Linear Algebra/Direct Sum

From Wikibooks, open books for an open world
Jump to navigation Jump to search

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

  1. For every x within V, there is xn within Hn such that

  2. When xn and yn is within Hn



    implies 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[edit]

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[edit]

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[edit]

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[edit]

  1. Prove that the second condition is equivalent to the following statement:

    When xn are elements of Hn, then

  2. 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.