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 4.5 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 1E
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
Given Subspace:\[S = \left\{ {\left[ {\begin{array}{*{20}{c}}{s - 2t}\\{s + t}\\{3t}\end{array}} \right];s,t \in R} \right\}\] We have to find the basis for the given subspace
Step-1: Parametric form of the subspace
Separate the parameters: \[\begin{array}{l}x = \left[ {\begin{array}{*{20}{c}}{s - 2t}\\{s + t}\\{3t}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{l}}s\\s\\0\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{ - 2t}\\t\\{3t}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{l}}1\\1\\0\end{array}} \right]s + \left[ {\begin{array}{*{20}{c}}{ - 2}\\1\\3\end{array}} \right]t\end{array}\]Every vector in the set S, is a linear combination of the the vectors: $\left\{ {\left[ {\begin{array}{*{20}{l}}1\\1\\0\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{ - 2}\\1\\3\end{array}} \right]} \right\}$
Step 2: The basis
Consider the vectors obtained above:
\[\left\{ {\left[ {\begin{array}{*{20}{l}}1\\1\\0\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{ - 2}\\1\\3\end{array}} \right]} \right\}\]The vectors span S
The vectors are linearly independent
Hence the form a basis for S
Step 3: Dimension of subspace
Since there are 2 non zero vectors in the basis, the dimension of the subspace is 2
ANSWERS
The Basis for S: \[\left\{ {\left[ {\begin{array}{*{20}{l}}1\\1\\0\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{ - 2}\\1\\3\end{array}} \right]} \right\}\]The Dimension
The basis has a dimension of 2
Given Subspace:\[S = \left\{ {\left[ {\begin{array}{*{20}{c}}{s - 2t}\\{s + t}\\{3t}\end{array}} \right];s,t \in R} \right\}\] We have to find the basis for the given subspace
Step-1: Parametric form of the subspace
Separate the parameters: \[\begin{array}{l}x = \left[ {\begin{array}{*{20}{c}}{s - 2t}\\{s + t}\\{3t}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{l}}s\\s\\0\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{ - 2t}\\t\\{3t}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{l}}1\\1\\0\end{array}} \right]s + \left[ {\begin{array}{*{20}{c}}{ - 2}\\1\\3\end{array}} \right]t\end{array}\]Every vector in the set S, is a linear combination of the the vectors: $\left\{ {\left[ {\begin{array}{*{20}{l}}1\\1\\0\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{ - 2}\\1\\3\end{array}} \right]} \right\}$
Step 2: The basis
Consider the vectors obtained above:
\[\left\{ {\left[ {\begin{array}{*{20}{l}}1\\1\\0\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{ - 2}\\1\\3\end{array}} \right]} \right\}\]The vectors span S
The vectors are linearly independent
Hence the form a basis for S
Step 3: Dimension of subspace
Since there are 2 non zero vectors in the basis, the dimension of the subspace is 2
ANSWERS
The Basis for S: \[\left\{ {\left[ {\begin{array}{*{20}{l}}1\\1\\0\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{ - 2}\\1\\3\end{array}} \right]} \right\}\]The Dimension
The basis has a dimension of 2