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 21E from Chapter 6.3 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 21E
Chapter:
Problem:
In Exercises 21 and 22, all vectors and subspaces are in Rn. Mark each statement True or False. Justify each answer.
Step-by-Step Solution
Given Information
We are given with some statements that we have to find whether they are True or False.
Step-1: (a)
If a vector z is orthogonal to the vectors in the basis {$u_1,u_2$} for, then z is in $W ^ { \perp }$. Hence
The given statement is True
Step-2: (b)
By the orthogonal decomposition theorem, for the vector subspace $W$ of $R^n$. each y in $R^n$ can be written as $\mathbf { y } = \operatorname { proj } _ { \mathbf { w } } \mathbf { y } + \mathbf { z }$. Here \[ \operatorname { proj } _ { W } \mathbf { y } = \sum _ { i = 1 } ^ { p } \dfrac { \mathbf { y } \cdot \mathbf { u } _ { i } } { \mathbf { u } _ { i } \cdot \mathbf { u } _ { i } } \mathbf { u } _ { i } \in W \] and \[ \mathbf { z } = \mathbf { y } - \operatorname { proj } _ { W } \mathbf { y } \in W ^ { \perp } \] So, the $\mathbf { y } - \text { proj } _ { y } \mathbf { y }$ is the orthogonal complement of projection of vector y.
The given statement is True
Step-3: (c)
By the uniqueness of orthogonal decomposition of y $\mathbf { y } = \operatorname { proj } _ { W } \mathbf { y } + \mathbf { z } $, it shows that the orthogonal projection depends only on W and not on the particular basis used to compute it
The given statement is False
Step-4: (d)
By the orthogonal projection theorem, for a vector y in $W = \operatorname { Span } \left\{ \mathbf { u } _ { 1 } , \mathbf { u } _ { 2 } , \cdots , \mathbf { u } _ { p } \right\} $, we have \[pro{j_W}y = y\] Therefore,
The given statement is true
Step-5: (e)
If U has orthogonal columns given by: \[ U = \left[ \begin{array} { l l l l } { \mathbf { u } _ { 1 } } & { \mathbf { u } _ { 2 } } & { \cdots } & { \mathbf { u } _ { p } } \end{array} \right] \] Then, \[ \begin{aligned} U U ^ { T } \mathbf { y } & = \operatorname { proj } _ { W } \mathbf { y } \\ & = \left( \mathbf { y } \cdot \mathbf { u } _ { 1 } \right) \mathbf { u } _ { 1 } + \left( y \cdot \mathbf { u } _ { 2 } \right) \mathbf { u } _ { 2 } + \cdots + \left( y \cdot \mathbf { u } _ { p } \right) \mathbf { u } _ { p } \end{aligned} \] Therefore,
The given statement is True
We are given with some statements that we have to find whether they are True or False.
Step-1: (a)
If a vector z is orthogonal to the vectors in the basis {$u_1,u_2$} for, then z is in $W ^ { \perp }$. Hence
The given statement is True
Step-2: (b)
By the orthogonal decomposition theorem, for the vector subspace $W$ of $R^n$. each y in $R^n$ can be written as $\mathbf { y } = \operatorname { proj } _ { \mathbf { w } } \mathbf { y } + \mathbf { z }$. Here \[ \operatorname { proj } _ { W } \mathbf { y } = \sum _ { i = 1 } ^ { p } \dfrac { \mathbf { y } \cdot \mathbf { u } _ { i } } { \mathbf { u } _ { i } \cdot \mathbf { u } _ { i } } \mathbf { u } _ { i } \in W \] and \[ \mathbf { z } = \mathbf { y } - \operatorname { proj } _ { W } \mathbf { y } \in W ^ { \perp } \] So, the $\mathbf { y } - \text { proj } _ { y } \mathbf { y }$ is the orthogonal complement of projection of vector y.
The given statement is True
Step-3: (c)
By the uniqueness of orthogonal decomposition of y $\mathbf { y } = \operatorname { proj } _ { W } \mathbf { y } + \mathbf { z } $, it shows that the orthogonal projection depends only on W and not on the particular basis used to compute it
The given statement is False
Step-4: (d)
By the orthogonal projection theorem, for a vector y in $W = \operatorname { Span } \left\{ \mathbf { u } _ { 1 } , \mathbf { u } _ { 2 } , \cdots , \mathbf { u } _ { p } \right\} $, we have \[pro{j_W}y = y\] Therefore,
The given statement is true
Step-5: (e)
If U has orthogonal columns given by: \[ U = \left[ \begin{array} { l l l l } { \mathbf { u } _ { 1 } } & { \mathbf { u } _ { 2 } } & { \cdots } & { \mathbf { u } _ { p } } \end{array} \right] \] Then, \[ \begin{aligned} U U ^ { T } \mathbf { y } & = \operatorname { proj } _ { W } \mathbf { y } \\ & = \left( \mathbf { y } \cdot \mathbf { u } _ { 1 } \right) \mathbf { u } _ { 1 } + \left( y \cdot \mathbf { u } _ { 2 } \right) \mathbf { u } _ { 2 } + \cdots + \left( y \cdot \mathbf { u } _ { p } \right) \mathbf { u } _ { p } \end{aligned} \] Therefore,
The given statement is True