# Collection of all linear transformation is a vector space

• Feb 1st 2008, 06:18 PM
Collection of all linear transformation is a vector space
Let V and W be vector space over a field F, and let T,U: V -> W be linear. Prove that the collection of all linear transformation from V to W is a vector space over F.

Proof.

Let F denote the set of all linear transformations from V to W.
Let T,U,S be in F, and let t,u,s be in V.

Is this the right way to do it? Thanks
• Feb 2nd 2008, 02:46 PM
ThePerfectHacker
Quote:

Let V and W be vector space over a field F, and let T,U: V -> W be linear. Prove that the collection of all linear transformation from V to W is a vector space over F.

Proof.

Let F denote the set of all linear transformations from V to W.
Let T,U,S be in F, and let t,u,s be in V.

There is a lot of stuff to show here. Let $\displaystyle T_1,T_2$ be linear transformations $\displaystyle V\mapsto W$. Define $\displaystyle T_1 + T_2$ to be the sum (as a function sum) of the linear transformations. Then $\displaystyle (T_1 + T_2 )(\bold{u}+\bold{v}) = T_1 (\bold{u}) + T_1(\bold{v}) + T_2(\bold{u})+T(\bold{v}) = (T_1+T_2)(\bold{u}) + (T_1+T_2)(\bold{v})$. Futhermore, $\displaystyle (T_1+T_2)(k\bold{u}) = T_1(k\bold{u}) + T_2(k\bold{u}) = kT_1(\bold{u})+kT_2(\bold{u}) = k(T_1+T_2)(\bold{u})$. Thus, the set of all linear transformations is closed under this definition. There are more things we need to prove. We need to show that this set $\displaystyle S$ forms an abelian group under this operation $\displaystyle +$. We know that $\displaystyle T_1 + T_2 = T_2 + T_1$ and $\displaystyle (T_1+T_2)+T_3 = T_1 + (T_2 + T_3)$ and $\displaystyle T_1 + \hat{0} = \hat{0}+T_1 = T_1$ (where $\displaystyle \hat{0}$ is the trivial linear transformation). And if $\displaystyle -T_1$ is the negative linear transformation then $\displaystyle T_1 + (-T_1) = \hat{0}$. Which shows $\displaystyle S$ is an abelian group. To show that $\displaystyle S$ is a vector space we need to confirm more things. Can you finish that?