# Vector Space Decomposition

The decomposition of a vector space involves breaking down each element of the space into the sum of two vector subspaces.

Given a vector space V over the field K, and two vector subspaces A and B, if A+B=V, then for every v∈V, there exist two elements a∈A and b∈B such that a+b=v.

$$ \text{If } A+B=V $$

$$ \text{then } \forall v \in V \; \exists \; a \in A , b \in B : a+b=v. $$

This decomposition is unique if the vector subspaces are complementary subspaces.

$$ A \oplus B = V $$

## Proof

__Uniqueness of Decomposition__

Assume, for the sake of contradiction, that the decomposition is not unique and the subspaces A and B are complementary.

Let v be any element of the vector space.

$$ v \in V $$

By contradiction, assume there are two equal sums to v:

$$ a + b = v $$

$$ a' + b' = v $$

where

$$ a,a' \in A $$

$$ b,b' \in B $$

If both sums equal v, then they are also equal to each other:

$$ a+b = a'+b' $$

Group the elements of A on the left and those of B on the right:

$$ a+a' = b+b' $$

Since A is a vector subspace, the sum a+a' is an element of A.

Similarly, as B is a vector subspace, the sum b+b' is an element of B.

$$ a+a' \in A $$

$$ b+b' \in B $$

Thus, v is within the intersection A ∩ B.

$$ v \in A \cap B $$

Knowing that in complementary subspaces, the intersection A ∩ B is equal to the null vector { 0 }:

$$ A \cap B = \{ 0 \} $$

It follows that:

$$ v = 0 $$

$$ a+a'=0 $$

$$ b+b'=0 $$

Therefore:

$$ a'=a $$

$$ b'=b $$

In the two sums a+b and a'+b', the elements are identical.

Hence, the decomposition is unique.