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 1E from Chapter 4.1 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 1E
Chapter:
Problem:
Let V be the first quadrant in the xy-plane; that is, let...
Step-by-Step Solution
Step 1
Given vector:\[V = \left\{ {\left[ {\begin{array}{*{20}{l}}x\\y\end{array}} \right]:x \ge 0,y \ge 0} \right\}\]
Step 2: (a) Prove that ${\bf{u}} + {\bf{v}}$ is in V
We are given that ${\bf{u}}$ and ${\bf{v}}$ are in V, then we have to prove that ${\bf{u}} + {\bf{v}}$ is in V.
As per the condition, V is set of positive real numbers.
If ${\bf{u}}$ and ${\bf{v}}$ are in V then, they contain positive numbers.
Since sum of two positive numbers is also a positive number, then ${\bf{u}} + {\bf{v}}$ consists of positive numbers. hence ${\bf{u}} + {\bf{v}}$ is in V .
Step 3: (b)
let us take a case of vector\[{\bf{u}} = \left[ {\begin{array}{*{20}{c}}6\\7\end{array}} \right]\]and scaler \[c = - 1\]So,\[\begin{array}{l}c{\bf{u}} = - 1\left[ {\begin{array}{*{20}{c}}6\\7\end{array}} \right]\\\\c{\bf{u}} = \left[ {\begin{array}{*{20}{c}}{ - 3}\\{ - 5}\end{array}} \right]\end{array}\]The matrix found above does not have all positive entries, hence it does not lie in V. hence, the set V is not a vector space.
Given vector:\[V = \left\{ {\left[ {\begin{array}{*{20}{l}}x\\y\end{array}} \right]:x \ge 0,y \ge 0} \right\}\]
Step 2: (a) Prove that ${\bf{u}} + {\bf{v}}$ is in V
We are given that ${\bf{u}}$ and ${\bf{v}}$ are in V, then we have to prove that ${\bf{u}} + {\bf{v}}$ is in V.
As per the condition, V is set of positive real numbers.
If ${\bf{u}}$ and ${\bf{v}}$ are in V then, they contain positive numbers.
Since sum of two positive numbers is also a positive number, then ${\bf{u}} + {\bf{v}}$ consists of positive numbers. hence ${\bf{u}} + {\bf{v}}$ is in V .
Step 3: (b)
let us take a case of vector\[{\bf{u}} = \left[ {\begin{array}{*{20}{c}}6\\7\end{array}} \right]\]and scaler \[c = - 1\]So,\[\begin{array}{l}c{\bf{u}} = - 1\left[ {\begin{array}{*{20}{c}}6\\7\end{array}} \right]\\\\c{\bf{u}} = \left[ {\begin{array}{*{20}{c}}{ - 3}\\{ - 5}\end{array}} \right]\end{array}\]The matrix found above does not have all positive entries, hence it does not lie in V. hence, the set V is not a vector space.