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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.
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Week 8
Date Section Assignment 3/7 4.1 1, 2, 6, 9, 12, 21, 24, 24, 26, 28, 31
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
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?