For example, the set A in Example 1 above could not be a subspace of R 2 because it does not contain the vector 0 = (0, 0).
If the set does not contain the zero vector, then it cannot be a subspace. This result can provide a quick way to conclude that a particular set is not a Euclidean space. By selecting 0 as the scalar, the vector 0 v, which equals 0, must be in V. Since V is a subspace, it must be closed under scalar multiplication. This proves that C is a subspace of R 4.Įxample 4: Show that if V is a subspace of R n, then V must contain the zero vector.įirst, choose any vector v in V. Is in C, establishing closure under scalar multiplication. Satisfies the conditions for membership in C, verifying closure under addition. So let u = ( u 1, 0, u 3, −5 u 1) and v = ( v 1, 0, v 3, −5 v 1) be arbitrary vectors in C. 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. Choosing particular vectors in C and checking closure under addition and scalar multiplication would lead you to conjecture that C is indeed a subspace. Since B is not closed under addition, B is not a subspace of R 3.Įxample 3: Is the following set a subspace of R 4?įor a 4‐vector to be in C, exactly two conditions must be satisfied: Namely, its second component must be zero, and its fourth component must be −5 times the first. However, note that while u = (1, 1, 1) and v = (2, 4, 8) are both in B, their sum, (3, 5, 9), clearly is not. In order for a sub set of R 3 to be a sub space of R 3, both closure properties (1) and (2) must be satisfied. Įxample 2: Is the following set a subspace of R 3? Therefore, the set A is not closed under addition, so A cannot be a subspace. In order for a vector v = ( v 1, v 2 to be in A, the second component ( v 2) must be 1 more than three times the first component ( v 1). For instance, both u = (1, 4) and v = (2, 7) are in A, but their sum, u + v = (3, 11), is not. In the present case, it is very easy to find such a counterexample. If a counterexample to even one of these properties can be found, then the set is not a subspace. To establish that A is a subspace of R 2, it must be shown that A is closed under addition and scalar multiplication. The set V = is a Euclidean vector space, a subspace of R 2.Įxample 1: Is the following set a subspace of R 2?
The sum of any two elements in V is an element of V.Įvery scalar multiple of an element in V is an element of V.Īny subset of R nthat satisfies these two properties-with the usual operations of addition and scalar multiplication-is called a subspace of R nor a Euclidean vector space. Thus, the elements in V enjoy the following two properties: The set V is therefore said to be closed under scalar multiplication. In fact, every scalar multiple of any vector in V is itself an element of V. Next, consider a scalar multiple of u, say, The set V is therefore said to be closed under addition.
In fact, it can be easily shown that the sum of any two vectors in V will produce a vector that again lies in V. Is also a vector in V, because its second component is three times the first. Now, choose any two vectors from V, say, u = (1, 3) and v = (‐2, ‐6). Therefore, ?v_1? and ?v_2? are in ?V?, but ?v_1+v_2? is not in ?V?, which proves that ?V? is not closed under addition, which means that ?V? is not a subspace.The endpoints of all such vectors lie on the line y = 3 x in the x‐y plane. The components of ?v_1+v_2=(1,1)? do not have a product of ?0?, because the product of its components are ?(1)(1)=1?. And we know about three-dimensional space, ?\mathbb?
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