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 34E from Chapter 1.3 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 34E
Chapter:
Problem:
Use the vectors u = (u1, … , un), to verify the following algebraic properties of ℝn.
Step-by-Step Solution
Given Information
We are given with following vectors: \[ \mathbf { u } = \left( u _ { 1 } , \ldots , u _ { n } \right) \] We have to verify \[ \mathbf { u } + ( - \mathbf { u } ) = ( - \mathbf { u } ) + \mathbf { u } = \mathbf { 0 } \] and \[ c ( d \mathbf { u } ) = ( c d ) \mathbf { u } \]
Step-1: (a)
Apply the scalar multiple -1. \[ \begin{aligned} - \mathbf { u } & = - \left( u _ { 1 } , \ldots , u _ { n } \right) \\ & = \left( - u _ { 1 } , \ldots , - u _ { n } \right) \end{aligned} \] Addition of u and -u: \[ \begin{aligned} \mathbf { u } + ( - \mathbf { u } ) & = \left( u _ { 1 } , \ldots , u _ { n } \right) + \left( - u _ { 1 } , \ldots , - u _ { n } \right) \\ & = \left( u _ { 1 } + \left( - u _ { 1 } \right) , \ldots , u _ { n } + \left( - u _ { n } \right) \right) \\ & = \left( u _ { 1 } - u _ { 1 } , \ldots , u _ { n } - u _ { n } \right) \\ & = ( 0 , \ldots , 0 ) \\ & = \mathbf { 0 } \end{aligned} \] Addition of -u and u: \[ \begin{aligned} ( - \mathbf { u } ) + \mathbf { u } & = \left( - u _ { 1 } , \ldots , - u _ { n } \right) + \left( u _ { 1 } , \ldots , u _ { n } \right) \\ & = \left( - u _ { 1 } + u _ { 1 } , \ldots , - u _ { n } + u _ { n } \right) \\ & = ( 0 , \ldots , 0 ) \\ & = \mathbf { 0 } \end{aligned} \]
\[ \mathbf { u } + ( - \mathbf { u } ) = ( - \mathbf { u } ) + \mathbf { u } = \mathbf { 0 } \]
Step-2: (b)
Let $c$ and $d$ be any scalars. Multiple $d$ with $u$ \[ d \mathbf { u } = \left( d u _ { 1 } , \ldots , d u _ { n } \right) \] Multiple $c$ with $du$ \[ \begin{aligned} c ( d \mathbf { u } ) & = c \left( d u _ { 1 } , \ldots , d u _ { n } \right) \\ & = c \left( d \left( u _ { 1 } , \ldots , u _ { n } \right) \right) \\ & = c d \left( u _ { 1 } , \ldots , u _ { n } \right) \\ & = ( c d ) \left( u _ { 1 } , \ldots , u _ { n } \right) \\ & = ( c d ) \mathbf { u } \end{aligned} \] Hence,
\[ \mathbf { u } + ( - \mathbf { u } ) = ( - \mathbf { u } ) + \mathbf { u } = \mathbf { 0 } \]
We are given with following vectors: \[ \mathbf { u } = \left( u _ { 1 } , \ldots , u _ { n } \right) \] We have to verify \[ \mathbf { u } + ( - \mathbf { u } ) = ( - \mathbf { u } ) + \mathbf { u } = \mathbf { 0 } \] and \[ c ( d \mathbf { u } ) = ( c d ) \mathbf { u } \]
Step-1: (a)
Apply the scalar multiple -1. \[ \begin{aligned} - \mathbf { u } & = - \left( u _ { 1 } , \ldots , u _ { n } \right) \\ & = \left( - u _ { 1 } , \ldots , - u _ { n } \right) \end{aligned} \] Addition of u and -u: \[ \begin{aligned} \mathbf { u } + ( - \mathbf { u } ) & = \left( u _ { 1 } , \ldots , u _ { n } \right) + \left( - u _ { 1 } , \ldots , - u _ { n } \right) \\ & = \left( u _ { 1 } + \left( - u _ { 1 } \right) , \ldots , u _ { n } + \left( - u _ { n } \right) \right) \\ & = \left( u _ { 1 } - u _ { 1 } , \ldots , u _ { n } - u _ { n } \right) \\ & = ( 0 , \ldots , 0 ) \\ & = \mathbf { 0 } \end{aligned} \] Addition of -u and u: \[ \begin{aligned} ( - \mathbf { u } ) + \mathbf { u } & = \left( - u _ { 1 } , \ldots , - u _ { n } \right) + \left( u _ { 1 } , \ldots , u _ { n } \right) \\ & = \left( - u _ { 1 } + u _ { 1 } , \ldots , - u _ { n } + u _ { n } \right) \\ & = ( 0 , \ldots , 0 ) \\ & = \mathbf { 0 } \end{aligned} \]
\[ \mathbf { u } + ( - \mathbf { u } ) = ( - \mathbf { u } ) + \mathbf { u } = \mathbf { 0 } \]
Step-2: (b)
Let $c$ and $d$ be any scalars. Multiple $d$ with $u$ \[ d \mathbf { u } = \left( d u _ { 1 } , \ldots , d u _ { n } \right) \] Multiple $c$ with $du$ \[ \begin{aligned} c ( d \mathbf { u } ) & = c \left( d u _ { 1 } , \ldots , d u _ { n } \right) \\ & = c \left( d \left( u _ { 1 } , \ldots , u _ { n } \right) \right) \\ & = c d \left( u _ { 1 } , \ldots , u _ { n } \right) \\ & = ( c d ) \left( u _ { 1 } , \ldots , u _ { n } \right) \\ & = ( c d ) \mathbf { u } \end{aligned} \] Hence,
\[ \mathbf { u } + ( - \mathbf { u } ) = ( - \mathbf { u } ) + \mathbf { u } = \mathbf { 0 } \]