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). The endpoints of all such vectors lie on the line y = 3 x in the x‐y plane.
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