Then p + q = ( p 1+ q 1, p 2+ q 2) since p 1 = p 2 and q 1 = q 2, then p 1 + q 1 = p 2 + q 2, so p + q is an element of W. Let p = ( p 1, p 2) and q = ( q 1, q 2) be elements of W, that is, points in the plane such that p 1 = p 2 and q 1 = q 2.Take W to be the set of points ( x, y) of R 2 such that x = y. Thus, c u is an element of W too.Įxample II: Let the field be R again, but now let the vector space be the Euclidean geometry R 2. Given u in W and a scalar c in R, if u = ( u 1, u 2,0) again, then c u = ( cu 1, cu 2, c0) = ( cu 1, cu 2,0).Given u and v in W, then they can be expressed as u = ( u 1, u 2,0) and v = ( v 1, v 2,0).Take W to be the set of all vectors in V whose last component is 0. One can also argue that since W is nonempty, there is an element x in W, and 0 is in the field K so and therefore property 1 is satisfied.Įxample I: Let the field K be the set R of real numbers, and let the vector space V be the Euclidean space R 3. Since W is nonempty, there is an element x in W, and is in W, so property 1 is satisfied. By property 3, -w is in W whenever w is, and it follows that W is closed under subtraction as well. By the closure of W under scalar multiplication (specifically by 0 and -1), axioms 3 and 4 of a vector space are satisfied.Ĭonversely, if W is subspace of V, then W is itself a vector space under the operations induced by V, so properties 2 and 3 are satisfied. Since elements of W are necessarily elements of V, axioms 1, 2 and 5-8 of a vector space are satisfied. Looking at the definition of a vector space, we see that properties 2 and 3 above assure closure of W under addition and scalar multiplication, so the vector space operations are well defined.
Proof: Firstly, property 1 ensures W is nonempty. If u is an element of W and c is a scalar from K, then the scalar product c u is an element of W.If u and v are elements of W, then any linear combination of u and v is an element of W.Then W is a subspace if and only if it satisfies the following three conditions: Theorem: Let V be a vector space over the field K, and let W be a subset of V.
Instead, we can prove a theorem that gives us an easier way to show that a subset of a vector space is a subspace. To use this definition, we don't have to prove that all the properties of a vector space hold for W. If W is a vector space itself, with the same vector space operations as V has, then it is a subspace of V. As usual, we call elements of V vectors and call elements of K scalars. Let K be a field (such as the field of real numbers), and let V be a vector space over K.
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