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 15E from Chapter 4.4 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 15E
Chapter:
Problem:
In Exercises 15 and 16, mark each statement True or False. Justify each answer. Unless stated otherwise, is a basis for a vector space V. a. If x is in V and if contains n vectors, then the - coordinate vector of x is in . b. If PB is the change-of-coordinates matrix, then . c. The vector spaces and are isomorphic.
Step-by-Step Solution
Given Information
We are given with some statements that we have to prove True or False.
Step-1: (a)
Let $ B = \left[ \begin{array} { l l l } { \mathbf { b } _ { 1 } } & { \dots } & { \mathbf { b } _ { n } } \end{array} \right] $ is a basis for V and x is in V. The vector x satisfies following relationship. \[ \mathbf { x } = c _ { 1 } \mathbf { b } _ { 1 } + \ldots + c _ { n } \mathbf { b } _ { n } \] Hence the B-coordinate vector of x is in $R^n$ Therefore
The statement is True.
Step-2: (b)
If B is a basis for the vector space $V$ and $P_B$ is the change of coordinate matrix, then \[ \mathbf { x } = P _ { B } [ \mathbf { x } ] _ { B } \] Therefore
The statement is False.
Step-3: (c)
Let $B = \left\{ 1 , t , t ^ { 2 } , t ^ { 3 } \right\}$ is the standard basis of the space $P_3$. So it can be written as. \[ p ( t ) = a _ { 0 } + a _ { 1 } t + a _ { 2 } t ^ { 2 } + a _ { 3 } t ^ { 3 } \] Therefore, \[ [ p ] _ { B } = \left[ \begin{array} { l } { a _ { 0 } } \\ { a _ { 1 } } \\ { a _ { 2 } } \\ { a _ { 3 } } \end{array} \right] \] Hence, the mapping is an isomorphism from $P_3$ to $R^4$ not to $R^3$
The statement is False.
We are given with some statements that we have to prove True or False.
Step-1: (a)
Let $ B = \left[ \begin{array} { l l l } { \mathbf { b } _ { 1 } } & { \dots } & { \mathbf { b } _ { n } } \end{array} \right] $ is a basis for V and x is in V. The vector x satisfies following relationship. \[ \mathbf { x } = c _ { 1 } \mathbf { b } _ { 1 } + \ldots + c _ { n } \mathbf { b } _ { n } \] Hence the B-coordinate vector of x is in $R^n$ Therefore
The statement is True.
Step-2: (b)
If B is a basis for the vector space $V$ and $P_B$ is the change of coordinate matrix, then \[ \mathbf { x } = P _ { B } [ \mathbf { x } ] _ { B } \] Therefore
The statement is False.
Step-3: (c)
Let $B = \left\{ 1 , t , t ^ { 2 } , t ^ { 3 } \right\}$ is the standard basis of the space $P_3$. So it can be written as. \[ p ( t ) = a _ { 0 } + a _ { 1 } t + a _ { 2 } t ^ { 2 } + a _ { 3 } t ^ { 3 } \] Therefore, \[ [ p ] _ { B } = \left[ \begin{array} { l } { a _ { 0 } } \\ { a _ { 1 } } \\ { a _ { 2 } } \\ { a _ { 3 } } \end{array} \right] \] Hence, the mapping is an isomorphism from $P_3$ to $R^4$ not to $R^3$
The statement is False.