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 20E from Chapter 4.3 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 20E
Chapter:
Problem:
Let It can be verified that v1 -3v2 + 5v3 = 0. Use this information to find a basis for H = Span {v1, v2, v3}.
Step-by-Step Solution
Given Information
We are given with a set of vectors: \[ \mathbf { v } _ { 1 } = \left[ \begin{array} { r } { 7 } \\ { 4 } \\ { - 9 } \\ { - 5 } \end{array} \right] , \mathbf { v } _ { 2 } = \left[ \begin{array} { r } { 4 } \\ { - 7 } \\ { 2 } \\ { 5 } \end{array} \right] , \mathbf { v } _ { 3 } = \left[ \begin{array} { r } { 1 } \\ { - 5 } \\ { 3 } \\ { 4 } \end{array} \right] \] We also given that ${{\bf{v}}_1} - 3{{\bf{v}}_2} + 5{{\bf{v}}_3} = {\bf{0}}$. We have to find a basis for H.
Step 1: The basis for H
We can see that the three vectors are linearly dependent. However, none of the pair of vectors are linearly dependent. Hence,
Either of the pair of vectors form a basis for H.
We are given with a set of vectors: \[ \mathbf { v } _ { 1 } = \left[ \begin{array} { r } { 7 } \\ { 4 } \\ { - 9 } \\ { - 5 } \end{array} \right] , \mathbf { v } _ { 2 } = \left[ \begin{array} { r } { 4 } \\ { - 7 } \\ { 2 } \\ { 5 } \end{array} \right] , \mathbf { v } _ { 3 } = \left[ \begin{array} { r } { 1 } \\ { - 5 } \\ { 3 } \\ { 4 } \end{array} \right] \] We also given that ${{\bf{v}}_1} - 3{{\bf{v}}_2} + 5{{\bf{v}}_3} = {\bf{0}}$. We have to find a basis for H.
Step 1: The basis for H
We can see that the three vectors are linearly dependent. However, none of the pair of vectors are linearly dependent. Hence,
Either of the pair of vectors form a basis for H.