Suppose V and W are subspaces of a vector space
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Suppose V and W are subspaces of a vector space

[From: ] [author: ] [Date: 11-07-01] [Hit: ]
and x and y are in W.Since V and W are each closed under addition, x + y is in V and x + y is in W.So V and W have the element x + y in common, and so x + y is in V intersect W.Therefore,......
Show that the intersection of vector spaces V and W are also a subspace

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By definition of subspace, we are given that V and W each contain 0 and are closed under addition and scalar multiplication, and we need to prove that these same facts are true for V intersect W.

i) Since V and W each contain 0, they have the element 0 in common. Therefore, V intersect W contains 0.

ii) Suppose x and y are both in V intersect W. This means that V and W have both x and y in common, so x and y are in V, and x and y are in W. Since V and W are each closed under addition, x + y is in V and x + y is in W. So V and W have the element x + y in common, and so x + y is in V intersect W. Therefore, V intersect W is closed under addition.

iii) Suppose c is any scalar, and x is in V intersect W. This means that V and W have x in common, so x is in V, and x is in W. Since V and W are each closed under scalar multiplication, cx is in V and cx is in W. So V and W have the element cx in common, and so cx is in V intersect W. Therefore, V intersect W is closed under scalar multiplication.

We conclude that V intersect W is a subspace. QED

By the way, the union of two subspaces might not be a subspace! Consider the coordinate plane. The x-axis and the y-axis are each subspaces, but their union is not a subspace: for example, (1, 0) and (0, 1) are in this union, but the sum (1, 0) + (0, 1) = (1, 1) is not in the union.

Lord bless you today!

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The condition of a set being a subspace is equivalent to:

1) The set is closed under vector addition
2) The set is closed under scalar multiplication (by any scalar)
3) The set contains the 0 vector

Checking 1), if u and v are in the intersection, then u and v are in both V and W. The above conditions hold for V and W, so u + v is in both V and W. Thus u + v is also in the intersection, and the intersection is closed under vector addition.

Checking 2) is very similar. Given u in V and W, and a scalar c, it follows from V and W being subspaces that cu is in both V and W, and hence their intersection.

Checking 3) is trivial; 0 lies in both V and W, and hence the intersection.
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