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 3E from Chapter 4.5 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 3E
Chapter:
Problem:
For each subspace in Exercises 1–8, (a) find a basis for the subspace, and (b) state the dimension.
Step-by-Step Solution
Given Information
We are given with a subspace \[ \left\{ \left[ \begin{array} { c } { 2 c } \\ { a - b } \\ { b - 3 c } \\ { a + 2 b } \end{array} \right] : a , b , c \text { in } \mathbb { R } \right\} \] We have to find the basis and state the dimension
Step-1:
The given set can be written in the parametric form \[\begin{array}{l} x = \left[ {\begin{array}{*{20}{c}} {2c}\\ {a - b}\\ {b - 3c}\\ {a + 2b} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{l}} 0\\ a\\ 0\\ a \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 0\\ { - b}\\ b\\ {2b} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} {2c}\\ 0\\ { - 3c}\\ 0 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{l}} 0\\ 1\\ 0\\ 1 \end{array}} \right]a + \left[ {\begin{array}{*{20}{c}} 0\\ { - 1}\\ 1\\ 2 \end{array}} \right]b + \left[ {\begin{array}{*{20}{c}} 2\\ 0\\ { - 3}\\ 0 \end{array}} \right]c \end{array}\] Therefore the set given below spans H: \[{\rm{S}} = \left\{ {\left[ {\begin{array}{*{20}{l}} 0\\ 1\\ 0\\ 1 \end{array}} \right],\left[ {\begin{array}{*{20}{c}} 0\\ { - 1}\\ 1\\ 2 \end{array}} \right],\left[ {\begin{array}{*{20}{c}} 2\\ 0\\ { - 3}\\ 0 \end{array}} \right]} \right\}\]
Step-2: Check for linear dependence
The Row-Reduced augmented matrix \[\begin{array}{l} M = \left[ {\begin{array}{*{20}{c}} 0&0&2\\ 1&{ - 1}&0\\ 0&1&{ - 3}\\ 1&2&0 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 0&0&2\\ 1&{ - 1}&0\\ 0&1&{ - 3}\\ 0&3&0 \end{array}} \right]::\,\,\left\{ {{R_4} = {R_4} - {R_2}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 0&0&2\\ 1&{ - 1}&0\\ 0&1&{ - 3}\\ 0&0&9 \end{array}} \right]::\,\,\left\{ {{R_4} = {R_4} - 3{R_3}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 0&0&2\\ 1&{ - 1}&0\\ 0&1&{ - 3}\\ 0&0&1 \end{array}} \right]::\,\,\left\{ {{R_4} = \dfrac{{{R_4}}}{9}} \right\} \end{array}\] As there are 3 pivot elements, the vectors are linearly independent and form a basis for H and has a dimension of 3. Hence the basis is given by:
\[B = \left\{ {\left[ {\begin{array}{*{20}{l}} 0\\ 1\\ 0\\ 1 \end{array}} \right],\left[ {\begin{array}{*{20}{c}} 0\\ { - 1}\\ 1\\ 2 \end{array}} \right],\left[ {\begin{array}{*{20}{c}} 2\\ 0\\ { - 3}\\ 0 \end{array}} \right]} \right\}\] The dimension is 3
We are given with a subspace \[ \left\{ \left[ \begin{array} { c } { 2 c } \\ { a - b } \\ { b - 3 c } \\ { a + 2 b } \end{array} \right] : a , b , c \text { in } \mathbb { R } \right\} \] We have to find the basis and state the dimension
Step-1:
The given set can be written in the parametric form \[\begin{array}{l} x = \left[ {\begin{array}{*{20}{c}} {2c}\\ {a - b}\\ {b - 3c}\\ {a + 2b} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{l}} 0\\ a\\ 0\\ a \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 0\\ { - b}\\ b\\ {2b} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} {2c}\\ 0\\ { - 3c}\\ 0 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{l}} 0\\ 1\\ 0\\ 1 \end{array}} \right]a + \left[ {\begin{array}{*{20}{c}} 0\\ { - 1}\\ 1\\ 2 \end{array}} \right]b + \left[ {\begin{array}{*{20}{c}} 2\\ 0\\ { - 3}\\ 0 \end{array}} \right]c \end{array}\] Therefore the set given below spans H: \[{\rm{S}} = \left\{ {\left[ {\begin{array}{*{20}{l}} 0\\ 1\\ 0\\ 1 \end{array}} \right],\left[ {\begin{array}{*{20}{c}} 0\\ { - 1}\\ 1\\ 2 \end{array}} \right],\left[ {\begin{array}{*{20}{c}} 2\\ 0\\ { - 3}\\ 0 \end{array}} \right]} \right\}\]
Step-2: Check for linear dependence
The Row-Reduced augmented matrix \[\begin{array}{l} M = \left[ {\begin{array}{*{20}{c}} 0&0&2\\ 1&{ - 1}&0\\ 0&1&{ - 3}\\ 1&2&0 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 0&0&2\\ 1&{ - 1}&0\\ 0&1&{ - 3}\\ 0&3&0 \end{array}} \right]::\,\,\left\{ {{R_4} = {R_4} - {R_2}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 0&0&2\\ 1&{ - 1}&0\\ 0&1&{ - 3}\\ 0&0&9 \end{array}} \right]::\,\,\left\{ {{R_4} = {R_4} - 3{R_3}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 0&0&2\\ 1&{ - 1}&0\\ 0&1&{ - 3}\\ 0&0&1 \end{array}} \right]::\,\,\left\{ {{R_4} = \dfrac{{{R_4}}}{9}} \right\} \end{array}\] As there are 3 pivot elements, the vectors are linearly independent and form a basis for H and has a dimension of 3. Hence the basis is given by:
\[B = \left\{ {\left[ {\begin{array}{*{20}{l}} 0\\ 1\\ 0\\ 1 \end{array}} \right],\left[ {\begin{array}{*{20}{c}} 0\\ { - 1}\\ 1\\ 2 \end{array}} \right],\left[ {\begin{array}{*{20}{c}} 2\\ 0\\ { - 3}\\ 0 \end{array}} \right]} \right\}\] The dimension is 3