A linear subspace of dimension 2 is a vector plane. If the answer to both of these questions is yes, then \(U\) is a vector space. A linear subspace of dimension 1 is a vector line. If we multiply any vector in \(U\) by any constant, do we end up with a vector in \(U\)? Suppose V is a vector space and U is a nonempty family of linear subspaces of V.Well, if A and -A are both in our subspace, then so must A (-A) which is of course, the zero vector. If we add any two vectors in \(U\), do we end up with a vector in \(U\)? After all, if A is a vector in our subspace, and so is -1A (from rule 2) then the subspace must also include a zero vector because if vector addition holds, then the sum of any two vectors in our subspace must ALSO be in our subspace.That is, if we have some set \(U\) of vectors that come from some bigger vector space \(V\), to check if \(U\) itself forms a smaller vector space we need check only two things: We can use this theorem to check if a set is a vector space. Note that the requirements of the subspace theorem are often referred to as "closure''. Definition: A Subspace of is any set H that contains the zero vector is closed under vector addition and is closed under scalar multiplication. The span of those vectors is the subspace. In summary, the vectors that define the subspace are not the subspace. All of the other eight properties is true in \(U\) because it is true in \(V\). The subspace defined by those two vectors is the span of those vectors and the zero vector is contained within that subspace as we can set c1 and c2 to zero. We know that the additive closure and multiplicative closure properties are satisfied. That is, we need to show that the ten properties of vector spaces are satisfied. \) in \(U\) and all constants \(\mu, \nu\), then \(U\) is a vector space.
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