Q: what is the dimension of the image of linear mapping $L(u,v,w)=(u-v,v-w,w-u)$? let $U,V,W$ be subspace of finite dimensional vectorspace P.
and define linear map $L$ like this.
$$L : U \times V \times W \rightarrow P^3$$
more specifically, $$L(u,v,w)=(u-v,v-w,w-u)$$
obviously, $$Im(L)=(U+V)\times (V+W)\times (W+U)$$ 
(when
$Im(L)=\{p\in P^3 : ~\exists$ $(u,v,w)$$\in U\times V\times W$  such that $L(u,v,w)=p\}$
$U+V=\{u+v : u\in U, v\in V\}$
$U\times V = \{(u,v) : u\in U, v\in V\}$)
so i thought following equation is true: $$\dim\text{Im}(L)= \dim(U+V)+\dim(V+W) +\dim(W+U)$$
simply using $dim(V\times W)=dim(V)+dim(W)$ two times.. however, it meets contradiction if i try to use the following relation.  $$dim(U \times V \times W)=dimkerL + dimImL$$
applying $dimkerL=dim(U\cap V  \cap W$), i get $$dim(U)+dim(V)+dim(W)-dim(U\cap V)-dim(V\cap W)-dim(W\cap U)+dim(U\cap V\cap W)=0$$
which doesn't make a sense at all. 
i guess the problem is from dimension of the image. (thanks for your help)

A: Obviously $$\text{Im}(L) \subseteq (U+V)\times (V+W)\times (W+U)$$ holds, but not necessarily
$$\text{Im}(L)=(U+V)\times (V+W)\times (W+U)$$
e.g: $U=V=W=P$ then $\text{Im(L)} \ne P^3$, because $(0,0,1)$ not in $\text{Im}(L)$.
