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 23E from Chapter 4.4 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 23E
Chapter:
Problem:
Exercises 23–26 concern a vector space V, a basis and the coordinate mapping Show that the coordinate mapping is one-to-one. (Hint: Suppose for some u and w in V, and show that u = w.)
Step-by-Step Solution
Given Information
We have to prove that the coordinate mapping is one-to-one.
Step-1:
We have to prove that $u=w$ if $ [ \mathbf { u } ] _ { B } = [ \mathbf { w } ] _ { B }$. Since u and v are in V and $ B = \left[ \mathbf { b } _ { 1 } , \dots , \mathbf { b } _ { n } \right] $ is the basis for V. So we can write. \[ \begin{aligned} \mathbf { u } & = p _ { 1 } \mathbf { b } _ { 1 } + \ldots + p _ { n } \mathbf { b } _ { n } \\ \mathbf { w } & = q _ { 1 } \mathbf { b } _ { 1 } + \ldots + q _ { n } \mathbf { b } _ { n } \end{aligned} \] Hence the change of coordinate matrices are: \[ [ \mathbf { u } ] _ { B } = \left[ \begin{array} { c } { p _ { 1 } } \\ { \vdots } \\ { p _ { n } } \end{array} \right] \text { and } [ \mathbf { w } ] _ { B } = \left[ \begin{array} { c } { q _ { 1 } } \\ { \vdots } \\ { q _ { n } } \end{array} \right] \]
Step-2:
If $ [ \mathbf { u } ] _ { B } = [ \mathbf { w } ] _ { B }$, then, \[\begin{array}{l} \left[ {\begin{array}{*{20}{c}} {{p_1}}\\ \vdots \\ {{p_n}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {{q_1}}\\ \vdots \\ {{q_n}} \end{array}} \right]\\ {p_1} = {q_1}\,\,{p_2} = {q_2},\,\,, \ldots ,{p_n} = {q_n}\\ {p_1}{b_1} = {q_1}{b_1}\,\,{p_2}{b_2} = {q_2},{b_2}\,\,, \ldots ,{p_n}{b_n} = {q_n}{b_n} \end{array}\] Hence, \[{\bf{u = w}}\] Therefore,
The coordinate mapping is one-to-one
We have to prove that the coordinate mapping is one-to-one.
Step-1:
We have to prove that $u=w$ if $ [ \mathbf { u } ] _ { B } = [ \mathbf { w } ] _ { B }$. Since u and v are in V and $ B = \left[ \mathbf { b } _ { 1 } , \dots , \mathbf { b } _ { n } \right] $ is the basis for V. So we can write. \[ \begin{aligned} \mathbf { u } & = p _ { 1 } \mathbf { b } _ { 1 } + \ldots + p _ { n } \mathbf { b } _ { n } \\ \mathbf { w } & = q _ { 1 } \mathbf { b } _ { 1 } + \ldots + q _ { n } \mathbf { b } _ { n } \end{aligned} \] Hence the change of coordinate matrices are: \[ [ \mathbf { u } ] _ { B } = \left[ \begin{array} { c } { p _ { 1 } } \\ { \vdots } \\ { p _ { n } } \end{array} \right] \text { and } [ \mathbf { w } ] _ { B } = \left[ \begin{array} { c } { q _ { 1 } } \\ { \vdots } \\ { q _ { n } } \end{array} \right] \]
Step-2:
If $ [ \mathbf { u } ] _ { B } = [ \mathbf { w } ] _ { B }$, then, \[\begin{array}{l} \left[ {\begin{array}{*{20}{c}} {{p_1}}\\ \vdots \\ {{p_n}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {{q_1}}\\ \vdots \\ {{q_n}} \end{array}} \right]\\ {p_1} = {q_1}\,\,{p_2} = {q_2},\,\,, \ldots ,{p_n} = {q_n}\\ {p_1}{b_1} = {q_1}{b_1}\,\,{p_2}{b_2} = {q_2},{b_2}\,\,, \ldots ,{p_n}{b_n} = {q_n}{b_n} \end{array}\] Hence, \[{\bf{u = w}}\] Therefore,
The coordinate mapping is one-to-one