# 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

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

${\displaystyle x=\sum _{i=1}^{n}x_{n}}$
2. When xn and yn is within Hn

${\displaystyle \sum _{i=1}^{n}x_{n}=\sum _{i=1}^{n}y_{n}}$

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

${\displaystyle x=\sum _{i=1}^{n}x_{n}=0}$

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

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

When xn are elements of Hn, then

${\displaystyle x=\sum _{i=1}^{n}x_{n}=0}$
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.