Each of the given sets below is not a subspace because at least one of the 3 properties of being a subspace is failing. • Property 1: the zero vector belongs to the set. • Property 2: the set is closed under vector addition. • Property 3: the set is closed under scalar multiplication. In each case indicate the number of the first condition that is not satisfied, starting checking from condition 1 For example, if a set contains the zero vector but it is not closed under vector addition , you should fill in 2 (because the first condition that is not satisfied is condition 2.) wi (a) W = w2 E R : wi = w2 } is not a subspace and the first property to fail is property 2 {E] Wi (b) W : E R : wi <0, w2 2 0 } is not a subspace and the first property to fail is property w2

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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Each of the given sets below is not a subspace because at least one of the 3 properties of being a subspace is
failing.
• Property 1: the zero vector belongs to the set.
• Property 2: the set is closed under vector addition.
Property 3: the set is closed under scalar multiplication.
In each case indicate the number of the first condition that is not satisfied, starting checking from condition 1.
For example, if a set contains the zero vector but it is not closed under vector addition , you should fill in 2
(because the first condition that is not satisfied is condition 2.)
Wi
(a) W =
E R : wi = w2 is not a subspace and the first property to fail is property
w2
2
Wi
(b) W =
E R : wi <0, w2 > 0 } is not a subspace and the first property to fail is property
Transcribed Image Text:Each of the given sets below is not a subspace because at least one of the 3 properties of being a subspace is failing. • Property 1: the zero vector belongs to the set. • Property 2: the set is closed under vector addition. Property 3: the set is closed under scalar multiplication. In each case indicate the number of the first condition that is not satisfied, starting checking from condition 1. For example, if a set contains the zero vector but it is not closed under vector addition , you should fill in 2 (because the first condition that is not satisfied is condition 2.) Wi (a) W = E R : wi = w2 is not a subspace and the first property to fail is property w2 2 Wi (b) W = E R : wi <0, w2 > 0 } is not a subspace and the first property to fail is property
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