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.

Commutativity of addition: (T+U)(t+u)=T(t)+U(u)=U(u)+T(t)=(U+T)(u+t)

Is this the right way to do it? Thanks

Quote:

Originally Posted by tttcomrader

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.

Commutativity of addition: (T+U)(t+u)=T(t)+U(u)=U(u)+T(t)=(U+T)(u+t)

Is this the right way to do it? Thanks

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

Yes, sorry I was lazy to type the other stuff out, I know that I have to prove all those, but I wasn't sure if I define the linear transformations right.

Thanks!