Every element of $U + V + W$ can be expressed uniquely in the form $u + v + w$ 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!
 A: People are really missing the point of the exercise. If I have two (finite-dimensional) subspaces $S,T$ in some larger vector space, the following are equivalent:
(1) $\dim (S + T) = \dim S + \dim T$
(2) $S \cap T = \{ \; \vec{0} \; \} $  
(3) every vector in $S+T$ has only one expression as $s+t,$ with $s \in S, \; t \in T.$ 
About points (2) and (3), if I have some nonzero vector $v \in S \cap T,$ then I can write it as either $v \in S$ or $v \in T,$ which could be written in order as $v + 0 = 0 + v,$ giving non-uniqueness for the expression. For that matter, we get an alternative expression $0 = v + (-v).$ So, given any $w = s+t,$ we get a second expression $w = (s + v) + (t - v).$ That is called nonuniqueness.
Your question is about three subspaces $U,V,W$ which have (pairwise) trivial intersections. 
The important words here are "uniquely" and "intersections."
EDIT, Monday, June 17. Notice that if you can prove the equivalence of the three properties above, a really good, careful proof, then the question for three subspaces is just proof by induction (just one such step), using some inequalities: 
$$  \dim (U+V) \leq \dim U + \dim V.  $$
$$  \dim (U+V+W) \leq \dim (U+V) + \dim W.  $$
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ANOTHER EDIT, same day: 
THEOREM 1: 
$$  \dim (U+V) \leq \dim U + \dim V,  $$ and equality holds if and only if $$ U \cap V = \{ \vec{0} \}. $$
PROOF: by @Sarunas
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THEOREM 2: 
$$  \dim (U+V + W) \leq \dim (U+V) + \dim W,  $$ and equality holds if and only if $$ (U+V) \cap W = \{ \vec{0} \}. $$
PROOF: apply Theorem 1.
CAUTION: the condition (for equality) in Theorem 2 above is sometimes stronger than just pairwise trivial intersections. The simplest example is three distinct one-dimensional spaces (lines through the origin) in the plane $\mathbb R^2.$ See MO_COMMON_FALSE_BELIEFS_TILMAN including the comment by @Willie Wong
A: 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 $$\dim(U\times V\times W ) = \dim(\text{im}(\phi)) \\ \Leftrightarrow \dim(U\times V\times W) = \dim(U+V+W)\\ \Leftrightarrow \dim(U) + \dim(V) + \dim(W) = \dim(U+V+W)$$
A: Your proof of one implication is correct (of course finite dimension is assumed).
Now assume an element can be written in two distinct ways
$$
u+v+w=u'+v'+w'
$$
with $u,u'\in U$, $v,v'\in V$ and $w,w'\in W$. This implies $(u-u')+(v-v')+(w-w')=0$, so we can assume $u+v+w=0$ with at least one among $u$, $v$ and $w$ non zero. It's not restrictive to assume $u\ne0$.
Set $u=u_1$ and complete it to a basis $\{u_1,\dots,u_m\}$ of $U$. Fix also a basis $\{v_1,\dots,v_n\}$ of $W$ and a basis $\{w_1,\dots,w_p\}$ of $W$.
The union of these bases is clearly a set of generators of $U+V+W$, but we know it's linearly dependent, because we can write
$$
(\alpha_1u_1+\dots+\alpha_mu_m)+(\beta_1v_1+\dots+\beta_nv_n)+(\gamma_1w_1+\dots+\gamma_pw_p)=0
$$
where
\begin{gather}
\alpha_1=1,\alpha_2=\dots=\alpha_m=0\\
\beta_1v_1+\dots+\beta_nv_n=v\\
\gamma_1w_1+\dots+\gamma_pw_p=w
\end{gather}
Thus from this set of generators we can remove a vector, obtaining still a set of generators; therefore 
$$
\dim(U+V+W)<m+n+p
$$
because from every set of generators we can extract a basis.
A: I think I understand now, what @WillJagy intended to hint, but I want to make sure, that I really do. So, the first part of my answer was correct, right? But then, for the converse:
Suppose $$\dim(U+V+W)=\dim(U)+\dim(V)+\dim(W)$$
Also, assume for contradiction, that one of the intersections contain at least one, nonzero vector, say $U\cap W=\{0,z_1\}$. Then I can write it as $z_1\in U$ or $z_1\in W$. But that means that one of the vectors in $U+W$ (or in $U+V+W$ for that matter) has a non-unique representation, meaning that 
$$\dim(U+V+W)<\dim(U)+\dim(V)+\dim(W)$$ and contradicting the initial assumption. Thus we have a pairwise trivial intersections between the given subspaces
$$U\cap V=V\cap W=U\cap W=\{0\}$$
But then to express a vector in $U+V+W$, we need some basis $u_1,...u_n,v_1,...,v_m,w_1,...,w_q$, where $u_i\in U$, $v_i\in V$, $w_i\in W$. Since, pairwise intersections contain only the zero vector, these elements are linearly independent and thus every element in $U+V+W$ has a unique representation.
Am I right?
