(a) Can you construct an example of a subset S ⊂ℝ^2 with the property that c 𝐯 ∈S for all c ∈ℝ,

(a) Can you construct an example of a subset S ⊂ℝ^2 with the...

Key Concepts

Counterexample Construction

Constructing counterexamples involves finding subsets that satisfy one of the closure properties (either under scalar multiplication or vector addition) but fail to satisfy the other. Such examples illustrate the necessity of both conditions for a subset to be a subspace. They help in understanding that satisfying only one of the closure conditions is insufficient to guarantee the structure of a subspace.

Subspace

A subspace of a vector space is a subset that is itself a vector space under the same operations. This means it must contain the zero vector and must be closed under both vector addition and scalar multiplication. Failure to meet any of these axioms means the subset is not a subspace, even if it satisfies one of the closure properties.

Closure under Scalar Multiplication

Closure under scalar multiplication means that if a vector is in the subset, then any scalar multiple of that vector is also in the subset. This property is one requirement for a set to be a subspace, but alone it is not sufficient since the set must also be closed under addition to be considered a subspace.

Closure under Addition

Closure under vector addition means that the sum of any two vectors in the subset is also contained in the subset. Like closure under scalar multiplication, this is a key property for subspaces; however, a subset that is closed under addition but not under scalar multiplication will also fail to qualify as a subspace.

Transcript

00:01 In part a, we're asked to show that the union of s1 and s2 is not a subspace of v.

00:08 So the union of s1 and s2 is the set which contains all vectors v, that are either in s1 or s2.

00:18 We can prove this by setting up a contradiction, and we'll define two vectors v1 and v2, and we'll let b1 be in s1, but not s2.

00:31 So v1 is in s1, but not s2.

00:39 And v2, let that be in s2, but not s1.

00:49 Now, if the union of s1 and s2 is a subspace, it should be closed under addition and scatter multiplication.

00:58 So then we should be able to write that the addition of v1 and v2 is in s1.

01:09 And we should be able to write that multiplying v1 by a scalar, say minus 1 times v1, that should also be an s1.

01:24 So then we can combine these, and we can write that v1 plus v2 minus this v1, and then we can simplify that.

01:38 So we have v1 minus v1, which is the zero vector, plus v2, and that should be.

01:46 Just equal to v2 and based on our notation that it should also be in s1 if it's a subspace but we defined at the start that v2 is in s2 but not s1 and we just showed that it's in s1 so that's a contradiction which means that the union of s1 and s2 is not a subspace moving on to part b in part b we have to show that the inner section of s1 and s2 is a subspace of v.

02:26 So we have that the intersection of s1 and s2 is the set of all vectors v that are in both s1 and s2.

02:36 Again, we'll start by defining two vectors.

02:39 We'll say v1 and v2.

02:45 So if this v1 and v2 are in the intersection of s1 and s2, then by definition that v1 and v2 could also be in s1 and the v1 and v2 vectors should be in s2.

03:12 And since s1 and s2 are subspaces, they should be closed under addition, meaning that v1 plus v2, v1 plus v2 is in s1, and that v1 plus v2 is in s2, which implies that v1 plus v2 is in the intersection of s1 and s2.

03:58 And that's what we needed to show that it's closed under addition.

04:04 It should also be closed under scalar multiplication.

04:08 And so we can have, again, our v1 vector, which is in the intersection of s1 and s2, so that we have some scalar k that we can multiply by v1.

04:23 And that will be in s1 and then again multiplying that scalar by v1 and that should be in s2 and so then we can write that the scalar k times the vector v1 is in the intersection of s1 and s2 so it is closed under scalar multiplication so it's closed under scalar multiplication and addition so the intersection of s1 and s2 is a subspace of v...

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