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F1 To be a subspace H of V, the following must all be true i. 0 (the zero vector) is in H ii. It is closed under addition. i.e. if u, v are in H, then u + v are in H iii. It is closed under scalar multiplication. i.e if u is in H, then for any constant c, cu is in H Show the following IS a subspace using Theorem 1. The set of all vectors of the form : {(2a, a - b, 4b): a, b = R}: 2a A a-b 46 This can also be written as To help you through this follow these steps. Note: You COULD prove all three conditions are met above (0 vector is in set, closed under vector addition, closed under scalar multiplication)...but this is a LOT OF WORK. It is EASIER to use the theorem used in class. Show this is in the span of a set of vectors. -- ■ 0 W DELL

F1 To be a subspace H of V, the following must all be true i. 0 (the zero vector) is in H ii. It is closed under addition. i.e. if u, v are in H, then u + v are in H iii. It is closed under scalar multiplication. i.e if u is in H, then for any constant c, cu is in H Show the following IS a subspace using Theorem 1. The set of all vectors of the form : {(2a, a - b, 4b): a, b = R}: 2a A a-b 46 This can also be written as To help you through this follow these steps. Note: You COULD prove all three conditions are met above (0 vector is in set, closed under vector addition, closed under scalar multiplication)...but this is a LOT OF WORK. It is EASIER to use the theorem used in class. Show this is in the span of a set of vectors. -- ■ 0 W DELL

Elementary Linear Algebra (MindTap Course List)
Elementary Linear Algebra (MindTap Course List)
8th Edition
ISBN:9781305658004
Author:Ron Larson
Publisher:Ron Larson
Chapter5: Inner Product Spaces
Section5.CR: Review Exercises
Problem 41CR: Let B={(0,2,2),(1,0,2)} be a basis for a subspace of R3, and consider x=(1,4,2), a vector in the...
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F1 ! 1 о A To be a subspace H of V, the following must all be true i. 0 (the zero vector) is in H ii. It is closed under addition. i.e. if u, v are in H, then u + v are in H iii. It is closed under scalar multiplication. i.e if u is in H, then for any constant c, cu is in H Show the following IS a subspace using Theorem 1. The set of all vectors of the form :{(2a, a - b, 4b): a, b = R}: 2a A a b 4b This can also be written as To help you through this follow these steps. Note: You COULD prove all three conditions are met above (0 vector is in set, closed under vector addition, closed under scalar multiplication)...but this is a LOT OF WORK. It is EASIER to use the theorem used in class. Show this is in the span of a set of vectors. To do this bal. F2 NP @ 2 W S F3 A # 3 E D I F4 KA $ 4 F5 R F % 5 F6 T G A 6 F7 DELL L F8 & 7 a U F9 * 8 00 F10 ( 9 F11 O O F12 P T
Transcribed Image Text:F1 ! 1 о A To be a subspace H of V, the following must all be true i. 0 (the zero vector) is in H ii. It is closed under addition. i.e. if u, v are in H, then u + v are in H iii. It is closed under scalar multiplication. i.e if u is in H, then for any constant c, cu is in H Show the following IS a subspace using Theorem 1. The set of all vectors of the form :{(2a, a - b, 4b): a, b = R}: 2a A a b 4b This can also be written as To help you through this follow these steps. Note: You COULD prove all three conditions are met above (0 vector is in set, closed under vector addition, closed under scalar multiplication)...but this is a LOT OF WORK. It is EASIER to use the theorem used in class. Show this is in the span of a set of vectors. To do this bal. F2 NP @ 2 W S F3 A # 3 E D I F4 KA $ 4 F5 R F % 5 F6 T G A 6 F7 DELL L F8 & 7 a U F9 * 8 00 F10 ( 9 F11 O O F12 P T
=1 X This can also be written as To help you through this follow these steps. Note: You COULD prove all three conditions are met above (0 vector is in set, closed under vector addition, closed under scalar multiplication)...but this is a LOT OF WORK. It is EASIER to use the theorem used in class. Show this is in the span of a set of vectors. To do this I will help you through the process. The vector can be written as 2a b 4-8-8 a 46 b F2 N 4 @ 2 MY F3 #m 2a A b a 4b 3 2 W E F4 KA $ 4 F5 R ► 11 % 5 F6 T A 6 F7 DELL Y F8 & 7 F9 * 00 8 S F10 9 F11 F1
Transcribed Image Text:=1 X This can also be written as To help you through this follow these steps. Note: You COULD prove all three conditions are met above (0 vector is in set, closed under vector addition, closed under scalar multiplication)...but this is a LOT OF WORK. It is EASIER to use the theorem used in class. Show this is in the span of a set of vectors. To do this I will help you through the process. The vector can be written as 2a b 4-8-8 a 46 b F2 N 4 @ 2 MY F3 #m 2a A b a 4b 3 2 W E F4 KA $ 4 F5 R ► 11 % 5 F6 T A 6 F7 DELL Y F8 & 7 F9 * 00 8 S F10 9 F11 F1
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