# Solution space of linear homogeneous PDE forms a vector space?

1. Sep 14, 2009

### kingwinner

1. The problem statement, all variables and given/known data
Claim:
The solution space of a linear homogeneous PDE Lu=0 (where L is a linear operator) forms a "vector space".

Proof:
Assume Lu=0 and Lv=0 (i.e. have two solutions)
(i) By linearity, L(u+v)=Lu+Lv=0
(ii) By linearity, L(au)=a(Lu)=(a)(0)=0
=> any linear combination of the solutions of a linear homoegenous PDE solves the PDE
=> it forms a vector space

2. Relevant equations
N/A

3. The attempt at a solution and comments
Now, I don't understand why ONLY by proving (i) and (ii) alone would lead us to conclude that it is a vector space. There are like TEN properties that we have to prove before we can say that it is a vector sapce, am I not right?
Are there any theorem or alternative definition that they have been using?

Thanks!

Last edited: Sep 14, 2009
2. Sep 14, 2009

### lanedance

maybe consider the other axioms and test them, some may have been considered more bovious, but always worth checking

3. Sep 14, 2009

### Office_Shredder

Staff Emeritus
Other things are satisfied by the definition of functions, for example u+v=v+u is obvious and not a property of being a solution of a PDE. It may be implicitly assumed that you're talking about the subset of the vector space of all functions from U to V (where U and V are Rm and Rn such that you'd be looking for solutions with that domain and codomain). In that case you only have to prove the two subspace properties