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 24E from Chapter 1.3 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 24E
Chapter:
Problem:
In Exercises 23 and 24, mark each statement True or False. Justify each answer. a. When u and v are nonzero vectors, Span {u, v} contains only the line through u and the origin, and the line through v and the origin. b. Any list of five real numbers is a vector in ℝ5 . c. Asking whether the linear system corresponding to an augmented matrix [a1 a2 a3 b] has a solution amounts to asking whether b is in Span {a1 a2 a3} d. The vector v results when a vector u – v is added to the vector v. e. The weights c1, … , c2 in a linear combination c1 v1+ … + cp vp cannot all be zero.
Step-by-Step Solution
Given Information
We are given with some statements, we have to prove whether they are True or False
Step-1: (a)
A collection of 5 real numbers can be written as a column vector of size $5 \times 1$. Therefore, it represents a vector in $R^5$. Hence
The statement is TRUE
Step-2: (b)
Write the vector addition of ($u-v$) and ($v$)
\[\begin{array}{l} \left( {{\bf{u - v}}} \right) + {\bf{v = u}} + \left( { - {\bf{v}}} \right) + {\bf{v}}\\ = {\bf{u}} + \left( {{\bf{v - v}}} \right)\\ = {\bf{u}} \end{array}\]
The statement is TRUE
Step-3: (c)
A vector $b$ in $R^n$ is said to be the linear combination of set of vectors $\left\{ {{x_1},{x_2},{x_3}...{x_k}} \right\}$ if there exists scalars $\left\{ {{a_1},{a_2},{a_3}...{a_k}} \right\}$ such that $b = {x_1}{a_1} + {x_2}{a_2}...{x_k}{a_k}$
If $b$ is a zero vector, then
$${a_1} = 0,\,\,{a_2} = 0.....{a_k} = 0$$ Therefore, all the weights can also be zero.
The statement is FALSE
Step-4: (d)
The Span{u,v} contains line through v and the origin as well as the line through u and the origin. If there is a vector b that belongs to the span of {u,v},then:
\[b = {a_1}u + {a_2}v\] Take the case when $a_2$ is 0. Then the line becomes: \[b = {a_1}u\] This is the equation of line passing through u and the origin. .
The statement is TRUE
Step-5: (e)
The augmented matrix given by ${{{\bf{a}}_{\bf{1}}}\,\,{{\bf{a}}_{\bf{2}}}\,\,{{\bf{a}}_{\bf{3}}}\,\,{\bf{b}}}$ has a solution only if the column $b$ is a linear combination of other columns. So, if the last column $b$ can be written as a linear combination of vectors $\left[ {{{\bf{a}}_{\bf{1}}}\,\,{{\bf{a}}_{\bf{2}}}\,\,{{\bf{a}}_{\bf{3}}}} \right]$, then b is in the span of $\left\{ {{{\bf{a}}_{\bf{1}}},{{\bf{a}}_{\bf{2}}},{{\bf{a}}_{\bf{3}}}} \right\}$
The statement is TRUE
We are given with some statements, we have to prove whether they are True or False
Step-1: (a)
A collection of 5 real numbers can be written as a column vector of size $5 \times 1$. Therefore, it represents a vector in $R^5$. Hence
The statement is TRUE
Step-2: (b)
Write the vector addition of ($u-v$) and ($v$)
\[\begin{array}{l} \left( {{\bf{u - v}}} \right) + {\bf{v = u}} + \left( { - {\bf{v}}} \right) + {\bf{v}}\\ = {\bf{u}} + \left( {{\bf{v - v}}} \right)\\ = {\bf{u}} \end{array}\]
The statement is TRUE
Step-3: (c)
A vector $b$ in $R^n$ is said to be the linear combination of set of vectors $\left\{ {{x_1},{x_2},{x_3}...{x_k}} \right\}$ if there exists scalars $\left\{ {{a_1},{a_2},{a_3}...{a_k}} \right\}$ such that $b = {x_1}{a_1} + {x_2}{a_2}...{x_k}{a_k}$
If $b$ is a zero vector, then
$${a_1} = 0,\,\,{a_2} = 0.....{a_k} = 0$$ Therefore, all the weights can also be zero.
The statement is FALSE
Step-4: (d)
The Span{u,v} contains line through v and the origin as well as the line through u and the origin. If there is a vector b that belongs to the span of {u,v},then:
\[b = {a_1}u + {a_2}v\] Take the case when $a_2$ is 0. Then the line becomes: \[b = {a_1}u\] This is the equation of line passing through u and the origin. .
The statement is TRUE
Step-5: (e)
The augmented matrix given by ${{{\bf{a}}_{\bf{1}}}\,\,{{\bf{a}}_{\bf{2}}}\,\,{{\bf{a}}_{\bf{3}}}\,\,{\bf{b}}}$ has a solution only if the column $b$ is a linear combination of other columns. So, if the last column $b$ can be written as a linear combination of vectors $\left[ {{{\bf{a}}_{\bf{1}}}\,\,{{\bf{a}}_{\bf{2}}}\,\,{{\bf{a}}_{\bf{3}}}} \right]$, then b is in the span of $\left\{ {{{\bf{a}}_{\bf{1}}},{{\bf{a}}_{\bf{2}}},{{\bf{a}}_{\bf{3}}}} \right\}$
The statement is TRUE