Linear Algebra/Direct Sum

Let V be a vector space, and let H1, H2, H3, ..., Hn are all subspaces of V. V is defined to be a direct sum of H1, H2, H3, ..., Hn when For every x within V, there is xn within Hn such that When xn and yn is within Hnimplies that xn=yn. The second condition can easily be proven to be equivalent to the following statement: When xn are elements of Hn, then implies that all xn are also equal to 0. Because of the second condition, the intersection of any two of the subspaces involved in a direct sum is the single element {0} where 0 is the 0 vector. This implies that all n dimension spaces are a direct sum of n one-dimension subspaces. Theorem If V is a vector space, then for any subspace H, there exists a subspace G such that V is a direct sum of H and G. Proof Let {e1, e2,..., ek} be a basis of H. This can be extended to a basis of V, say, {e1, e2,..., en}. Then the space G spanned by {ek+1, ek+2,..., en} is such that V is a direct sum of H and G. General Sums Given a vector space V, and H1, H2, H3, ..., Hn are all subspaces of V, then V is a sum of H1, H2, H3, ..., Hn are all subspaces of V when all elements of V can be expressed as a sum of elements in H1, H2, H3, ..., Hn. Exercises Prove that the second condition is equivalent to the following statement:When xn are elements of Hn, then Prove that the intersection of of any two of the subspaces involved in a direct sum is the single element {0} where 0 is the 0 vector.