Therefore the intersection is a subspace. Sign up to join this community. The best answers are voted up and rise to the top. Stack Overflow for Teams — Collaborate and share knowledge with a private group. Create a free Team What is Teams? Learn more. Ask Question. Asked 9 years, 8 months ago. Active 5 years, 5 months ago. Viewed 67k times. Here's my logic - A. Closed under multiplication and addition so it is a subspace. So where am I going wrong? Community Bot 1.
Add a comment. Active Oldest Votes. F Not a subspace. Bill Cook Bill Cook Patrick Patrick 1, 8 8 silver badges 10 10 bronze badges. Bumblebee Bumblebee Sign up or log in Sign up using Google. Sign up using Facebook. Sign up using Email and Password. Post as a guest Name. Email Required, but never shown. Since B is not closed under addition, B is not a subspace of R 3.
Choosing particular vectors in C and checking closure under addition and scalar multiplication would lead you to conjecture that C is indeed a subspace. However, no matter how many specific examples you provide showing that the closure properties are satisfied, the fact that C is a subspace is established only when a general proof is given. Then their sum,. Finally, if k is a scalar, then. This proves that C is a subspace of R 4.
Example 4 : Show that if V is a subspace of R n , then V must contain the zero vector. First, choose any vector v in V. Since V is a subspace, it must be closed under scalar multiplication.
By selecting 0 as the scalar, the vector 0 v , which equals 0 , must be in V. This result can provide a quick way to conclude that a particular set is not a Euclidean space. If the set does not contain the zero vector, then it cannot be a subspace.
It is important to realize that containing the zero vector is a necessary condition for a set to be a Euclidean space, not a sufficient one. That is, just because a set contains the zero vector does not guarantee that it is a Euclidean space for example, consider the set B in Example 2 ; the guarantee is that if the set does not contain 0 , then it is not a Euclidean vector space. As always, the distinction between vectors and points can be blurred, and sets consisting of points in R n can be considered for classification as subspaces.
Example 5 : Is the following set a subspace of R 2? Figure 1. The set D is closed under addition since the sum of nonnegative numbers is nonnegative. This implies that. However, D is not closed under scalar multiplication.
If x and y are both positive, then x, y is in D , but for any negative scalar k ,. Therefore, D is not a subspace of R 2. As illustrated in Figure , this set consists of all points in the first and third quadrants, including the axes:.
Figure 2. The proof of this last statement follows immediately from the condition for membership in E. A point is in E if the product of its two coordinates is nonnegative.
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