Intersection and Union of Subspaces in Linear Algebra, Assignments of Linear Algebra

A proof that the intersection of two subspaces is a subspace of the vector space, using the properties of subspaces. It also includes an example showing that the union of two subspaces is not necessarily a subspace, with counterexamples given in r2. Students of linear algebra will find this document useful for understanding the concepts of subspaces and their operations.

Typology: Assignments

Pre 2010

Uploaded on 08/19/2009

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MATH 214
Linear Algebra
Week 8
Homework (Due 3/24)
Date Section Assignment
3/7 4.1 1, 2, 6, 9, 12, 21, 24, 24, 26, 28, 31
Bonus Question
1. Let Hand Kbe subspaces of a vector space V. The intersection of Hand K, written as
HK, is the set of vectors that belong to both Hand K. Show that HKis a subspace
of V. Give an example in R2to show that the union of two subspaces is not, in general, a
subspace.
Proof. We have that 0Hbecause His a subspace of V, and 0Kbecause Kis a subspace
of V. Hence 0HK.
Suppose u,vHK. Then u,vH. Hence u+vHbecause His a subspace of V.
Similarly, u,vK. Hence u+vKbecause Kis a subspace of V. Thus u+vHK.
Suppose uHKand cis a scalar. Then uHand uK. Hence cuHand cuK
because Hand Kare subspaces of V. Thus cuHK.
Therefore HKis a subspace of V.
To show that the union of two subspaces in not necessarily a subspace, let H= Span  1
0
and K= Span  0
1. Then 1
0HKand 0
1HK, but
1
0+0
1=1
1/HK.
Other counterexamples will work, but Hand Kmust be subspaces of R2, and you must find
specific vectors uand vsuch that u,vHKand u+v/HK. Is it possible to find a
vector uand a scalar csuch that uHKbut cu/HK?
1

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MATH 214

Linear Algebra

Week 8

Homework (Due 3/24)

Date Section Assignment 3/7 4.1 1, 2, 6, 9, 12, 21, 24, 24, 26, 28, 31

Bonus Question

  1. Let H and K be subspaces of a vector space V. The intersection of H and K, written as H ∩ K, is the set of vectors that belong to both H and K. Show that H ∩ K is a subspace of V. Give an example in R^2 to show that the union of two subspaces is not, in general, a subspace.

Proof. We have that 0 ∈ H because H is a subspace of V , and 0 ∈ K because K is a subspace of V. Hence 0 ∈ H ∩ K. Suppose u, v ∈ H ∩ K. Then u, v ∈ H. Hence u + v ∈ H because H is a subspace of V. Similarly, u, v ∈ K. Hence u + v ∈ K because K is a subspace of V. Thus u + v ∈ H ∩ K. Suppose u ∈ H ∩ K and c is a scalar. Then u ∈ H and u ∈ K. Hence cu ∈ H and cu ∈ K because H and K are subspaces of V. Thus cu ∈ H ∩ K. Therefore H ∩ K is a subspace of V.

To show that the union of two subspaces in not necessarily a subspace, let H = Span

{[

]}

and K = Span

{[

]}

. Then

[

]

∈ H ∪ K and

[

]

∈ H ∪ K, but

[ 1 0

]

[

]

[

]

∈ / H ∪ K.

Other counterexamples will work, but H and K must be subspaces of R^2 , and you must find specific vectors u and v such that u, v ∈ H ∪ K and u + v ∈/ H ∪ K. Is it possible to find a vector u and a scalar c such that u ∈ H ∪ K but cu ∈/ H ∪ K?