Linear Algebra/Direct Sum

Let V be a vector space, and let H₁, H₂, H₃, ..., H_n are all subspaces of V. V is defined to be a direct sum of H₁, H₂, H₃, ..., H_n when

1. For every x within V, there is x_n within H_n such that
   
   $x=\sum _{i=1}^{n}x_{n}$
2. When x_n and y_n is within H_n
   
   $\sum _{i=1}^{n}x_{n}=\sum _{i=1}^{n}y_{n}$
   
   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

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

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

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,..., e_k} be a basis of H. This can be extended to a basis of V, say, {e₁, e₂,..., 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

Given a vector space V, and H₁, H₂, H₃, ..., H_n are all subspaces of V, then V is a sum of H₁, H₂, H₃, ..., H_n are all subspaces of V when all elements of V can be expressed as a sum of elements in H₁, H₂, H₃, ..., H_n.

Exercises

1. Prove that the second condition is equivalent to the following statement:
   
   When x_n are elements of H_n, then
   
   $x=\sum _{i=1}^{n}x_{n}=0$
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.