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 7E from Chapter 4.2 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 7E
Chapter:
Problem:
In Exercises 7-14, either use an appropriate theorem to show that the given set, W, is a vector space, or find a specific example to the contrary...
Step-by-Step Solution
Step 1
Given set\[W = \left\{ {\left[ {\begin{array}{*{20}{l}}a\\b\\c\end{array}} \right]:a + b + c = 2} \right\}\]We have prove or disprove whether the given set is a subspace or not.
Step 2: The alternate form
We can write the above set as:\[\begin{array}{l}W = \left\{ {\left[ {\begin{array}{*{20}{l}}a\\b\\c\end{array}} \right]:a + b + c = 2} \right\}\\ = \left\{ {\left[ {\begin{array}{*{20}{l}}a\\b\\c\end{array}} \right]:a = 2 - b - c} \right\}\\ = \left\{ {\left[ {\begin{array}{*{20}{l}}{2 - b - c}\\b\\c\end{array}} \right]:a,b,c \in R} \right\}\end{array}\]As, we have three independent rows, the above set is a subset of $R^3$. Now let us check if the set is a vector space or not.
Step 3: The zero vbector condition
Let us see if the zero vector lies in given set W\[\left[ {\begin{array}{*{20}{l}}{2 - b - c}\\b\\c\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right]\]From the last two rows, \[\begin{array}{l}b = 0\\c = 0\end{array}\]Putting above values in first row:\[\begin{array}{l}2 - b - c = 0\\2 - 0 - 0 = 0\\2 = 0\end{array}\]
ANSWER
As the last 2=0 is not possible, the zero vector does not lie in given set W.
W is not a vector space
Given set\[W = \left\{ {\left[ {\begin{array}{*{20}{l}}a\\b\\c\end{array}} \right]:a + b + c = 2} \right\}\]We have prove or disprove whether the given set is a subspace or not.
Step 2: The alternate form
We can write the above set as:\[\begin{array}{l}W = \left\{ {\left[ {\begin{array}{*{20}{l}}a\\b\\c\end{array}} \right]:a + b + c = 2} \right\}\\ = \left\{ {\left[ {\begin{array}{*{20}{l}}a\\b\\c\end{array}} \right]:a = 2 - b - c} \right\}\\ = \left\{ {\left[ {\begin{array}{*{20}{l}}{2 - b - c}\\b\\c\end{array}} \right]:a,b,c \in R} \right\}\end{array}\]As, we have three independent rows, the above set is a subset of $R^3$. Now let us check if the set is a vector space or not.
Step 3: The zero vbector condition
Let us see if the zero vector lies in given set W\[\left[ {\begin{array}{*{20}{l}}{2 - b - c}\\b\\c\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right]\]From the last two rows, \[\begin{array}{l}b = 0\\c = 0\end{array}\]Putting above values in first row:\[\begin{array}{l}2 - b - c = 0\\2 - 0 - 0 = 0\\2 = 0\end{array}\]
ANSWER
As the last 2=0 is not possible, the zero vector does not lie in given set W.
W is not a vector space