1}}&{{{\bf{v}}_2}}&{{{\bf{v}}_3}} \end{array}} \right]\left[ {{{\left[ {{{\bf{u}}_1}} \right]}_c}\quad {{\left[ {{{\bf{u}}_2}} \right]}_c}\quad {{\left[ {{{\bf{u}}_2}} \right]}_c}} \right]\\ = \left[ {\begin{array}{*{20}{l}} {{{\bf{v}}_1}}&{{{\bf{v}}_2}}&{{{\bf{v}}_3}} \end{array}} \right]P\\ = \left[ {\begin{array}{*{20}{c}} { - 2}&{ - 8}&{ - 7}\\ 2&5&2\\ 3&2&6 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&2&{ - 1}\\ { - 3}&{ - 5}&0\\ 4&6&1 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} { - 6}&{ - 6}&{ - 5}\\ { - 5}&{ - 9}&0\\ {21}&{32}&3 \end{array}} \right] \end{array}\] The basis is given by the columns:
\[ \left\{ \mathbf { u } _ { 1 } = \left[ \begin{array} { c } { - 6 } \\ { - 5 } \\ { 21 } \end{array} \right] , \mathbf { u } _ { 2 } = \left[ \begin{array} { c } { - 6 } \\ { - 9 } \\ { 32 } \end{array} \right] , \mathbf { u } _ { 3 } = \left[ \begin{array} { c } { - 5 } \\ { 0 } \\ { 3 } \end{array} \right] \right\} \]
Step-2: (b)
Using the definition of C-coordinate vectors: \[\begin{array}{l} \left[ {\begin{array}{*{20}{l}} {{{\bf{w}}_1}}&{{{\bf{w}}_2}}&{{{\bf{w}}_3}} \end{array}} \right]\left[ {{{\left[ {{{\bf{v}}_1}} \right]}_c}\,\,{{\left[ {{{\bf{v}}_2}} \right]}_c}\,\,{{\left[ {{{\bf{v}}_3}} \right]}_c}} \right] = \left[ {\begin{array}{*{20}{l}} {{{\bf{v}}_1}}&{{{\bf{v}}_2}}&{{{\bf{v}}_3}} \end{array}} \right]\\ \left[ {\begin{array}{*{20}{l}} {{{\bf{w}}_1}}&{{{\bf{w}}_2}}&{{{\bf{w}}_3}} \end{array}} \right]P = \left[ {\begin{array}{*{20}{l}} {{{\bf{v}}_1}}&{{{\bf{v}}_2}}&{{{\bf{v}}_3}} \end{array}} \right]\\ \left[ {\begin{array}{*{20}{l}} {{{\bf{w}}_1}}&{{{\bf{w}}_2}}&{{{\bf{w}}_3}} \end{array}} \right]P{P^{ - 1}} = \left[ {\begin{array}{*{20}{l}} {{{\bf{v}}_1}}&{{{\bf{v}}_2}}&{{{\bf{v}}_3}} \end{array}} \right]{P^{ - 1}}\\ \left[ {\begin{array}{*{20}{l}} {{{\bf{w}}_1}}&{{{\bf{w}}_2}}&{{{\bf{w}}_3}} \end{array}} \right] = \left[ {\begin{array}{*{20}{l}} {{{\bf{v}}_1}}&{{{\bf{v}}_2}}&{{{\bf{v}}_3}} \end{array}} \right]{P^{ - 1}}\\ = \left[ {\begin{array}{*{20}{c}} { - 2}&{ - 8}&{ - 7}\\ 2&5&2\\ 3&2&6 \end{array}} \right]{\left[ {\begin{array}{*{20}{c}} 1&2&{ - 1}\\ { - 3}&{ - 5}&0\\ 4&6&1 \end{array}} \right]^{ - 1}}\\ = \left[ {\begin{array}{*{20}{c}} { - 2}&{ - 8}&{ - 7}\\ 2&5&2\\ 3&2&6 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 5&8&5\\ { - 3}&{ - 5}&{ - 3}\\ { - 2}&{ - 2}&{ - 1} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {28}&{38}&{21}\\ { - 9}&{ - 13}&{ - 7}\\ { - 3}&2&3 \end{array}} \right] \end{array}\] The basis is given by the columns:
\[\left\{ {{{\bf{w}}_1} = \left[ {\begin{array}{*{20}{c}} {28}\\ { - 9}\\ { - 3} \end{array}} \right],{{\bf{w}}_2} = \left[ {\begin{array}{*{20}{c}} {38}\\ { - 13}\\ 2 \end{array}} \right],{{\bf{w}}_3} = \left[ {\begin{array}{*{20}{c}} {21}\\ { - 7}\\ 3 \end{array}} \right]} \right\}\]