# Linear Algebra/Direct Sum

Let V be a vector space, and let H_{1}, H_{2}, H_{3}, ..., H_{n} are all subspaces of V. V is defined to be a direct sum of H_{1}, H_{2}, H_{3}, ..., H_{n} when

- For every x within V, there is x
_{n}within H_{n}such that - When x
_{n}and y_{n}is within H_{n}

implies that x_{n}=y_{n}.

The second condition can easily be proven to be equivalent to the following statement:

When x_{n} are elements of H_{n}, then

implies that all x_{n} 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 | edit source]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 | edit source]Let {e_{1}, e_{2},..., e_{k}} be a basis of H. This can be extended to a basis of V, say, {e_{1}, e_{2},..., e_{n}}.
Then the space G spanned by {e_{k+1}, e_{k+2},..., e_{n}} is such that V is a direct sum of H and G.

## General Sums

[edit | edit source]Given a vector space V, and H_{1}, H_{2}, H_{3}, ..., H_{n} are all subspaces of V, then V is a sum of H_{1}, H_{2}, H_{3}, ..., H_{n} are all subspaces of V when all elements of V can be expressed as a sum of elements in H_{1}, H_{2}, H_{3}, ..., H_{n}.

## Exercises

[edit | edit source]- Prove that the second condition is equivalent to the following statement:

When x_{n}are elements of H_{n}, 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.