TITLE: Every element of $U + V + W$ can be expressed uniquely in the form $u + v + w$
QUESTION [5 upvotes]: Suppose that $U$, $V$ and $W$ are subspaces of some given vector space. With the obvious definition of $U + V + W$, show that every element of $U + V + W$ can be expressed uniquely in the form $u + v + w$, where $u ∈ U$, $v ∈ V$ and $w ∈ W$, if and only if 
$$\dim(U + V + W) = \dim(U) + \dim(V) + \dim(W)$$
1) Say every element of $U + V + W$ can be expressed uniquely in the form $u + v + w$. Let $u_1,...,u_n$, $v_1,...,v_m$, $w_1,...,w_q$ be respective bases. Then every element in $U + V + W$ can be expressed uniquely as
$$(\alpha_1u_1+...+\alpha_nu_n)+(\beta_1v_1+...+\beta_mv_m)+(\gamma_1w_1+...+\gamma_qw_q)$$
showing that 
$$\dim(U + V + W) = \dim(U) + \dim(V) + \dim(W)$$
2) How do I prove the converse? Hints are appreciated (also verification, that what I did, is not useless :) ). Thanks!

REPLY [3 votes]: This is a different approach, using a classical idea in linear algebra : expressing subspaces as images or kernels of linear maps.
First i will recall a few things about the cartesian product of vector spaces.

We know we can define $U\times V\times W $ to be the set of triples
  with first element in $U$, second in $V$, and third in $W$.  This set
  can be given a vector space structure by stating $$(u,v,w)+(u',v',w')
> = (u+u', v+v', w+w') ; \lambda(u,v,w) = (\lambda u,\lambda v,\lambda w)$$ We can show that if $(u_1, ..., u_m) ; (v_1, ..., v_n); (w_1,..., w_p)$ are basis of $U,V,W$ respectively, then $[(u_1,0,0), ...,(u_m,0,0) ; (0,v_1,0), ..., (0,v_n,0); (0,0,w_1), ..., (0,0,w_p)]$ is
  a basis of $U\times V\times W $.
Thus $$\dim(U\times V\times W) = \dim(U) + \dim(V) + \dim(W)$$

Let $$\phi : \cases{ U\times V\times W \rightarrow E \\(u,v,w) \mapsto u+v+w}$$
This is a linear map, it's the function that "builds" all the elements of $U + V + W$. Actually, we have $\text{im}(\phi) = U + V + W$. Remark that it automatically means that $U + V + W$ is a subspace of $E$ !
Asking for the uniqueness of the expression of elements of $U+V+W$ is the same as asking this function to be injective. This is the case if and only if $\ker(\phi) = {0}$, which by the fundamental theorem of linear algebra, is the cas if and only if $$\dim(U\times V\times W ) = \dim(\text{im}(\phi)) \\ \Leftrightarrow \dim(U\times V\times W) = \dim(U+V+W)\\ \Leftrightarrow \dim(U) + \dim(V) + \dim(W) = \dim(U+V+W)$$